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How to Solve Systems of Algebraic Equations Containing Two Variables

Last Updated: July 30, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,049,750 times.

In a "system of equations," you are asked to solve two or more equations at the same time. When these have two different variables in them, such as x and y, or a and b, it can be tricky at first glance to see how to solve them. [1] X Research source Fortunately, once you know what to do, all you need is basic algebra skills (and sometimes some knowledge of fractions) to solve the problem. If you are a visual learner or if your teacher requires it, learn how to graph the equations as well. Graphing can be useful to "see what's going on" or to check your work, but it can be slower than the other methods, and doesn't work well for all systems of equations.

Using the Substitution Method

Step 1 Move the variables to different sides of the equation.

  • This method often uses fractions later on. You can try the elimination method below instead if you don't like fractions.

Step 2 Divide both sides of the equation to

  • 4x = 8 - 2y
  • (4x)/4 = (8/4) - (2y/4)

Step 3 Plug this back into the other equation.

  • You know that x = 2 - ½y .
  • Your second equation, that you haven't yet altered, is 5x + 3y = 9 .
  • In the second equation, replace x with "2 - ½y": 5(2 - ½y) + 3y = 9 .

Step 4 Solve for the remaining variable.

  • 5(2 - ½y) + 3y = 9
  • 10 – (5/2)y + 3y = 9
  • 10 – (5/2)y + (6/2)y = 9 (If you don't understand this step, learn how to add fractions . This is often, but not always, necessary for this method.)
  • 10 + ½y = 9

Step 5 Use the answer to solve for the other variable.

  • You know that y = -2
  • One of the original equations is 4x + 2y = 8 . (You can use either equation for this step.)
  • Plug in -2 instead of y: 4x + 2(-2) = 8 .

Step 6 Know what to do when both variables cancel out.

  • If you end up with an equation that has no variables and isn't true (for instance, 3 = 5), the problem has no solution . (If you graphed both of the equations, you'd see they were parallel and never intersect.)
  • If you end up with an equation without variables that is true (such as 3 = 3), the problem has infinite solutions . The two equations are exactly equal to each other. (If you graphed the two equations, you'd see they were the same line.)

Using the Elimination Method

Step 1 Find the variable that cancels out.

  • You have the system of equations 3x - y = 3 and -x + 2y = 4 .
  • Let's change the first equation so that the y variable will cancel out. (You can choose x instead, and you'll get the same answer in the end.)
  • The - y on the first equation needs to cancel with the + 2y in the second equation. We can make this happen by multiplying - y by 2.
  • Multiply both sides of the first equation by 2, like this: 2(3x - y)=2(3) , so 6x - 2y = 6 . Now the - 2y will cancel out with the +2y in the second equation.

Step 3 Combine the two equations.

  • Your equations are 6x - 2y = 6 and -x + 2y = 4 .
  • Combine the left sides: 6x - 2y - x + 2y = ?
  • Combine the right sides: 6x - 2y - x + 2y = 6 + 4 .

Step 4 Solve for the last variable.

  • You have 6x - 2y - x + 2y = 6 + 4 .
  • Group the x and y variables together: 6x - x - 2y + 2y = 6 + 4 .
  • Simplify: 5x = 10
  • Solve for x: (5x)/5 = 10/5 , so x = 2 .

Step 5 Solve for the other variable.

  • You know that x = 2 , and one of your original equations is 3x - y = 3 .
  • Plug in 2 instead of x: 3(2) - y = 3 .
  • Solve for y in the equation: 6 - y = 3
  • 6 - y + y = 3 + y , so 6 = 3 + y

Step 6 Know what to do when both variables cancel out.

  • If your combined equation has no variables and is not true (like 2 = 7), there is no solution that will work on both equations. (If you graph both equations, you'll see they're parallel and never cross.)
  • If your combined equation has no variables and is true (like 0 = 0), there are infinite solutions . The two equations are actually identical. (If you graph them, you'll see that they're the same line.)

Graphing the Equations

Step 1 Only use this method when told to do so.

  • The basic idea is to graph both equations, and find the point where they intersect. The x and y values at this point will give us the value of x and the value of y in the system of equations.

Step 2 Solve both equations for y.

  • Your first equation is 2x + y = 5 . Change this to y = -2x + 5 .
  • Your second equation is -3x + 6y = 0 . Change this to 6y = 3x + 0 , then simplify to y = ½x + 0 .
  • If both equations are identical , the entire line will be an "intersection". Write infinite solutions .

Step 3 Draw coordinate axes.

  • If you don't have graph paper, use a ruler to make sure the numbers are spaced precisely apart.
  • If you are using large numbers or decimals, you may need to scale your graph differently. (For example, 10, 20, 30 or 0.1, 0.2, 0.3 instead of 1, 2, 3).

Step 4 Draw the y-intercept for each line.

  • In our examples from earlier, one line ( y = -2x + 5 ) intercepts the y-axis at 5 . The other ( y = ½x + 0 ) intercepts at 0 . (These are points (0,5) and (0,0) on the graph.)
  • Use different colored pens or pencils if possible for the two lines.

Step 5 Use the slope to continue the lines.

  • In our example, the line y = -2x + 5 has a slope of -2 . At x = 1, the line moves down 2 from the point at x = 0. Draw the line segment between (0,5) and (1,3).
  • The line y = ½x + 0 has a slope of ½ . At x = 1, the line moves up ½ from the point at x=0. Draw the line segment between (0,0) and (1,½).
  • If the lines have the same slope , the lines will never intersect, so there is no answer to the system of equations. Write no solution .

Step 6 Continue plotting the lines until they intersect.

  • If the lines are moving toward each other, keep plotting points in that direction.
  • If the lines are moving away from each other, move back and plot points in the other direction, starting at x = -1.
  • If the lines are nowhere near each other, try jumping ahead and plotting more distant points, such as at x = 10.

Step 7 Find the answer at the intersection.

