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problem solving on linear function

Linear Functions Problems with Solutions

Linear functions are highly used throughout mathematics and are therefore important to understand. A set of problems involving linear functions , along with detailed solutions, are presented. The problems are designed with emphasis on the meaning of the slope and the y intercept.

Problem 1: f is a linear function. Values of x and f(x) are given in the table below; complete the table.

Problem 2: A family of linear functions is given by

Solution to Problem 2: a)

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Problem 3: A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population P ; grows as a linear function of time t, where t is the number of years after 2003. a) How many students will be enrolled in the school in 2010? b) Find a linear function that relates the student population to the time t. Solution to Problem 3: a) The given information may be written as ordered pairs (t , P). The year 2003 correspond to t = 0 and the year 2006 corresponds to t = 3, hence the 2 ordered pairs (0, 1200) and (3, 1500) Since the population grows linearly with the time t, we use the two ordered pairs to find the slope m of the graph of P as follows m = (1500 - 1200) / (6 - 3) = 100 students / year The slope m = 100 means that the students population grows by 100 students every year. From 2003 to 2010 there are 7 years and the students population in 2010 will be P(2010) = P(2003) + 7 * 100 = 1200 + 700 = 1900 students. b) We know the slope and two points, we may use the point slope form to find an equation for the population P as a function of t as follows P - P1 = m (t - t1) P - 1200 = 100 (t - 0) P = 100 t + 1200

Problem 4: The graph shown below is that of the linear function that relates the value V (in $) of a car to its age t, where t is the number of years after 2000.

Problem 5: The cost of producing x tools by a company is given by

Problem 6: A 500-liter tank full of oil is being drained at the constant rate of 20 liters par minute. a) Write a linear function V for the number of liters in the tank after t minutes (assuming that the drainage started at t = 0). b) Find the V and the t intercepts and interpret them. e) How many liters are in the tank after 11 minutes and 45 seconds? Solution to Problem 6: After each minute the amount of oil in the tank deceases by 20 liters. After t minutes, the amount of oil in the tank decreases by 20*t liters. Hence if at the start there 500 liters, after t minute the amount V of oil left in the tank is given by V = 500 - 20 t b) To find the V intercept, set t = 0 in the equation V = 500 - 20 t. V = 500 liters : it is the amount of oil at the start of the drainage. To find the t intercept, set V = 0 in the equation V = 500 - 20 t and solve for t. 0 = 500 - 20 t t = 500 / 20 = 25 minutes : it is the total time it takes to drain the 500 liters of oil. c) Convert 11 minutes 45 seconds in decimal form. t = 11 minutes 45 seconds = 11.75 minutes Calculate V at t = 11.75 minutes. V(11.75) = 500 - 20*11.75 = 265 liters are in the tank after 11 minutes 45 seconds of drainage.

Problem 7: A 50-meter by 70-meter rectangular garden is surrounded by a walkway of constant width x meters.

Problem 8: A driver starts a journey with 25 gallons in the tank of his car. The car burns 5 gallons for every 100 miles. Assuming that the amount of gasoline in the tank decreases linearly, a) write a linear function that relates the number of gallons G left in the tank after a journey of x miles. b) What is the value and meaning of the slope of the graph of G? c) What is the value and meaning of the x intercept? Solution to Problem 8: a) If 5 gallons are burnt for 100 miles then (5 / 100) gallons are burnt for 1 mile. Hence for x miles, x * (5 / 100) gallons are burnt. G is then equal to the initial amount of gasoline decreased by the amount gasoline burnt by the car. Hence G = 25 - (5 / 100) x b) The slope of G is equal to 5 / 1000 and it represent the amount of gasoline burnt for a distance of 1 mile. c) To find the x intercept, we set G = 0 and solve for x. 25 - (5 / 100) x = 0 x = 500 miles : it is the distance x for which all 25 gallons of gasoline will be burnt.

Problem 9: A rectangular wire frame has one of its dimensions moving at the rate of 0.5 cm / second. Its width is constant and equal to 4 cm. If at t = 0 the length of the rectangle is 10 cm,

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Learning Objectives

In this section, you will:

  • Represent a linear function.
  • Determine whether a linear function is increasing, decreasing, or constant.
  • Interpret slope as a rate of change.
  • Write and interpret an equation for a linear function.
  • Graph linear functions.
  • Determine whether lines are parallel or perpendicular.
  • Write the equation of a line parallel or perpendicular to a given line.

Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( Figure 1 ). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes 2 .

Suppose a maglev train travels a long distance, and maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.

Representing Linear Functions

The function describing the train’s motion is a linear function , which is defined as a function with a constant rate of change. This is a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.

Representing a Linear Function in Word Form

Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.

  • The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.

The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

Representing a Linear Function in Function Notation

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x x is the input value, m m is the rate of change, and b b is the initial value of the dependent variable.

In the example of the train, we might use the notation D ( t ) D ( t ) where the total distance D D is a function of the time t . t . The rate, m , m , is 83 meters per second. The initial value of the dependent variable b b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.

Representing a Linear Function in Tabular Form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2 . From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Can the input in the previous example be any real number?

No. The input represents time so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.

Representing a Linear Function in Graphical Form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D ( t ) = 83 t + 250 , D ( t ) = 83 t + 250 , to draw a graph as represented in Figure 3 . Notice the graph is a line . When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y -intercept , of the line. We can see from the graph that the y -intercept in the train example we just saw is ( 0 , 250 ) ( 0 , 250 ) and represents the distance of the train from the station when it began moving at a constant speed.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f ( x ) = 2 x + 1. f ( x ) = 2 x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.

Linear Function

A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

where b b is the initial or starting value of the function (when input, x = 0 x = 0 ), and m m is the constant rate of change, or slope of the function. The y -intercept is at ( 0 , b ) . ( 0 , b ) .

Using a Linear Function to Find the Pressure on a Diver

The pressure, P , P , in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d , d , in feet. This relationship may be modeled by the equation, P ( d ) = 0.434 d + 14.696. P ( d ) = 0.434 d + 14.696. Restate this function in words.

To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.

Determining Whether a Linear Function Is Increasing, Decreasing, or Constant

The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function , as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5 (a) . For a decreasing function , the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5 (b) . If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5 (c) .

Increasing and Decreasing Functions

The slope determines if the function is an increasing linear function , a decreasing linear function , or a constant function.

  • f ( x ) = m x + b f ( x ) = m x + b is an increasing function if m > 0. m > 0.
  • f ( x ) = m x + b f ( x ) = m x + b is a decreasing function if m < 0. m < 0.
  • f ( x ) = m x + b f ( x ) = m x + b is a constant function if m = 0. m = 0.

Deciding Whether a Function Is Increasing, Decreasing, or Constant

Studies from the early 2010s indicated that teens sent about 60 texts a day, while more recent data indicates much higher messaging rates among all users, particularly considering the various apps with which people can communicate. 3 . For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.

  • ⓐ The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
  • ⓑ A person has a limit of 500 texts per month in their data plan. The input is the number of days, and output is the total number of texts remaining for the month.
  • ⓒ A person has an unlimited number of texts in their data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.

Analyze each function.

  • ⓐ The function can be represented as f ( x ) = 60 x f ( x ) = 60 x where x x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.
  • ⓑ The function can be represented as f ( x ) = 500 − 60 x f ( x ) = 500 − 60 x where x x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x x days.
  • ⓒ The cost function can be represented as f ( x ) = 50 f ( x ) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant.

Interpreting Slope as a Rate of Change

In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x 1 x 1 and x 2 , x 2 , and two corresponding values for the output, y 1 y 1 and y 2 y 2 —which can be represented by a set of points, ( x 1 ,  y 1 ) ( x 1 ,  y 1 ) and ( x 2 ,  y 2 ) ( x 2 ,  y 2 ) —we can calculate the slope m . m .

Note that in function notation we can obtain two corresponding values for the output y 1 y 1 and y 2 y 2 for the function f , f , y 1 = f ( x 1 ) y 1 = f ( x 1 ) and y 2 = f ( x 2 ) , y 2 = f ( x 2 ) , so we could equivalently write

Figure 6 indicates how the slope of the line between the points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is.

Are the units for slope always units for the output units for the input ? units for the output units for the input ?

Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.

Calculate Slope

The slope, or rate of change, of a function m m can be calculated according to the following:

where x 1 x 1 and x 2 x 2 are input values, y 1 y 1 and y 2 y 2 are output values.

Given two points from a linear function, calculate and interpret the slope.

  • Determine the units for output and input values.
  • Calculate the change of output values and change of input values.
  • Interpret the slope as the change in output values per unit of the input value.

Finding the Slope of a Linear Function

If f ( x ) f ( x ) is a linear function, and ( 3 , −2 ) ( 3 , −2 ) and ( 8 , 1 ) ( 8 , 1 ) are points on the line, find the slope. Is this function increasing or decreasing?

The coordinate pairs are ( 3 , −2 ) ( 3 , −2 ) and ( 8 , 1 ) . ( 8 , 1 ) . To find the rate of change, we divide the change in output by the change in input.

We could also write the slope as m = 0.6. m = 0.6. The function is increasing because m > 0. m > 0.

As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y -coordinate, used corresponds with the first input value, or x -coordinate, used. Note that if we had reversed them, we would have obtained the same slope.

If f ( x ) f ( x ) is a linear function, and ( 2 , 3 ) ( 2 , 3 ) and ( 0 , 4 ) ( 0 , 4 ) are points on the line, find the slope. Is this function increasing or decreasing?

Finding the Population Change from a Linear Function

The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.

The rate of change relates the change in population to the change in time. The population increased by 27 , 800 − 23 , 400 = 4400 27 , 800 − 23 , 400 = 4400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.

So the population increased by 1,100 people per year.

Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.

The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.

Writing and Interpreting an Equation for a Linear Function

Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form . Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f f in Figure 7 .

We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose ( 0 , 7 ) ( 0 , 7 ) and ( 4 , 4 ) . ( 4 , 4 ) .

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

If we want to rewrite the equation in the slope-intercept form, we would find

If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y -axis when the output value is 7. Therefore, b = 7. b = 7. We now have the initial value b b and the slope m m so we can substitute m m and b b into the slope-intercept form of a line.

So the function is f ( x ) = − 3 4 x + 7 , f ( x ) = − 3 4 x + 7 , and the linear equation would be y = − 3 4 x + 7. y = − 3 4 x + 7.

Given the graph of a linear function, write an equation to represent the function.

  • Identify two points on the line.
  • Use the two points to calculate the slope.
  • Determine where the line crosses the y -axis to identify the y -intercept by visual inspection.
  • Substitute the slope and y -intercept into the slope-intercept form of a line equation.

Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of f f shown in Figure 8 .

Identify two points on the line, such as ( 0 , 2 ) ( 0 , 2 ) and ( −2 , −4 ) . ( −2 , −4 ) . Use the points to calculate the slope.

Substitute the slope and the coordinates of one of the points into the point-slope form.

We can use algebra to rewrite the equation in the slope-intercept form.

This makes sense because we can see from Figure 9 that the line crosses the y -axis at the point ( 0 , 2 ) , ( 0 , 2 ) , which is the y -intercept, so b = 2. b = 2.

Writing an Equation for a Linear Cost Function

Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C C where C ( x ) C ( x ) is the cost for x x items produced in a given month.

The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50. The variable cost, called the marginal cost, is represented by 37.5. 37.5. The cost Ben incurs is the sum of these two costs, represented by C ( x ) = 1250 + 37.5 x . C ( x ) = 1250 + 37.5 x .

If Ben produces 100 items in a month, his monthly cost is found by substituting 100 for x . x .

So his monthly cost would be $5,000.

Writing an Equation for a Linear Function Given Two Points

If f f is a linear function, with f ( 3 ) = −2 , f ( 3 ) = −2 , and f ( 8 ) = 1 , f ( 8 ) = 1 , find an equation for the function in slope-intercept form.

We can write the given points using coordinates.

We can then use the points to calculate the slope.

If f ( x ) f ( x ) is a linear function, with f ( 2 ) = –11 , f ( 2 ) = –11 , and f ( 4 ) = −25 , f ( 4 ) = −25 , write an equation for the function in slope-intercept form.

Modeling Real-World Problems with Linear Functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

Given a linear function f f and the initial value and rate of change, evaluate f ( c ) . f ( c ) .

  • Determine the initial value and the rate of change (slope).
  • Substitute the values into f ( x ) = m x + b . f ( x ) = m x + b .
  • Evaluate the function at x = c . x = c .

Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N , N , in his collection as a function of time, t , t , the number of months. How many songs will he own at the end of one year?

The initial value for this function is 200 because he currently owns 200 songs, so N ( 0 ) = 200 , N ( 0 ) = 200 , which means that b = 200. b = 200.

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.

We can write the formula N ( t ) = 15 t + 200. N ( t ) = 15 t + 200.

With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at t = 12. t = 12.

Marcus will have 380 songs in 12 months.

Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Using a Linear Function to Calculate Salary Based on Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income I , I , depends on the number of new policies, n , n , he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I ( n ) , I ( n ) , and interpret the meaning of the components of the equation.

The given information gives us two input-output pairs: ( 3 , 760 ) ( 3 , 760 ) and ( 5 , 92 0 ) . ( 5 , 92 0 ) . We start by finding the rate of change.

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

We can then solve for the initial value.

The value of b b is the starting value for the function and represents Ilya’s income when n = 0 , n = 0 , or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.

We can now write the final equation.

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Using Tabular Form to Write an Equation for a Linear Function

Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

We can see from the table that the initial value for the number of rats is 1000, so b = 1000. b = 1000.

Rather than solving for m , m , we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using ( 2 , 1080 ) ( 2 , 1080 ) and ( 6 , 1240 ) ( 6 , 1240 )

Is the initial value always provided in a table of values like Table 1 ?