Practice Problems and Answers

how to solve algebraic equations with 2 variables

Community Q&A

Donagan

  • You can check your work by plugging the answers back into the original equations. If the equations end up true (for instance, 3 = 3), your answer is correct. Thanks Helpful 3 Not Helpful 1
  • In the elimination method, you will sometimes have to multiply one equation by a negative number in order to get a variable to cancel out. Thanks Helpful 1 Not Helpful 1

how to solve algebraic equations with 2 variables

  • These methods cannot be used if there is a variable raised to an exponent, such as x 2 . For more information on equations of this type, look up a guide to factoring quadratics with two variables. [11] X Research source Thanks Helpful 0 Not Helpful 0

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  • ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
  • ↑ https://calcworkshop.com/systems-equations/substitution-method/
  • ↑ https://www.cuemath.com/algebra/substitution-method/
  • ↑ https://tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx
  • ↑ http://www.purplemath.com/modules/systlin2.htm
  • ↑ http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
  • ↑ https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-quadratics-in-two-vari/v/factoring-quadratics-with-two-variables

About This Article

Grace Imson, MA

To solve systems of algebraic equations containing two variables, start by moving the variables to different sides of the equation. Then, divide both sides of the equation by one of the variables to solve for that variable. Next, take that number and plug it into the formula to solve for the other variable. Finally, take your answer and plug it into the original equation to solve for the other variable. To learn how to solve systems of algebraic equations using the elimination method, scroll down! Did this summary help you? Yes No

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  • 7.1 Systems of Linear Equations: Two Variables
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Solve systems of equations by graphing.
  • Solve systems of equations by substitution.
  • Solve systems of equations by addition.
  • Identify inconsistent systems of equations containing two variables.
  • Express the solution of a system of dependent equations containing two variables.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.

Introduction to Systems of Equations

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.

In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y -intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.

Another type of system of linear equations is an inconsistent system , which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system.

Types of Linear Systems

There are three types of systems of linear equations in two variables, and three types of solutions.

  • An independent system has exactly one solution pair ( x , y ) . ( x , y ) . The point where the two lines intersect is the only solution.
  • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
  • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.

Figure 2 compares graphical representations of each type of system.

Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.

  • Substitute the ordered pair into each equation in the system.
  • Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.

Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair ( 5 , 1 ) ( 5 , 1 ) is a solution to the given system of equations.

Substitute the ordered pair ( 5 , 1 ) ( 5 , 1 ) into both equations.

The ordered pair ( 5 , 1 ) ( 5 , 1 ) satisfies both equations, so it is the solution to the system.

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 3 .

Determine whether the ordered pair ( 8 , 5 ) ( 8 , 5 ) is a solution to the following system.

Solving Systems of Equations by Graphing

There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.

Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

Solve the first equation for y . y .

Solve the second equation for y . y .

Graph both equations on the same set of axes as in Figure 4 .

The lines appear to intersect at the point ( −3, −2 ) . ( −3, −2 ) . We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

The solution to the system is the ordered pair ( −3, −2 ) , ( −3, −2 ) , so the system is independent.

Solve the following system of equations by graphing.

Can graphing be used if the system is inconsistent or dependent?

Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.

Solving Systems of Equations by Substitution

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method , in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

Given a system of two equations in two variables, solve using the substitution method.

  • Solve one of the two equations for one of the variables in terms of the other.
  • Substitute the expression for this variable into the second equation, then solve for the remaining variable.
  • Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
  • Check the solution in both equations.

Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

First, we will solve the first equation for y . y .

Now we can substitute the expression x −5 x −5 for y y in the second equation.

Now, we substitute x = 8 x = 8 into the first equation and solve for y . y .

Our solution is ( 8 , 3 ) . ( 8 , 3 ) .

Check the solution by substituting ( 8 , 3 ) ( 8 , 3 ) into both equations.

Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.

Solving Systems of Equations in Two Variables by the Addition Method

A third method of solving systems of linear equations is the addition method . In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.

Given a system of equations, solve using the addition method.

  • Write both equations with x - and y -variables on the left side of the equal sign and constants on the right.
  • Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
  • Solve the resulting equation for the remaining variable.
  • Substitute that value into one of the original equations and solve for the second variable.
  • Check the solution by substituting the values into the other equation.

Solving a System by the Addition Method

Solve the given system of equations by addition.

Both equations are already set equal to a constant. Notice that the coefficient of x x in the second equation, –1, is the opposite of the coefficient of x x in the first equation, 1. We can add the two equations to eliminate x x without needing to multiply by a constant.

Now that we have eliminated x , x , we can solve the resulting equation for y . y .

Then, we substitute this value for y y into one of the original equations and solve for x . x .

The solution to this system is ( − 7 3 , 2 3 ) . ( − 7 3 , 2 3 ) .

Check the solution in the first equation.

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method .

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3 x 3 x in it and the second equation has x . x . So if we multiply the second equation by −3 , −3 , the x -terms will add to zero.

Now, let’s add them.

For the last step, we substitute y = −4 y = −4 into one of the original equations and solve for x . x .

Our solution is the ordered pair ( 3 , −4 ) . ( 3 , −4 ) . See Figure 6 . Check the solution in the original second equation.

Solve the system of equations by addition.

Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition.

One equation has 2 x 2 x and the other has 5 x . 5 x . The least common multiple is 10 x 10 x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x x by multiplying the first equation by −5 −5 and the second equation by 2. 2.

Then, we add the two equations together.

Substitute y = −4 y = −4 into the original first equation.

The solution is ( −2 , −4 ) . ( −2 , −4 ) . Check it in the other equation.

See Figure 7 .

Using the Addition Method in Systems of Equations Containing Fractions

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

Now multiply the second equation by −1 −1 so that we can eliminate the x -variable.

Add the two equations to eliminate the x -variable and solve the resulting equation.

Substitute y = 7 y = 7 into the first equation.

The solution is ( 11 2 , 7 ) . ( 11 2 , 7 ) . Check it in the other equation.

Identifying Inconsistent Systems of Equations Containing Two Variables

Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y y -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12 = 0. 12 = 0.

Solving an Inconsistent System of Equations

Solve the following system of equations.

We can approach this problem in two ways. Because one equation is already solved for x , x , the most obvious step is to use substitution.

Clearly, this statement is a contradiction because 9 ≠ 13. 9 ≠ 13. Therefore, the system has no solution.

The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.

We then convert the second equation expressed to slope-intercept form.

Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 8 .

Solve the following system of equations in two variables.

Expressing the Solution of a System of Dependent Equations Containing Two Variables

Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0 = 0. 0 = 0.

Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method .

With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x . x . If we multiply both sides of the first equation by −3 , −3 , then we will be able to eliminate the x x -variable.

Now add the equations.

We can see that there will be an infinite number of solutions that satisfy both equations.

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

See Figure 9 . Notice the results are the same. The general solution to the system is ( x , − 1 3 x + 2 3 ) . ( x , − 1 3 x + 2 3 ) .