No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f ( x ) = m x + b , f ( x ) = m x + b , and solve for b . b .

A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x x months since the measurements began. Write a linear function, H ( x ) , H ( x ) , where x x is the number of months since the start of the experiment.

Graphing Linear Functions

Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y- intercept and slope. And the third method is by using transformations of the identity function f ( x ) = x . f ( x ) = x .

Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f ( x ) = 2 x , f ( x ) = 2 x , we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point ( 1 , 2 ) . ( 1 , 2 ) . Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point ( 2 , 4 ) . ( 2 , 4 ) . Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

Given a linear function, graph by plotting points.

  • Choose a minimum of two input values.
  • Evaluate the function at each input value.
  • Use the resulting output values to identify coordinate pairs.
  • Plot the coordinate pairs on a grid.
  • Draw a line through the points.

Graphing by Plotting Points

Graph f ( x ) = − 2 3 x + 5 f ( x ) = − 2 3 x + 5 by plotting points.

Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

Evaluate the function at each input value, and use the output value to identify coordinate pairs.

Plot the coordinate pairs and draw a line through the points. Figure 11 represents the graph of the function f ( x ) = − 2 3 x + 5. f ( x ) = − 2 3 x + 5.

The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function.

Graph f ( x ) = − 3 4 x + 6 f ( x ) = − 3 4 x + 6 by plotting points.

Graphing a Function Using y- intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y- intercept, which is the point at which the input value is zero. To find the y- intercept, we can set x = 0 x = 0 in the equation.

The other characteristic of the linear function is its slope .

Let’s consider the following function.

The slope is 1 2 . 1 2 . Because the slope is positive, we know the graph will slant upward from left to right. The y- intercept is the point on the graph when x = 0. x = 0. The graph crosses the y -axis at ( 0 , 1 ) . ( 0 , 1 ) . Now we know the slope and the y -intercept. We can begin graphing by plotting the point ( 0 , 1 ) . ( 0 , 1 ) . We know that the slope is the change in the y -coordinate over the change in the x -coordinate. This is commonly referred to as rise over run, m = rise run . m = rise run . From our example, we have m = 1 2 , m = 1 2 , which means that the rise is 1 and the run is 2. So starting from our y -intercept ( 0 , 1 ) , ( 0 , 1 ) , we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12 .

Graphical Interpretation of a Linear Function

In the equation f ( x ) = m x + b f ( x ) = m x + b

  • b b is the y -intercept of the graph and indicates the point ( 0 , b ) ( 0 , b ) at which the graph crosses the y -axis.
  • m m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

Do all linear functions have y -intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line is parallel to the y-axis does not have a y-intercept, but it is not a function .)

Given the equation for a linear function, graph the function using the y -intercept and slope.

  • Evaluate the function at an input value of zero to find the y- intercept.
  • Identify the slope as the rate of change of the input value.
  • Plot the point represented by the y- intercept.
  • Use rise run rise run to determine at least two more points on the line.
  • Sketch the line that passes through the points.

Graphing by Using the y- intercept and Slope

Graph f ( x ) = − 2 3 x + 5 f ( x ) = − 2 3 x + 5 using the y- intercept and slope.

Evaluate the function at x = 0 x = 0 to find the y- intercept. The output value when x = 0 x = 0 is 5, so the graph will cross the y -axis at ( 0 , 5 ) . ( 0 , 5 ) .

According to the equation for the function, the slope of the line is − 2 3 . − 2 3 . This tells us that for each vertical decrease in the “rise” of – 2 – 2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y -intercept on the graph in Figure 13 . From the initial value ( 0 , 5 ) ( 0 , 5 ) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then drawing a line through the points.

The graph slants downward from left to right, which means it has a negative slope as expected.

Find a point on the graph we drew in Example 12 that has a negative x -value.

Graphing a Function Using Transformations

Another option for graphing is to use a transformation of the identity function f ( x ) = x . f ( x ) = x . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.

Vertical Stretch or Compression

In the equation f ( x ) = m x , f ( x ) = m x , the m m is acting as the vertical stretch or compression of the identity function. When m m is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of f ( x ) = x f ( x ) = x by m m stretches the graph of f f by a factor of m m units if m > 1 m > 1 and compresses the graph of f f by a factor of m m units if 0 < m < 1. 0 < m < 1. This means the larger the absolute value of m , m , the steeper the slope.

Vertical Shift

In f ( x ) = m x + b , f ( x ) = m x + b , the b b acts as the vertical shift , moving the graph up and down without affecting the slope of the line. Notice in Figure 15 that adding a value of b b to the equation of f ( x ) = x f ( x ) = x shifts the graph of f f a total of b b units up if b b is positive and | b | | b | units down if b b is negative.

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

Given the equation of a linear function, use transformations to graph the linear function in the form f ( x ) = m x + b . f ( x ) = m x + b .

  • Graph f ( x ) = x . f ( x ) = x .
  • Vertically stretch or compress the graph by a factor m . m .
  • Shift the graph up or down b b units.

Graphing by Using Transformations

Graph f ( x ) = 1 2 x − 3 f ( x ) = 1 2 x − 3 using transformations.

The equation for the function shows that m = 1 2 m = 1 2 so the identity function is vertically compressed by 1 2 . 1 2 . The equation for the function also shows that b = − 3 b = − 3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 16 .

Then show the vertical shift as in Figure 17 .

Graph f ( x ) = 4 + 2 x f ( x ) = 4 + 2 x using transformations.

In Example 15 , could we have sketched the graph by reversing the order of the transformations?

No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.

Writing the Equation for a Function from the Graph of a Line

Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18 . We can see right away that the graph crosses the y -axis at the point ( 0 , 4 ) ( 0 , 4 ) so this is the y -intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point ( – 2 , 0 ) . ( – 2 , 0 ) . To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

Substituting the slope and y- intercept into the slope-intercept form of a line gives

Given a graph of linear function, find the equation to describe the function.

  • Identify the y- intercept of an equation.
  • Choose two points to determine the slope.
  • Substitute the y- intercept and slope into the slope-intercept form of a line.

Matching Linear Functions to Their Graphs

Match each equation of the linear functions with one of the lines in Figure 19 .

  • ⓐ f ( x ) = 2 x + 3 f ( x ) = 2 x + 3
  • ⓑ g ( x ) = 2 x − 3 g ( x ) = 2 x − 3
  • ⓒ h ( x ) = − 2 x + 3 h ( x ) = − 2 x + 3
  • ⓓ j ( x ) = 1 2 x + 3 j ( x ) = 1 2 x + 3

Analyze the information for each function.

  • ⓐ This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g g has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through ( 0 , 3 ) ( 0 , 3 ) so f f must be represented by line I.
  • ⓑ This function also has a slope of 2, but a y -intercept of −3. −3. It must pass through the point ( 0 , −3 ) ( 0 , −3 ) and slant upward from left to right. It must be represented by line III.
  • ⓒ This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
  • ⓓ This function has a slope of 1 2 1 2 and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through ( 0 , 3 ) , ( 0 , 3 ) , but the slope of j j is less than the slope of f f so the line for j j must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 20 .

Finding the x -intercept of a Line

So far we have been finding the y- intercepts of a function: the point at which the graph of the function crosses the y -axis. Recall that a function may also have an x -intercept , which is the x -coordinate of the point where the graph of the function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set a function f ( x ) f ( x ) equal to zero and solve for the value of x . x . For example, consider the function shown.

Set the function equal to 0 and solve for x . x .

The graph of the function crosses the x -axis at the point ( 2 , 0 ) . ( 2 , 0 ) .

Do all linear functions have x -intercepts?

No. However, linear functions of the form y = c , y = c , where c c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21 .

x -intercept

The x -intercept of the function is value of x x when f ( x ) = 0. f ( x ) = 0. It can be solved by the equation 0 = m x + b . 0 = m x + b .

Finding an x -intercept

Find the x -intercept of f ( x ) = 1 2 x − 3. f ( x ) = 1 2 x − 3.

Set the function equal to zero to solve for x . x .

The graph crosses the x -axis at the point ( 6 , 0 ) . ( 6 , 0 ) .

A graph of the function is shown in Figure 22 . We can see that the x -intercept is ( 6 , 0 ) ( 6 , 0 ) as we expected.

Find the x -intercept of f ( x ) = 1 4 x − 4. f ( x ) = 1 4 x − 4.

Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y -value. In Figure 23 , we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 m = 0 in the equation f ( x ) = m x + b , f ( x ) = m x + b , the equation simplifies to f ( x ) = b . f ( x ) = b . In other words, the value of the function is a constant. This graph represents the function f ( x ) = 2. f ( x ) = 2.

A vertical line indicates a constant input, or x -value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

A vertical line, such as the one in Figure 25 , has an x -intercept, but no y- intercept unless it’s the line x = 0. x = 0. This graph represents the line x = 2. x = 2.

Horizontal and Vertical Lines

Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation in the form f ( x ) = b . f ( x ) = b .

A vertical line is a line defined by an equation in the form x = a . x = a .

Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 26 .

For any x -value, the y -value is − 4 , − 4 , so the equation is y = − 4. y = − 4.

Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 27 .

The constant x -value is 7 , 7 , so the equation is x = 7. x = 7.

Determining Whether Lines are Parallel or Perpendicular

The two lines in Figure 28 are parallel lines : they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y -intercept. If we shifted one line vertically toward the other, they would become coincident.

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y -intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. 1. So, if m 1 and  m 2 m 1 and  m 2 are negative reciprocals of one another, they can be multiplied together to yield –1. –1.

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 1 8 , 1 8 , and the reciprocal of 1 8 1 8 is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

The product of the slopes is –1.

Parallel and Perpendicular Lines

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

If and only if b 1 = b 2 b 1 = b 2 and m 1 = m 2 , m 1 = m 2 , we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines if they intersect to form a right angle.

Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

Parallel lines have the same slope. Because the functions f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 and j ( x ) = 2 x − 6 j ( x ) = 2 x − 6 each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1 2 1 2 are negative reciprocals, the functions g ( x ) = 1 2 x − 4 g ( x ) = 1 2 x − 4 and h ( x ) = −2 x + 2 h ( x ) = −2 x + 2 represent perpendicular lines.

A graph of the lines is shown in Figure 30 .

The graph shows that the lines f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 and j ( x ) = 2 x – 6 j ( x ) = 2 x – 6 are parallel, and the lines g ( x ) = 1 2 x – 4 g ( x ) = 1 2 x – 4 and h ( x ) = − 2 x + 2 h ( x ) = − 2 x + 2 are perpendicular.

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing Equations of Parallel Lines

Suppose for example, we are given the equation shown.

We know that the slope of the line formed by the function is 3. We also know that the y- intercept is ( 0 , 1 ) . ( 0 , 1 ) . Any other line with a slope of 3 will be parallel to f ( x ) . f ( x ) . So the lines formed by all of the following functions will be parallel to f ( x ) . f ( x ) .

Suppose then we want to write the equation of a line that is parallel to f f and passes through the point ( 1 , 7 ) . ( 1 , 7 ) . This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of b b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

So g ( x ) = 3 x + 4 g ( x ) = 3 x + 4 is parallel to f ( x ) = 3 x + 1 f ( x ) = 3 x + 1 and passes through the point ( 1 , 7 ) . ( 1 , 7 ) .

Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

  • Find the slope of the function.
  • Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.

Finding a Line Parallel to a Given Line

Find a line parallel to the graph of f ( x ) = 3 x + 6 f ( x ) = 3 x + 6 that passes through the point ( 3 , 0 ) . ( 3 , 0 ) .

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3 , x = 3 , m = 3 , x = 3 , and f ( x ) = 0 f ( x ) = 0 into the slope-intercept form to find the y- intercept.

The line parallel to f ( x ) f ( x ) that passes through ( 3 , 0 ) ( 3 , 0 ) is g ( x ) = 3 x − 9. g ( x ) = 3 x − 9.

We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect.

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown.

The slope of the line is 2, and its negative reciprocal is − 1 2 . − 1 2 . Any function with a slope of − 1 2 − 1 2 will be perpendicular to f ( x ) . f ( x ) . So the lines formed by all of the following functions will be perpendicular to f ( x ) . f ( x ) .

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f ( x ) f ( x ) and passes through the point ( 4 , 0 ) . ( 4 , 0 ) . We already know that the slope is − 1 2 . − 1 2 . Now we can use the point to find the y -intercept by substituting the given values into the slope-intercept form of a line and solving for b . b .

The equation for the function with a slope of − 1 2 − 1 2 and a y- intercept of 2 is

So g ( x ) = − 1 2 x + 2 g ( x ) = − 1 2 x + 2 is perpendicular to f ( x ) = 2 x + 4 f ( x ) = 2 x + 4 and passes through the point ( 4 , 0 ) . ( 4 , 0 ) . Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

  • Determine the negative reciprocal of the slope.
  • Substitute the new slope and the values for x x and y y from the coordinate pair provided into g ( x ) = m x + b . g ( x ) = m x + b .
  • Solve for b . b .
  • Write the equation of the line.

Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to f ( x ) = 3 x + 3 f ( x ) = 3 x + 3 that passes through the point ( 3 , 0 ) . ( 3 , 0 ) .

The original line has slope m = 3 , m = 3 , so the slope of the perpendicular line will be its negative reciprocal, or − 1 3 . − 1 3 . Using this slope and the given point, we can find the equation of the line.

The line perpendicular to f ( x ) f ( x ) that passes through ( 3 , 0 ) ( 3 , 0 ) is g ( x ) = − 1 3 x + 1. g ( x ) = − 1 3 x + 1.

A graph of the two lines is shown in Figure 32 .

Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely.