Using Systems of Equations to Investigate Profits

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R = x p , R = x p , where x = x = quantity and p = p = price. The revenue function is shown in orange in Figure 10 .

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 10 . The x x -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P ( x ) = R ( x ) − C ( x ) . P ( x ) = R ( x ) − C ( x ) . Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function C ( x ) = 0.85 x + 35,000 C ( x ) = 0.85 x + 35,000 and the revenue function R ( x ) = 1.55 x , R ( x ) = 1.55 x , find the break-even point and the profit function.

Write the system of equations using y y to replace function notation.

Substitute the expression 0.85 x + 35,000 0.85 x + 35,000 from the first equation into the second equation and solve for x . x .

Then, we substitute x = 50,000 x = 50,000 into either the cost function or the revenue function.

The break-even point is ( 50,000 , 77,500 ) . ( 50,000 , 77,500 ) .

The profit function is found using the formula P ( x ) = R ( x ) − C ( x ) . P ( x ) = R ( x ) − C ( x ) .

The profit function is P ( x ) = 0.7 x −35,000. P ( x ) = 0.7 x −35,000.

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11 .

We see from the graph in Figure 12 that the profit function has a negative value until x = 50,000 , x = 50,000 , when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $ 25.00 $ 25.00 for children and $ 50.00 $ 50.00 for adults. On a certain day, attendance at the circus is 2,000 2,000 and the total gate revenue is $ 70,000. $ 70,000. How many children and how many adults bought tickets?

Let c = the number of children and a = the number of adults in attendance.

The total number of people is 2,000. 2,000. We can use this to write an equation for the number of people at the circus that day.

The revenue from all children can be found by multiplying $ 25.00 $ 25.00 by the number of children, 25 c . 25 c . The revenue from all adults can be found by multiplying $ 50.00 $ 50.00 by the number of adults, 50 a . 50 a . The total revenue is $ 70,000. $ 70,000. We can use this to write an equation for the revenue.

We now have a system of linear equations in two variables.

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either c c or a . a . We will solve for a . a .

Substitute the expression 2,000 − c 2,000 − c in the second equation for a a and solve for c . c .

Substitute c = 1,200 c = 1,200 into the first equation to solve for a . a .

We find that 1,200 1,200 children and 800 800 adults bought tickets to the circus that day.

Meal tickets at the circus cost $ 4.00 $ 4.00 for children and $ 12.00 $ 12.00 for adults. If 1,650 1,650 meal tickets were bought for a total of $ 14,200 , $ 14,200 , how many children and how many adults bought meal tickets?

Access these online resources for additional instruction and practice with systems of linear equations.

  • Solving Systems of Equations Using Substitution
  • Solving Systems of Equations Using Elimination
  • Applications of Systems of Equations

Can a system of linear equations have exactly two solutions? Explain why or why not.

If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.

If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?

If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?

Given a system of equations, explain at least two different methods of solving that system.

For the following exercises, determine whether the given ordered pair is a solution to the system of equations.

5 x − y = 4 x + 6 y = 2 5 x − y = 4 x + 6 y = 2 and ( 4 , 0 ) ( 4 , 0 )

−3 x − 5 y = 13 − x + 4 y = 10 −3 x − 5 y = 13 − x + 4 y = 10 and ( −6 , 1 ) ( −6 , 1 )

3 x + 7 y = 1 2 x + 4 y = 0 3 x + 7 y = 1 2 x + 4 y = 0 and ( 2 , 3 ) ( 2 , 3 )

−2 x + 5 y = 7 2 x + 9 y = 7 −2 x + 5 y = 7 2 x + 9 y = 7 and ( −1 , 1 ) ( −1 , 1 )

x + 8 y = 43 3 x −2 y = −1 x + 8 y = 43 3 x −2 y = −1 and ( 3 , 5 ) ( 3 , 5 )

For the following exercises, solve each system by substitution.

x + 3 y = 5 2 x + 3 y = 4 x + 3 y = 5 2 x + 3 y = 4

3 x −2 y = 18 5 x + 10 y = −10 3 x −2 y = 18 5 x + 10 y = −10

4 x + 2 y = −10 3 x + 9 y = 0 4 x + 2 y = −10 3 x + 9 y = 0

2 x + 4 y = −3.8 9 x −5 y = 1.3 2 x + 4 y = −3.8 9 x −5 y = 1.3

− 2 x + 3 y = 1.2 − 3 x − 6 y = 1.8 − 2 x + 3 y = 1.2 − 3 x − 6 y = 1.8

x −0.2 y = 1 −10 x + 2 y = 5 x −0.2 y = 1 −10 x + 2 y = 5

3 x + 5 y = 9 30 x + 50 y = −90 3 x + 5 y = 9 30 x + 50 y = −90

−3 x + y = 2 12 x −4 y = −8 −3 x + y = 2 12 x −4 y = −8

1 2 x + 1 3 y = 16 1 6 x + 1 4 y = 9 1 2 x + 1 3 y = 16 1 6 x + 1 4 y = 9

− 1 4 x + 3 2 y = 11 − 1 8 x + 1 3 y = 3 − 1 4 x + 3 2 y = 11 − 1 8 x + 1 3 y = 3

For the following exercises, solve each system by addition.

−2 x + 5 y = −42 7 x + 2 y = 30 −2 x + 5 y = −42 7 x + 2 y = 30

6 x −5 y = −34 2 x + 6 y = 4 6 x −5 y = −34 2 x + 6 y = 4

5 x − y = −2.6 −4 x −6 y = 1.4 5 x − y = −2.6 −4 x −6 y = 1.4

7 x −2 y = 3 4 x + 5 y = 3.25 7 x −2 y = 3 4 x + 5 y = 3.25

−x + 2 y = −1 5 x −10 y = 6 −x + 2 y = −1 5 x −10 y = 6

7 x + 6 y = 2 −28 x −24 y = −8 7 x + 6 y = 2 −28 x −24 y = −8

5 6 x + 1 4 y = 0 1 8 x − 1 2 y = − 43 120 5 6 x + 1 4 y = 0 1 8 x − 1 2 y = − 43 120

1 3 x + 1 9 y = 2 9 − 1 2 x + 4 5 y = − 1 3 1 3 x + 1 9 y = 2 9 − 1 2 x + 4 5 y = − 1 3

−0.2 x + 0.4 y = 0.6 x −2 y = −3 −0.2 x + 0.4 y = 0.6 x −2 y = −3

−0.1 x + 0.2 y = 0.6 5 x −10 y = 1 −0.1 x + 0.2 y = 0.6 5 x −10 y = 1

For the following exercises, solve each system by any method.