Given the function h ( x ) = 2 x − 4 , h ( x ) = 2 x − 4 , write an equation for the line passing through ( 0 , 0 ) ( 0 , 0 ) that is

  • ⓐ parallel to h ( x ) h ( x )
  • ⓑ perpendicular to h ( x ) h ( x )

Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

  • Determine the slope of the line passing through the points.

Find the negative reciprocal of the slope.

  • Use the slope-intercept form or point-slope form to write the equation by substituting the known values.

Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

A line passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 4 , 5 ) . ( 4 , 5 ) . Find the equation of a perpendicular line that passes through the point ( 4 , 5 ) . ( 4 , 5 ) .

From the two points of the given line, we can calculate the slope of that line.

We can then solve for the y- intercept of the line passing through the point ( 4 , 5 ) . ( 4 , 5 ) .

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point ( 4 , 5 ) ( 4 , 5 ) is

A line passes through the points, ( −2 , −15 ) ( −2 , −15 ) and ( 2 , −3 ) . ( 2 , −3 ) . Find the equation of a perpendicular line that passes through the point, ( 6 , 4 ) . ( 6 , 4 ) .

Access this online resource for additional instruction and practice with linear functions.

  • Linear Functions
  • Finding Input of Function from the Output and Graph
  • Graphing Functions using Tables

4.1 Section Exercises

Terry is skiing down a steep hill. Terry's elevation, E ( t ) , E ( t ) , in feet after t t seconds is given by E ( t ) = 3000 − 70 t . E ( t ) = 3000 − 70 t . Write a complete sentence describing Terry’s starting elevation and how it is changing over time.

Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?

A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.

If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y -intercepts.

If a horizontal line has the equation f ( x ) = a f ( x ) = a and a vertical line has the equation x = a , x = a , what is the point of intersection? Explain why what you found is the point of intersection.

For the following exercises, determine whether the equation of the curve can be written as a linear function.

y = 1 4 x + 6 y = 1 4 x + 6

y = 3 x − 5 y = 3 x − 5

y = 3 x 2 − 2 y = 3 x 2 − 2

3 x + 5 y = 15 3 x + 5 y = 15

3 x 2 + 5 y = 15 3 x 2 + 5 y = 15

3 x + 5 y 2 = 15 3 x + 5 y 2 = 15

− 2 x 2 + 3 y 2 = 6 − 2 x 2 + 3 y 2 = 6

− x − 3 5 = 2 y − x − 3 5 = 2 y

For the following exercises, determine whether each function is increasing or decreasing.

f ( x ) = 4 x + 3 f ( x ) = 4 x + 3

g ( x ) = 5 x + 6 g ( x ) = 5 x + 6

a ( x ) = 5 − 2 x a ( x ) = 5 − 2 x

b ( x ) = 8 − 3 x b ( x ) = 8 − 3 x

h ( x ) = −2 x + 4 h ( x ) = −2 x + 4

k ( x ) = −4 x + 1 k ( x ) = −4 x + 1

j ( x ) = 1 2 x − 3 j ( x ) = 1 2 x − 3

p ( x ) = 1 4 x − 5 p ( x ) = 1 4 x − 5

n ( x ) = − 1 3 x − 2 n ( x ) = − 1 3 x − 2

m ( x ) = − 3 8 x + 3 m ( x ) = − 3 8 x + 3

For the following exercises, find the slope of the line that passes through the two given points.

( 2 , 4 ) ( 2 , 4 ) and ( 4 , 10 ) ( 4 , 10 )

( 1 , 5 ) ( 1 , 5 ) and ( 4 , 11 ) ( 4 , 11 )

( –1 , 4 ) ( –1 , 4 ) and ( 5 , 2 ) ( 5 , 2 )

( 8 , –2 ) ( 8 , –2 ) and ( 4 , 6 ) ( 4 , 6 )

( 6 , 11 ) ( 6 , 11 ) and ( –4 , 3 ) ( –4 , 3 )

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

f ( − 5 ) = −4 , f ( − 5 ) = −4 , and f ( 5 ) = 2 f ( 5 ) = 2

f ( −1 ) = 4 , f ( −1 ) = 4 , and f ( 5 ) = 1 f ( 5 ) = 1

Passes through ( 2 , 4 ) ( 2 , 4 ) and ( 4 , 10 ) ( 4 , 10 )

Passes through ( 1 , 5 ) ( 1 , 5 ) and ( 4 , 11 ) ( 4 , 11 )

Passes through ( −1 , 4 ) ( −1 , 4 ) and ( 5 , 2 ) ( 5 , 2 )

Passes through ( −2 , 8 ) ( −2 , 8 ) and ( 4 , 6 ) ( 4 , 6 )

x intercept at ( −2 , 0 ) ( −2 , 0 ) and y intercept at ( 0 , −3 ) ( 0 , −3 )

x intercept at ( −5 , 0 ) ( −5 , 0 ) and y intercept at ( 0 , 4 ) ( 0 , 4 )

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither.

4 x − 7 y = 10 7 x + 4 y = 1 4 x − 7 y = 10 7 x + 4 y = 1

3 y + x = 12 − y = 8 x + 1 3 y + x = 12 − y = 8 x + 1

3 y + 4 x = 12 − 6 y = 8 x + 1 3 y + 4 x = 12 − 6 y = 8 x + 1

6 x − 9 y = 10 3 x + 2 y = 1 6 x − 9 y = 10 3 x + 2 y = 1

For the following exercises, find the x - and y- intercepts of each equation.

f ( x ) = − x + 2 f ( x ) = − x + 2

g ( x ) = 2 x + 4 g ( x ) = 2 x + 4

h ( x ) = 3 x − 5 h ( x ) = 3 x − 5

k ( x ) = −5 x + 1 k ( x ) = −5 x + 1

− 2 x + 5 y = 20 − 2 x + 5 y = 20

7 x + 2 y = 56 7 x + 2 y = 56

For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

Line 1: Passes through ( 0 , 6 ) ( 0 , 6 ) and ( 3 , −24 ) ( 3 , −24 )

Line 2: Passes through ( −1 , 19 ) ( −1 , 19 ) and ( 8 , −71 ) ( 8 , −71 )

Line 1: Passes through ( −8 , −55 ) ( −8 , −55 ) and ( 10 , 89 ) ( 10 , 89 )

Line 2: Passes through ( 9 , − 44 ) ( 9 , − 44 ) and ( 4 , − 14 ) ( 4 , − 14 )

Line 1: Passes through ( 2 , 3 ) ( 2 , 3 ) and ( 4 , −1 ) ( 4 , −1 )

Line 2: Passes through ( 6 , 3 ) ( 6 , 3 ) and ( 8 , 5 ) ( 8 , 5 )

Line 1: Passes through ( 1 , 7 ) ( 1 , 7 ) and ( 5 , 5 ) ( 5 , 5 )

Line 2: Passes through ( −1 , −3 ) ( −1 , −3 ) and ( 1 , 1 ) ( 1 , 1 )

Line 1: Passes through ( 2 , 5 ) ( 2 , 5 ) and ( 5 , − 1 ) ( 5 , − 1 )

Line 2: Passes through ( −3 , 7 ) ( −3 , 7 ) and ( 3 , −5 ) ( 3 , −5 )

For the following exercises, write an equation for the line described.