5 x + 9 y = 16 x + 2 y = 4 5 x + 9 y = 16 x + 2 y = 4

6 x −8 y = −0.6 3 x + 2 y = 0.9 6 x −8 y = −0.6 3 x + 2 y = 0.9

5 x −2 y = 2.25 7 x −4 y = 3 5 x −2 y = 2.25 7 x −4 y = 3

x − 5 12 y = − 55 12 −6 x + 5 2 y = 55 2 x − 5 12 y = − 55 12 −6 x + 5 2 y = 55 2

7 x −4 y = 7 6 2 x + 4 y = 1 3 7 x −4 y = 7 6 2 x + 4 y = 1 3

3 x + 6 y = 11 2 x + 4 y = 9 3 x + 6 y = 11 2 x + 4 y = 9

7 3 x − 1 6 y = 2 − 21 6 x + 3 12 y = −3 7 3 x − 1 6 y = 2 − 21 6 x + 3 12 y = −3

1 2 x + 1 3 y = 1 3 3 2 x + 1 4 y = − 1 8 1 2 x + 1 3 y = 1 3 3 2 x + 1 4 y = − 1 8

2.2 x + 1.3 y = −0.1 4.2 x + 4.2 y = 2.1 2.2 x + 1.3 y = −0.1 4.2 x + 4.2 y = 2.1

0.1 x + 0.2 y = 2 0.35 x −0.3 y = 0 0.1 x + 0.2 y = 2 0.35 x −0.3 y = 0

For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.

3 x − y = 0.6 x −2 y = 1.3 3 x − y = 0.6 x −2 y = 1.3

− x + 2 y = 4 2 x −4 y = 1 − x + 2 y = 4 2 x −4 y = 1

x + 2 y = 7 2 x + 6 y = 12 x + 2 y = 7 2 x + 6 y = 12

3 x −5 y = 7 x −2 y = 3 3 x −5 y = 7 x −2 y = 3

3 x −2 y = 5 −9 x + 6 y = −15 3 x −2 y = 5 −9 x + 6 y = −15

For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.

0.1 x + 0.2 y = 0.3 −0.3 x + 0.5 y = 1 0.1 x + 0.2 y = 0.3 −0.3 x + 0.5 y = 1

−0.01 x + 0.12 y = 0.62 0.15 x + 0.20 y = 0.52 −0.01 x + 0.12 y = 0.62 0.15 x + 0.20 y = 0.52

0.5 x + 0.3 y = 4 0.25 x −0.9 y = 0.46 0.5 x + 0.3 y = 4 0.25 x −0.9 y = 0.46

0.15 x + 0.27 y = 0.39 −0.34 x + 0.56 y = 1.8 0.15 x + 0.27 y = 0.39 −0.34 x + 0.56 y = 1.8

−0.71 x + 0.92 y = 0.13 0.83 x + 0.05 y = 2.1 −0.71 x + 0.92 y = 0.13 0.83 x + 0.05 y = 2.1

For the following exercises, solve each system in terms of A , B , C , D , E , A , B , C , D , E , and F F where A – F A – F are nonzero numbers. Note that A ≠ B A ≠ B and A E ≠ B D . A E ≠ B D .

x + y = A x − y = B x + y = A x − y = B

x + A y = 1 x + B y = 1 x + A y = 1 x + B y = 1

A x + y = 0 B x + y = 1 A x + y = 0 B x + y = 1

A x + B y = C x + y = 1 A x + B y = C x + y = 1

A x + B y = C D x + E y = F A x + B y = C D x + E y = F

Real-World Applications

For the following exercises, solve for the desired quantity.

A stuffed animal business has a total cost of production C = 12 x + 30 C = 12 x + 30 and a revenue function R = 20 x . R = 20 x . Find the break-even point.

An Ethiopian restaurant has a cost of production C ( x ) = 11 x + 120 C ( x ) = 11 x + 120 and a revenue function R ( x ) = 5 x . R ( x ) = 5 x . When does the company start to turn a profit?

A cell phone factory has a cost of production C ( x ) = 150 x + 10 , 000 C ( x ) = 150 x + 10 , 000 and a revenue function R ( x ) = 200 x . R ( x ) = 200 x . What is the break-even point?

A musician charges C ( x ) = 64 x + 20,000 C ( x ) = 64 x + 20,000 where x x is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point?

A guitar factory has a cost of production C ( x ) = 75 x + 50,000. C ( x ) = 75 x + 50,000. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function.

For the following exercises, use a system of linear equations with two variables and two equations to solve.

Find two numbers whose sum is 28 and difference is 13.

A number is 9 more than another number. Twice the sum of the two numbers is 10. Find the two numbers.

The startup cost for a restaurant is $120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even?

A moving company charges a flat rate of $150, and an additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost?

A total of 1,595 first- and second-year college students gathered at a pep rally. The number of first-years exceeded the number of second-years by 15. How many students from each year group were in attendance?

276 students enrolled in an introductory chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed.

There were 130 faculty at a conference. If there were 18 more women than men attending, how many of each gender attended the conference?

A jeep and a pickup truck enter a highway running east-west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the pickup did, and traveled 7 mph slower than the pickup. After 2 hours from the time the pickup entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control and retained the same speed.

If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed?

An investor earned triple the profits of what they earned last year. If they made $500,000.48 total for both years, how much did the investor earn in profits each year?

An investor invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals?

If an investor invests a total of $25,000 into two bonds, one that pays 3% simple interest, and the other that pays 2 7 8 % 2 7 8 % interest, and the investor earns $737.50 annual interest, how much was invested in each account?

If an investor invests $23,000 into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account?

Blu-rays cost $5.96 more than regular DVDs at All Bets Are Off Electronics. How much would 6 Blu-rays and 2 DVDs cost if 5 Blu-rays and 2 DVDs cost $127.73?

A store clerk sold 60 pairs of sneakers. The high-tops sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold?

A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold?

Admission into an amusement park for 4 children and 2 adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket?

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Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

  • For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
  • For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

  • Add or Subtract the same value from both sides
  • Clear out any fractions by Multiplying every term by the bottom parts
  • Divide every term by the same nonzero value
  • Combine Like Terms
  • Expanding (the opposite of factoring) may also help
  • Recognizing a pattern, such as the difference of squares
  • Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

  • solve Quadratic Equations
  • solve Radical Equations
  • solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

  • Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
  • Show all the steps , so it can be checked later (by you or someone else)

Algebraic Equations

Algebraic equations are two algebraic expressions that are joined together using an equal to ( = ) sign. An algebraic equation is also known as a polynomial equation because both sides of the equal sign contain polynomials. An algebraic equation is built up of variables, coefficients, constants as well as algebraic operations such as addition, subtraction, multiplication, division, exponentiation, etc.

If there is a number or a set of numbers that satisfy the algebraic equation then they are known as the roots or the solutions of that equation. In this article, we will learn more about algebraic equations, their types, examples, and how to solve algebraic equations.

What is Algebraic Equations?

An algebraic equation is a mathematical statement that contains two equated algebraic expressions. The general form of an algebraic equation is P = 0 or P = Q, where P and Q are polynomials . Algebraic equations that contain only one variable are known as univariate equations and those which contain more than one variable are known as multivariate equations. An algebraic equation will always be balanced. This means that the right-hand side of the equation will be equal to the left-hand side.

Algebraic Equations

Algebraic Expressions

A polynomial expression that contains variables, coefficients, and constants joined together using operations such as addition , subtraction, multiplication, division, and non-negative exponentiation is known as an algebraic expression . An algebraic expression should not be confused with an algebraic equation. When two algebraic expressions are merged together using an "equal to" sign then they form an algebraic equation. Thus, 5x + 1 is an expression while 5x + 1 = 0 will be an equation.

Algebraic Equations Examples

x 2 - 5x = 3 is a univariate algebraic equation while y 2 x - 5z = 3x is an example of a multivariate algebraic equation.

Types of Algebraic Equations

Algebraic equations can be classified into different types based on the degree of the equation. The degree can be defined as the highest exponent of a variable in an algebraic equation. Suppose there is an equation given by x 4 + y 3 = 3 5 then the degree will be 4. In determining the degree, the exponent of the constant or coefficient is not considered. The number of roots of an algebraic equation depends on its degree. An algebraic equation where the degree equals 5 will have a maximum of 5 roots. The various types of algebraic equations are as follows:

Linear Algebraic Equations

A linear algebraic equation is one in which the degree of the polynomial is 1. The general form of a linear equation is given as a 1 x 1 +a 2 x 2 +...+a n x n = 0 where at least one coefficient is a non-zero number. These linear equations are used to represent and solve linear programming problems.

Example: 3x + 5 = 5 is a linear equation in one variable . y = 2x - 6 is a linear equation in two variables .

Quadratic Algebraic Equations

An equation where the degree of the polynomial is 2 is known as a quadratic algebraic equation . The general form of such an equation is ax 2 + bx + c = 0, where a is not equal to 0.

Example: 3x 2 + 2x - 6 = 0 is a quadratic algebraic equation. This type of equation will have a maximum of two solutions.

Cubic Algebraic Equations

An algebraic equation where the degree equals 3 will be classified as a cubic algebraic equation . ax 3 + bx 2 + cx + d = 0 is the general form of a cubic algebraic equation (a ≠ 0).

Example: x 3 + x 2 - x - 1 = 0. A cubic algebraic equation will have a maximum of three roots as the degree is 3.

Higher-Order Polynomial Algebraic Equations

Algebraic equations that have a degree greater than 3 are known as higher-order polynomial algebraic equations. Quartic (degree = 4), quintic (5), sextic (6), septic (7) equations all fall under the category of higher algebraic equations. Such equations might not be solvable using a finite number of operations.

Algebraic Equations Formulas

Algebraic equations can be simplified using several formulas and identities. These help to expedite the process of solving a given equation. Given below are some important algebraic formulas :

  • (a + b) 2 = a 2 + 2ab + b 2
  • (a - b) 2 = a 2 - 2ab + b 2
  • (a + b)(a - b) = a 2 - b 2
  • (x + a)(x + b) = x 2 + x(a + b) + ab
  • (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
  • (a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
  • a 3 + b 3 = (a + b)(a 2 - ab + b 2 )
  • a 3 - b 3 = (a - b)(a 2  + ab + b 2 )
  • (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca
  • Quadratic Formula : [-b ± √(b² - 4ac)]/2a
  • Discriminant : b 2 - 4ac

How to Solve Algebraic Equations

There are many different methods that are available for solving algebraic equations depending upon the degree. If an algebraic equation has two variables then two equations will be required to find the solution. Thus, it can be said that the number of equations required to solve an algebraic equation will be equal to the number of variables present in the equation. Given below are the ways to solve algebraic equations.

Algebraic Equations Example

A linear algebraic equation in one variable can be solved by simply applying basic arithmetic operations  on both sides of the equation.

E.g: 4x + 1 = 5.

4x = 5 - 1 (Subtracted 1 from both sides).

4x = 4 (Solve the R.H.S using algebraic operations)

x = 1 (Divided both sides by 4)

Linear algebraic equations in more than one variable will be solved using the concept of simultaneous equations .

Algebraic Equations Types

A quadratic algebraic equation can be solved by using identities , factorizing , long division, splitting the middle term, completing the square , applying the quadratic formula, and using graphs . A quadratic equation will always have a maximum of two roots.

E.g: x 2 + 2x + 1 = 0

Using the identity (a + b) 2 = a 2 + 2ab + b 2 , we get

a = x and b = 1

(x + 1) 2 = 0

(x + 1)(x + 1) = 0

x = -1, -1.

The most effective way of solving higher-order algebraic polynomials in one variable is by using the long division method. This decomposes the higher-order polynomial into polynomials of a lower degree thus, making it easier to find the solutions.

☛ Related Articles:

  • Variable Expressions
  • Algebraic Formula Calculator
  • Solve For x Calculator
  • Equation Calculator

Important Notes on Algebraic Equations:

  • An algebraic equation is an equation where two algebraic expressions are joined together using an equal sign .
  • Polynomial equations are algebra equations.
  • Algebraic equations can be one-step, two-step , or multi-step equations .
  • Algebra equations are classified as linear, quadratic, cubic, and higher-order equations based on the degree.
  • Example 1: Solve the algebraic equation x + 3 = 2x Solution: Taking the variable terms on one side of the equation and keeping the constant terms on the other side we get, 3 = 2x - x 3 = x Answer: x = 3
  • Example 2: A total of 15 items can fit in a box. If the box contains 2 scales, 7 pencils, and 1 eraser then how many pens can fit in the box? Solution: Converting this problem statement in the form of an algebraic equation we get, 2 scales + 7 pencils + 1 eraser + x pens = 15 2 + 7 + 1 + x = 15 Solving the L.H.S 10 + x = 15 x = 15 - 10 x = 5 Answer: 5 pens can fit in the box
  • Example 3: Find the roots of the quadratic equation x 2 + x - 6 = 0 Solution: Using the quadratic formula x = [-b ± √(b² - 4ac)]/2a. a = 1, b = 1, c = - 6 x = [-1 ± √(1² - 4 · 1 · -6)] / (2 · 1) x = [-1 ± √(25)] / 2 x = [-1 + 5] / 2,  [-1 - 5] / 2 x = 2, -3 Answer: The roots of the given algebraic equation are 2 and -3.