Write an equation for a line parallel to f ( x ) = − 5 x − 3 f ( x ) = − 5 x − 3 and passing through the point ( 2 , – 12 ) . ( 2 , – 12 ) .

Write an equation for a line parallel to g ( x ) = 3 x − 1 g ( x ) = 3 x − 1 and passing through the point ( 4 , 9 ) . ( 4 , 9 ) .

Write an equation for a line perpendicular to h ( t ) = −2 t + 4 h ( t ) = −2 t + 4 and passing through the point ( −4 , –1 ) . ( −4 , –1 ) .

Write an equation for a line perpendicular to p ( t ) = 3 t + 4 p ( t ) = 3 t + 4 and passing through the point ( 3 , 1 ) . ( 3 , 1 ) .

For the following exercises, find the slope of the line graphed.

For the following exercises, write an equation for the line graphed.

For the following exercises, match the given linear equation with its graph in Figure 33 .

f ( x ) = − x − 1 f ( x ) = − x − 1

f ( x ) = −3 x − 1 f ( x ) = −3 x − 1

f ( x ) = − 1 2 x − 1 f ( x ) = − 1 2 x − 1

f ( x ) = 2 f ( x ) = 2

f ( x ) = 2 + x f ( x ) = 2 + x

f ( x ) = 3 x + 2 f ( x ) = 3 x + 2

For the following exercises, sketch a line with the given features.

An x -intercept of ( –4 , 0 ) ( –4 , 0 ) and y -intercept of ( 0 , –2 ) ( 0 , –2 )

An x -intercept ( –2 , 0 ) ( –2 , 0 ) and y -intercept of ( 0 , 4 ) ( 0 , 4 )

A y -intercept of ( 0 , 7 ) ( 0 , 7 ) and slope − 3 2 − 3 2

A y -intercept of ( 0 , 3 ) ( 0 , 3 ) and slope 2 5 2 5

Passing through the points ( –6 , –2 ) ( –6 , –2 ) and ( 6 , –6 ) ( 6 , –6 )

Passing through the points ( –3 , –4 ) ( –3 , –4 ) and ( 3 , 0 ) ( 3 , 0 )

For the following exercises, sketch the graph of each equation.

f ( x ) = −2 x − 1 f ( x ) = −2 x − 1

f ( x ) = −3 x + 2 f ( x ) = −3 x + 2

f ( x ) = 1 3 x + 2 f ( x ) = 1 3 x + 2

f ( x ) = 2 3 x − 3 f ( x ) = 2 3 x − 3

f ( t ) = 3 + 2 t f ( t ) = 3 + 2 t

p ( t ) = −2 + 3 t p ( t ) = −2 + 3 t

x = 3 x = 3

x = −2 x = −2

r ( x ) = 4 r ( x ) = 4

For the following exercises, write the equation of the line shown in the graph.

For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.

For the following exercises, use a calculator or graphing technology to complete the task.

If f f is a linear function, f ( 0.1 ) = 11.5 f ( 0.1 ) = 11.5 , and f ( 0.4 ) = –5.9 f ( 0.4 ) = –5.9 , find an equation for the function.

Graph the function f f on a domain of [ –10 , 10 ] : f ( x ) = 0.02 x − 0.01. [ –10 , 10 ] : f ( x ) = 0.02 x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x x to be −10 −10 and the maximum value of x x to be 10 10 .

Graph the function f f on a domain of [ –10 , 10 ] : f x ) = 2 , 500 x + 4 , 000 [ –10 , 10 ] : f x ) = 2 , 500 x + 4 , 000

Table 3 shows the input, w , w , and output, k , k , for a linear function k . k .

  • ⓐ Fill in the missing values of the table.
  • ⓑ Write the linear function

k , k , round to 3 decimal places.

Table 4 shows the input, p , p , and output, q , q , for a linear function q . q .

Graph the linear function f f on a domain of [ − 10 , 10 ] [ − 10 , 10 ] for the function whose slope is 1 8 1 8 and y -intercept is 31 16 . 31 16 . Label the points for the input values of −10 −10 and 10. 10.

Graph the linear function f f on a domain of [ − 0.1 , 0.1 ] [ − 0.1 , 0.1 ] for the function whose slope is 75 and y -intercept is −22.5. −22.5. Label the points for the input values of −0.1 −0.1 and 0.1. 0.1.

Graph the linear function f f where f ( x ) = a x + b f ( x ) = a x + b on the same set of axes on a domain of [ − 4 , 4 ] [ − 4 , 4 ] for the following values of a a and b . b .

  • ⓐ a = 2 ; b = 3 a = 2 ; b = 3
  • ⓑ a = 2 ; b = 4 a = 2 ; b = 4
  • ⓒ a = 2 ; b = –4 a = 2 ; b = –4
  • ⓓ a = 2 ; b = –5 a = 2 ; b = –5

Find the value of x x if a linear function goes through the following points and has the following slope: ( x , 2 ) , ( −4 , 6 ) , m = 3 ( x , 2 ) , ( −4 , 6 ) , m = 3

Find the value of y if a linear function goes through the following points and has the following slope: ( 10 , y ) , ( 25 , 100 ) , m = −5 ( 10 , y ) , ( 25 , 100 ) , m = −5

Find the equation of the line that passes through the following points:

( a , b ) ( a , b ) and ( a , b + 1 ) ( a , b + 1 )

( 2 a , b ) ( 2 a , b ) and ( a , b + 1 ) ( a , b + 1 )

( a , 0 ) ( a , 0 ) and ( c , d ) ( c , d )

Find the equation of the line parallel to the line g ( x ) = −0. 01 x +2 .01 g ( x ) = −0. 01 x +2 .01 through the point ( 1 , 2 ) . ( 1 , 2 ) .

Find the equation of the line perpendicular to the line g ( x ) = −0. 01 x +2 .01 g ( x ) = −0. 01 x +2 .01 through the point ( 1 , 2 ) . ( 1 , 2 ) .

For the following exercises, use the functions f ( x ) = −0. 1 x +200 and  g ( x ) = 20 x + 0.1. f ( x ) = −0. 1 x +200 and  g ( x ) = 20 x + 0.1.

Find the point of intersection of the lines f f and g . g .

Where is f ( x ) f ( x ) greater than g ( x ) ? g ( x ) ? Where is g ( x ) g ( x ) greater than f ( x ) ? f ( x ) ?

Real-World Applications

At noon, a barista notices that they have $20 in their tip jar. If the barista makes an average of $0.50 from each customer, how much will they have in the tip jar if they serve n n more customers during the shift?

A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?

A clothing business finds there is a linear relationship between the number of shirts, n , n , it can sell and the price, p , p , it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $ 30 , $ 30 , while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p ( n ) = m n + b p ( n ) = m n + b that gives the price p p they can charge for n n shirts.