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Practice Questions on Algebraic Equations

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FAQs on Algebraic Equations

What are algebraic equations.

Algebraic equations are polynomial equations where two algebraic expressions are equated. Both sides of the equation must be balanced. The general form of an algebraic equation is P = 0.

What is an Example of Algebraic Equation?

An algebraic equation can be linear, quadratic, etc. Hence, an example of an algebraic equation can be 3x 2  - 6 = 0.

How Do You Solve Algebraic Equations?

There are many methods available to solve algebraic equations depending on the degree. Some techniques include applying simple algebraic operations , solving simultaneous equations , splitting the middle term, quadratic formula, long division, and so on.

What are Algebraic Expressions and Algebraic Equations?

Mathematical statements that consist of variables , coefficients , constants , and algebraic operations are known as algebraic expressions. When two algebraic expressions are equated together, they are known as algebraic equations.

How Do You Write Algebraic Equation?

We can convert real-life statement involving numbers and conditions into algebraic equation. For example, if the problem says, "the length of a rectangular field is 5 more than twice the width", then it can be written as the algebraic equations l = 2w + 5, where 'l' and 'w' are the length and width of the rectangular field.

What are Linear Algebraic Equations?

An algebraic equation where the highest exponent of the variable term is 1 is a linear algebraic equation. In other words, algebraic equations with degree 1 will be linear. For example, 3y - 9 = 1

Are Quadratic Equations Algebraic Equations?

Yes, quadratic equations are algebraic equations. It consists of an algebraic expression of the second degree.

What are the Basic Formulas of Algebraic Equations?

Some of the basic formulas of algebraic equations are listed below:

  • Quadratic Formula: [-b ± √(b² - 4ac)]/2a
  • Discriminant: b 2 - 4ac

What are the Rules for Algebraic Equations?

There are 5 basic rules for algebraic equations. These are as follows:

  • Commutative Rule of Addition
  • Commutative Rule of Multiplication
  • Associative Rule of Addition
  • Associative Rule of Multiplication
  • Distributive Rule of Multiplication

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Solving systems of equations in two variables

  • System with two variables I
  • System with two variables II
  • System with two variables III

A system of a linear equation comprises two or more equations and one seeks a common solution to the equations. In a system of linear equations, each equation corresponds with a straight line corresponds and one seeks out the point where the two lines intersect.

Solve the following system of linear equations:

$$\left\{\begin{matrix} y=2x+4\\ y=3x+2\\ \end{matrix}\right.$$

Since we are seeking out the point of intersection, we may graph the equations:

Graph 6

We see here that the lines intersect each other at the point x = 2, y = 8. This is our solution and we may refer to it as a graphic solution to the task.

But how does one reach a solution if the lines never intersect? One cannot, the system of equations have no solution.

One may also arrive at the correct answer with the help of the elimination method (also called the addition method or the linear combination method) or the substitution method.

When using the substitution method we use the fact that if two expressions y and x are of equal value x=y, then x may replace y or vice versa in another expression without changing the value of the expression.

Solve the systems of equations using the substitution method

We substitute the y in the top equation with the expression for the second equation:

$$\begin{array}{lcl} 2x+4 & = & 3x+2\\ 4-2 & = & 3x-2x\\ 2 & = & x\\ \end{array}$$

To determine the y -value, we may proceed by inserting our x -value in any of the equations. We select the first equation:

We plug in x=2 and get

$$y=2\cdot 2+4=8$$

We have thus arrived at precisely the same answer as in the graphic solution.

The elimination method requires us to add or subtract the equations in order to eliminate either x or y , often one may not proceed with the addition directly without first multiplying either the first or second equation by some value.

$$2x-2y=8$$

We now wish to add the two equations but it will not result in either x or y being eliminated. Therefore we must multiply the second equation by 2 on both sides and get:

$$2x+2y=2$$

Now we attempt to add our system of equations. We commence with the x -terms on the left, and the y -terms thereafter and finally with the numbers on the right side:

$$(2x+2x)+(-2y+2y)=8+2$$

The y -terms have now been eliminated and we now have an equation with only one variable:

$$x=\frac{10}{4}=2.5$$

Thereafter, in order to determine the y -value we insert x =2.5 in one of the equations. We select the first:

$$\begin{array}{lcl} 2\cdot 2.5-2y & = & 8\\ 5-8 & = & 2y\\ -3 & = & 2y\\ \frac{-3}{2} & = & y\\ y & = & -1.5\\ \end{array}$$

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Solve the system of equations:

$$\left\{\begin{matrix} 2x-4y=0\\ -4x+4y=-4 \end{matrix}\right.$$

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Mathematics LibreTexts

4.2: Solving Linear Systems with Two Variables

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Learning Objectives

  • Solve linear systems using the substitution method.
  • Solve linear systems using the elimination method.
  • Identify the strengths and weaknesses of each method.

The Substitution Method

In this section, we review a completely algebraic technique for solving systems, the substitution method 11 . The idea is to solve one equation for one of the variables and substitute the result into the other equation. After performing this substitution step, we are left with a single equation with one variable, which can be solved using algebra.

Example \(\PageIndex{1}\):

Solve by substitution: \(\left\{ \begin{array} { l } { 2 x + y = - 3 } \\ { 3 x - 2 y = - 8 } \end{array} \right.\).

Solve for either variable in either equation. If you choose the first equation, you can isolate \(y\) in one step.

\(\begin{aligned} 2 x + y & = - 3 \\ y & = - 2 x - 3 \end{aligned}\)

Substitute the expression \(-2x-3\) for the variable \(y\) in the other equation.

c4ebb8396d6fda10444882fe32dc0b8e.png

\(3 x - 2 ( - \color{OliveGreen}{2 x - 3}\color{Black}{ )} = - 8\)

This leaves us with an equivalent equation with one variable, which can be solved using the techniques learned up to this point. Solve for the remaining variable.

\(\begin{aligned} 3 x - 2 ( \color{OliveGreen}{- 2 x - 3}\color{Black}{ )} & = - 8 \\ 3 x + 4 x + 6 & = - 8 \\ 7 x + 6 & = - 8 \\ 7 x & = - 14 \\ x & = - 2 \end{aligned}\)

Back substitute 12 to find the other coordinate. Substitute \(x = −2\) into either of the original equations or their equivalents. Typically, we use the equivalent equation that we found when isolating a variable in the first step.

\(\begin{aligned} y & = - 2 x - 3 \\ & = - 2 ( \color{OliveGreen}{- 2}\color{Black}{ )} - 3 \\ & = 4 - 3 \\ & = 1 \end{aligned}\)

Remember to present the solution as an ordered pair: \((−2, 1)\). Verify that these coordinates solve both equations of the original system:

The graph of this linear system follows:

d96f1c69a85d456f86fcfedbddf0487c.png

The substitution method for solving systems is a completely algebraic method. Thus graphing the lines is not required.

\((-2, 1)\)

Example \(\PageIndex{2}\):

Solve by substitution: \(\left\{ \begin{array} { l } { 3 x - 5 y = 9 } \\ { 4 x + 2 y = - 1 } \end{array} \right.\).

It does not matter which variable we choose to isolate first. In this case, begin by solving for \(x\) in the first equation.

\(\begin{aligned} 3 x - 5 y & = 9 \\ 3 x & = 5 y + 9 \\ x & = \frac { 5 y + 9 } { 3 } \\ x & = \frac { 5 } { 3 } y + 3 \end{aligned}\)

\(\left\{ \begin{array} { c } { 3 x - 5 y = 9 \Longrightarrow \color{Cerulean}{x}\color{Black}{ =} \frac { 5 } { 3 } y + 3 } \\ { 4\color{Cerulean}{ x}\color{Black}{ +} 2 y = - 1 } \end{array} \right.\)

Next, substitute into the second equation and solve for \(y\).

\(\begin{aligned} 4 \left( \frac { 5 } { 3 } y + 3 \right) + 2 y & = - 1 \\ \frac { 20 } { 3 } y + 12 + 2 y & = - 1 \\ \frac { 26 } { 3 } y & = - 13 \\ y & = - 13 \left( \frac { 3 } { 26 } \right) \\ y & = - \frac { 3 } { 2 } \end{aligned}\)

Back substitute into the equation used in the substitution step:

\(\begin{aligned} x & = \frac { 5 } { 3 } y + 3 \\ & = \frac { 5 } { 3 } \left( \color{Cerulean}{- \frac { 3 } { 2 }} \right) + 3 \\ & = - \frac { 5 } { 2 } + 3 \\ & = \frac { 1 } { 2 } \end{aligned}\)

\(\left( \frac { 1 } { 2 } , - \frac { 3 } { 2 } \right)\)

Exercise \(\PageIndex{1}\)

Solve by substitution: \(\left\{ \begin{array} { l } { 5 x - 4 y = 3 } \\ { x + 2 y = 2 } \end{array} \right.\).

\(\left( 1 , \frac { 1 } { 2 } \right)\)

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As we know, not all linear systems have only one ordered pair solution. Next, we explore what happens when using the substitution method to solve a dependent system.

Example \(\PageIndex{3}\):

Solve by substitution: \(\left\{ \begin{array} { l } { - 5 x + y = - 1 } \\ { 10 x - 2 y = 2 } \end{array} \right.\).

Since the first equation has a term with coefficient \(1\), we choose to solve for that first.

\(\left\{ \begin{array} { l } { - 5 x + y = - 1 \quad \Rightarrow \quad \color{Cerulean}{y}\color{Black}{ =} 5 x - 1 } \\ { 10 x - 2\color{Cerulean}{ y}\color{Black}{ =} 2 } \end{array} \right.\)

Next, substitute this expression in for \(y\) in the second equation.

\(\begin{aligned} 10 x - 2 y & = 2 \\ 10 x - 2 ( \color{OliveGreen}{5 x - 1}\color{Black}{ )} & = 2 \\ 10 x - 10 x + 2 & = 2 \\ 2 & = 2\quad \color{Cerulean}{True} \end{aligned}\)

This process led to a true statement; hence the equation is an identity and any real number is a solution. This indicates that the system is dependent. The simultaneous solutions take the form \((x, mx + b)\), or in this case, \((x, 5x − 1)\), where \(x\) is any real number.

\(( x , 5 x - 1 )\)

To have a better understanding of the previous example, rewrite both equations in slope-intercept form and graph them on the same set of axes.

\(\left\{ \begin{array} { l } { - 5 x + y = - 1 } \\ { 10 x - 2 y = 2 } \end{array} \right. \Rightarrow \left\{ \begin{array} { l } { y = 5 x - 1 } \\ { y = 5 x - 1 } \end{array} \right.\)

4995c4539f20276813b2ee0d2514d06f.png

We can see that both equations represent the same line, and thus the system is dependent. Now explore what happens when solving an inconsistent system using the substitution method.

Example \(\PageIndex{4}\):

Solve by substitution: \(\left\{ \begin{array} { l } { - 7 x + 3 y = 3 } \\ { 14 x - 6 y = - 16 } \end{array} \right.\).

Solve for \(y\) in the first equation.

\(\begin{aligned} - 7 x + 3 y & = 3 \\ - 7 x + 3 y & = 3 \\ 3 y & = 7 x + 3 \\ y & = \frac { 7 x + 3 } { 3 } \\ y & = \frac { 7 } { 3 } x + 1 \end{aligned}\)

\(\left\{ \begin{array} { l l } { - 7 x + 3 y = 3 } & { \Rightarrow \quad \color{Cerulean}{y}\color{Black}{ =} \frac { 7 } { 3 } x + 1 } \\ { 14 x - 6\color{Cerulean}{ y}\color{Black}{ =} - 16 } \end{array} \right. \)

Substitute into the second equation and solve.