A phone company charges for service according to the formula: C ( n ) = 24 + 0.1 n , C ( n ) = 24 + 0.1 n , where n n is the number of minutes talked, and C ( n ) C ( n ) is the monthly charge, in dollars. Find and interpret the rate of change and initial value.

A farmer finds there is a linear relationship between the number of bean stalks, n , n , she plants and the yield, y , y , each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form y = m n + b y = m n + b that gives the yield when n n stalks are planted.

A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.

A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P ( t ) , P ( t ) , for the population t t years after 2003.

Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I ( x ) = 1054 x + 23,286 I ( x ) = 1054 x + 23,286 , where x x is the number of years after 1990. Which of the following interprets the slope in the context of the problem?

  • ⓐ As of 1990, average annual income was $23,286.
  • ⓑ In the ten-year period from 1990–1999, average annual income increased by a total of $1,054.
  • ⓒ Each year in the decade of the 1990s, average annual income increased by $1,054.
  • ⓓ Average annual income rose to a level of $23,286 by the end of 1999.

When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C , C , the Celsius temperature, F ( C ) . F ( C ) .

  • ⓐ Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
  • ⓑ Find and interpret F ( 28 ) . F ( 28 ) .
  • ⓒ Find and interpret F ( –40 ) . F ( –40 ) .
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Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Elimination Method of solving linear equations

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Graphical Method of Solving Linear Equations

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Cross Multiplication Method of solving linear equations

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

  • Linear Equations
  • Application of Linear Equations
  • Two-Variable Linear Equations
  • Linear Equations and Half Planes
  • One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Solving Linear Equations Example

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

  • Substitution method
  • Elimination method
  • Cross multiplication method
  • Graphical method

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Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

  • When x increases, y increases twice as fast , so we need 2x
  • When x is 0, y is already 1. So +1 is also needed
  • And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

  • Exponents (like the 2 in x 2 )
  • Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

  • Slope: m = 2
  • Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

  • Finding the Midpoint of a Line Segment
  • Finding Parallel and Perpendicular Lines
  • Finding the Equation of a Line from 2 Points
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Solving a Linear Function - Part 2

In the previous lesson on functions you learned how to find the slope and write an equation when given a function.

Linear functions are very much like linear equations, the only difference is you are using function notation "f(x)" instead of "y". Otherwise, the process is the same.

Ok, let's move on! In our first example, we are going to find the value of x when given a value for f(x). This is one of the trickier problems in the function unit. Watch carefully where we substitute the given number 4.

Let's take a look.

Example 1: Solving for x in a linear function

Pretty easy, right? This is really just a review of concepts that you've already learned. Once you figure out that you substitute 4 for f(x), you solve this as a regular two step equation.

In example 2, you will see how to write the equation of a function given slope and a point.

This process for this problem is exactly the same as you learned when writing equations. The only difference is how you state your "function" at the end. It must be written in function notation.

Example 2: Finding the Equation of a Function

This completes our lesson on Linear Functions. Hopefully you do not let the word "function" intimidate you. As you can see, you know how to solve all of these problems from studying equations. Now you just have to be a little fancier with how you name your equation.

In the next lesson, we will continue our study of functions by taking a look at quadratic functions .

  • Linear Functions (2)

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  • \frac{3}{4}x+\frac{5}{6}=5x-\frac{125}{3}
  • \sqrt{2}x-\sqrt{3}=\sqrt{5}
  • 7y+5-3y+1=2y+2
  • \frac{x}{3}+\frac{x}{2}=10
  • What is a linear equation?
  • A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
  • How do you find the linear equation?
  • To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
  • What are the 4 methods of solving linear equations?
  • There are four common methods to solve a system of linear equations: Graphing, Substitution, Elimination and Matrix.
  • How do you identify a linear equation?
  • Here are a few ways to identify a linear equation: Look at the degree of the equation, a linear equation is a first-degree equation. Check if the equation has two variables. Graph the equation.
  • What is the most basic linear equation?
  • The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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  • High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Read More

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About solving equations

A value is said to be a root of a polynomial if ..

The largest exponent of appearing in is called the degree of . If has degree , then it is well known that there are roots, once one takes into account multiplicity. To understand what is meant by multiplicity, take, for example, . This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

How Wolfram|Alpha solves equations

For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and reliability. Other operations rely on theorems and algorithms from number theory, abstract algebra and other advanced fields to compute results. These methods are carefully designed and chosen to enable Wolfram|Alpha to solve the greatest variety of problems while also minimizing computation time.

Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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5.4: Solve Applications with Systems of Equations

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Learning Objectives

By the end of this section, you will be able to:

  • Translate to a system of equations
  • Solve direct translation applications
  • Solve geometry applications
  • Solve uniform motion applications

Before you get started, take this readiness quiz.

  • The sum of twice a number and nine is 31. Find the number. If you missed this problem, review Exercise 3.1.10 .
  • Twins Jon and Ron together earned $96,000 last year. Ron earned $8,000 more than three times what Jon earned. How much did each of the twins earn? If you missed this problem, review Exercise 3.1.31 .
  • Alessio rides his bike \(3\frac{1}{2}\) hours at a rate of 10 miles per hour. How far did he ride? If you missed this problem, review Exercise 2.6.1 .

Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem Solving Strategy for Systems of Linear Equations.

USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models .

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

In that chapter we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?

Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

Exercise \(\PageIndex{1}\): How to Translate to a System of Equations

Translate to a system of equations:

This figure has four rows and three columns. The first row reads, “Step 1: Read the problem. Make sure you understand all the words and ideas. This is a number problem. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.”

Exercise \(\PageIndex{2}\)

The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

\(\left\{\begin{array}{l}{m+n=-23} \\ {m=n-7}\end{array}\right.\)

Exercise \(\PageIndex{3}\)

The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.

\(\left\{\begin{array}{l}{m+n=-18} \\ {m=n+40}\end{array}\right.\)

We’ll do another example where we stop after we write the system of equations.

Exercise \(\PageIndex{4}\)

\(\begin{array}{ll}{\text {We are looking for the amount that }} & {\text {Let } h=\text { the amount the husband earns. }} \\ {\text {the husband and wife each earn. }} & { w=\text { the amount the wife earns }} \\ {\text{Translate.}} & {\text{A married couple together earns \$110,000.} }\\ {} & {w+h=110000} \\ & \text{The wife earns \$16,000 less than twice what} \\ & \text{husband earns.} \\ & w=2h−16,000 \\ \text{The system of equations is:} & \left\{\begin{array}{l}{w+h=110,000} \\ {w=2 h-16,000}\end{array}\right.\end{array}\)

Exercise \(\PageIndex{5}\)

A couple has a total household income of $84,000. The husband earns $18,000 less than twice what the wife earns. How much does the wife earn?

\(\left\{\begin{array}{l}{w+h=84,000} \\ {h=2 w-18,000}\end{array}\right.\)

Exercise \(\PageIndex{6}\)

A senior employee makes $5 less than twice what a new employee makes per hour. Together they make $43 per hour. How much does each employee make per hour?

\(\left\{\begin{array}{l}{s=2 n-5} \\ {s+n=43}\end{array}\right.\)

Solve Direct Translation Applications

We set up, but did not solve, the systems of equations in Exercise \(\PageIndex{1}\) and Exercise \(\PageIndex{4}\) Now we’ll translate a situation to a system of equations and then solve it.

Exercise \(\PageIndex{7}\)

Translate to a system of equations and then solve:

Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

Exercise \(\PageIndex{8}\)

Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.

Ali is 26 and Jameela is 14.

Exercise \(\PageIndex{9}\)

Jake’s dad is 6 more than 3 times Jake’s age. The sum of their ages is 42. Find their ages.

Jake is 9 and his dad is 33.

Exercise \(\PageIndex{10}\)

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

Exercise \(\PageIndex{11}\)

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Exercise \(\PageIndex{12}\)

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

Solve Geometry Applications

When we learned about Math Models , we solved geometry applications using properties of triangles and rectangles. Now we’ll add to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

COMPLEMENTARY AND SUPPLEMENTARY ANGLES

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Exercise \(\PageIndex{13}\)

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

\(\begin{array}{ll}{\textbf {Step 1. Read}\text{ the problem. }} & {} \\ {\textbf {Step 2. Identify}\text{ what we are looking for.}} & {\text {We are looking for the measure of each angle.}} \\ \\ {\textbf{Step 3. Name}\text{ what we are looking for.}} & {\text{Let x = the measure of the first angle.} }\\ {} & \text{y = the measure of the second angle} \\ \textbf{Step 4. Translate}\text{ into a system of equations.}& \text{The angles are complementary.} \\ & \text{x+y=90} \\ & \text{The difference of the two angles is 26 degrees.} \\ & \text{x−y=26} \\ \\ \text{The system is} & {\left\{\begin{array}{l}{x+y=90} \\ {x-y=26}\end{array}\right.} \\ \textbf{Step 5. Solve}\text{ the system of equations by elimination.} \\& \left\{\begin{array}{l}{x+y=90} \\ \underline{x-y=26}\end{array}\right. \\ & \quad2x\quad=116 \\ \text{Substitute x = 58 into the first equation.}& \begin{array}{lrll} &x&=&58 \\ &x+y&=&90 \\ &58+y&=&90 \\ &y&=&32\end{array} \\ \textbf{Step 6. Check}\text{ the answer in the problem.} & \\ 58+32=90\checkmark\\ 58-32=36\checkmark \\ \\ \textbf{Step 7. Answer}\text{ the question.} & \text{The angle measures are 58 degrees and 32 degrees.}\end{array}\)

Exercise \(\PageIndex{14}\)

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

The angle measures are 55 degrees and 35 degrees.

Exercise \(\PageIndex{15}\)

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

The angle measures are 5 degrees and 85 degrees.

Exercise \(\PageIndex{16}\)

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Exercise \(\PageIndex{17}\)

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

The angle measures are 42 degrees and 138 degrees.

Exercise \(\PageIndex{18}\)

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

The angle measures are 66 degrees and 114 degrees.

Exercise \(\PageIndex{19}\)

Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Exercise \(\PageIndex{20}\)

Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

The length is 60 feet and the width is 35 feet.

Exercise \(\PageIndex{21}\)

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

The length is 60 feet and the width is 38 feet.

Solve Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Exercise \(\PageIndex{22}\)

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”

Exercise \(\PageIndex{23}\)

Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

It will take Clark 4 hours to catch Mitchell.

Exercise \(\PageIndex{24}\)

Translate to a system of equations and then solve: Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes (1/4 hour) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

It will take Sally \(1\frac{1}{2}\) hours to catch up to Charlie.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat traveling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

Figure \(\PageIndex{1}\) and Figure \(\PageIndex{2}\) show how a river current affects the speed at which a boat is actually traveling. We’ll call the speed of the boat in still water b and the speed of the river current c .

In Figure \(\PageIndex{1}\) the boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c .

This figure shows a boat floating in water. On the right, there is an arrow pointing towards the boat. It is labeled “c.” On the left, there is an arrow pointing away from the boat. It is labeled “b.”

In Figure \(\PageIndex{2}\) the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b−c.

This figure shows a boat floating in water. To the left is an arrow pointing away from the boat labeled “b,” and an arrow pointing towards the boat labeled “c.”

We’ll put some numbers to this situation in Exercise \(\PageIndex{25}\).

Exercise \(\PageIndex{25}\)

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Read the problem.

This is a uniform motion problem and a picture will help us visualize the situation.

This figure shows an arrow labeled “c” which continues to the right, representing the wave. Under the wave is a ray that points to the right and is labeled “four hours.” Under this ray is another ray pointing to the left labeled “five hours.” It is the same length as the ray labeled “four hours.” There is a bracket under the ray labeled “five hours.” The bracket is labeled “60 miles.”

Exercise \(\PageIndex{26}\)

Translate to a system of equations and then solve: A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

Exercise \(\PageIndex{27}\)

Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in Exercise \(\PageIndex{28}\). A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

Exercise \(\PageIndex{28}\)

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

This is a uniform motion problem and a picture will help us visualize.

This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

Exercise \(\PageIndex{29}\)

Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

Exercise \(\PageIndex{30}\)

Translate to a system of equations and then solve: A commercial jet can fly 1728 miles in 4 hours with a tailwind but only 1536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

Grade 8 Mathematics Module: “Solving Problems Involving Linear Functions”

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this module, you will be acquainted with solving problems involving linear functions. The scope of this module enables you to use it in many different learning situations. The lesson is arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

This module contains:

  • Lesson 1: Solving Problems Involving Linear Functions

After going through this module, you are expected to:

1. identify steps in modeling and solving word problems involving linear functions;

2. create linear functions that represent relation between quantities; and

3. apply the concepts of linear function in solving real-life problems.

Grade 8 Mathematics Quarter 2 Self-Learning Module: “Solving Problems Involving Linear Functions”

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  1. Linear Functions Problems with Solutions

    Solution to Problem 1: f is a linear function whose formula has the form f (x) = a x + b where a and b are constants to be found. Note that 2 ordered pairs (-3,17) and (4,-18) are given in the table. These two ordered pairs are used to write a system of linear equations as follows 17 = - 3 a + b and -18 = 4 a + b

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    Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.

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    Guidelines for Setting Up and Solving Word Problems. Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.; Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.; Step 3: Translate and set up an algebraic equation that models the problem.; Step 4: Solve the resulting algebraic equation.

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    Solving a Linear Function - Part 2. In the previous lesson on functions you learned how to find the slope and write an equation when given a function. Linear functions are very much like linear equations, the only difference is you are using function notation "f (x)" instead of "y". Otherwise, the process is the same.

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  26. Grade 8 Mathematics Module: "Solving Problems Involving Linear Functions"

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