\(\begin{aligned} 14 x - 6 y & = - 16 \\ 14x-6 \left( \color{OliveGreen}{\frac { 7 } { 3 } x + 1} \right) & = - 16 \\ 14x - \overset{\color{Cerulean}{2}}{\color{Black}{\cancel{6}}} \cdot \frac { 7 } { \underset{\color{Cerulean}{1}}{\cancel{3} }} x - 6 & = - 16 \\14 x - 14 x - 6 & = - 16 \\ - 6 &= - 16\quad\color{red}{False} \end{aligned}\)

Solving leads to a false statement. This indicates that the equation is a contradiction. There is no solution for \(x\) and hence no solution to the system.

\(\varnothing\)

A false statement indicates that the system is inconsistent, or in geometric terms, that the lines are parallel and do not intersect. To illustrate this, determine the slope-intercept form of each line and graph them on the same set of axes.

\(\left\{ \begin{array} { l l } { - 7 x + 3 y = 3 } \\ { 14 x - 6 y = - 16 } \end{array} \right. \Rightarrow \left\{ \begin{array} { l } { y = \frac { 7 } { 3 } x + 1 } \\ { y = \frac { 7 } { 3 } x + \frac { 8 } { 3 } } \end{array} \right.\)

29f0a74d32c346db54accdfa033095c0.png

In slope-intercept form, it is easy to see that the two lines have the same slope but different \(y\)-intercepts.

Exercise \(\PageIndex{2}\)

Solve by substitution: \(\left\{ \begin{array} { r } { 2 x - 5 y = 3 } \\ { 4 x - 10 y = 6 } \end{array} \right.\).

\(\left( x , \frac { 2 } { 5 } x - \frac { 3 } { 5 } \right)\)

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The Elimination Method

In this section, the goal is to review another completely algebraic method for solving a system of linear equations called the elimination method 13 or addition method 14 . This method depends on the addition property of equations 15 : given algebraic expressions A, B, C, and D we have

\(\text{If} \:A = B \text { and } C = D , \text { then } A + C = B + D\)

We can add the equations together to eliminate the variable \(y\).

\(\begin{aligned} x \color{red}{+ y}\color{Black}{ =} 5 \\ \pm \frac { x \color{red}{- y} = 1 } { 2 x \:\:\:= \:\:\: 6 }\end{aligned}\)

This leaves us with a linear equation with one variable that can be easily solved:

\(\begin{aligned} 2 x & = 6 \\ x & = 3 \end{aligned}\)

At this point, we have the \(x\)-coordinate of the simultaneous solution, so all that is left to do is back substitute to find the corresponding \(y\)-value.

\(\begin{array} { r } { x + y = 5 } \\ { \color{OliveGreen}{3} \color{Black}{+} y = 5 } \\ { y = 2 } \end{array}\)

The solution to the system is \((3, 2)\). Of course, the variable is not always so easily eliminated. Typically, we have to find an equivalent system by applying the multiplication property of equality to one or both of the equations as a means to line up one of the variables to eliminate. The goal is to arrange that either the \(x\) terms or the \(y\) terms are opposites, so that when the equations are added, the terms eliminate.

Example \(\PageIndex{5}\):

Solve by elimination: \(\left\{ \begin{array} { l } { 5 x - 3 y = - 1 } \\ { 3 x + 2 y = 7 } \end{array} \right.\).

We choose to eliminate the terms with variable \(y\) because the coefficients have different signs. To do this, we first determine the least common multiple of the coefficients; in this case, the \(LCM(3, 2)\) is \(6\). Therefore, multiply both sides of both equations by the appropriate values to obtain coefficients of \(−6\) and \(6\). This results in the following equivalent system:

\(\left\{ \begin{array} { l l } { 5 x - 3 y = - 1 } & { \stackrel { x 2 } { \Rightarrow } } \\ { 3 x + 2 y = 7 } & { \stackrel { x3 } { \Rightarrow } } \end{array} \right. \left\{ \begin{array} { c } { 10 x - 6 y = - 2 } \\ { 9 x + 6 y = 21 } \end{array} \right.\)

The terms involving \(y\) are now lined up to eliminate. Add the equations together and solve for \(x\).

\(\begin{aligned} 10 x \color{red}{- 6 y} & \color{Black}{=} - 2 \\ + \quad 9 x \color{red}{+ 6 y} & \color{Black}{=} 21 \\ \hline 19 x & = 19 \\ x & = 1 \end{aligned}\)

Back substitute.

\(\begin{aligned} 3 x + 2 y & = 7 \\ 3 ( \color{OliveGreen}{1}\color{Black}{ )} + 2 y & = 7 \\ 3 + 2 y & = 7 \\ 2 y & = 4 \\ y & = 2 \end{aligned}\)

Therefore the simultaneous solution is \((1, 2)\). The check follows.

Sometimes linear systems are not given in standard form \(ax + by = c\). When this is the case, it is best to rearrange the equations before beginning the steps to solve by elimination. Also, we can eliminate either variable. The goal is to obtain a solution for one of the variables and then back substitute to find a solution for the other.

Example \(\PageIndex{6}\):

Solve by elimination: \(\left\{ \begin{aligned} 12 x + 5 y & = 11 \\ 3 x & = 4 y + 1 \end{aligned} \right.\).

First, rewrite the second equation in standard form.

\(\begin{aligned} 3 x & = 4 y + 1 \\ 3 x - 4 y & = 1 \end{aligned}\)

This results in an equivalent system in standard form, where like terms are aligned in columns.

\(\left\{ \begin{array} { c c } { 12 x + 5 y = 11 } \\ { 3 x = 4 y + 1 } \end{array} \right. \Rightarrow \left\{ \begin{array} { c } { 12 x + 5 y = 11 } \\ { 3 x - 4 y = 1 } \end{array} \right.\)

We can eliminate the term with variable \(x\) if we multiply the second equation by \(−4\).

c1c845246da68b14739fc1ec5922f938.png

Next, we add the equations together,

\(\begin{aligned} \color{red}{12 x} \color{Black}{+5 y} & \color{Black}{=} 11 \\ + \quad \color{red}{-12x} \color{Black}{+ 16 y} & \color{Black}{=} -4 \\ \hline 21y & = 7 \\ y & = \frac{7}{21} = \frac{1}{3} \end{aligned}\)

\(\begin{array} { l } { 3 x = 4 y + 1 } \\ { 3 x = 4 \left( \color{OliveGreen}{\frac { 1 } { 3 }} \right) + 1 } \\ { 3 x = \frac { 4 } { 3 } + 1 } \\ { 3 x = \frac { 7 } { 3 } } \\ { x = \frac { 7 } { 3 } \cdot \frac { 1 } { 3 } } \\ { x = \frac { 7 } { 9 } } \end{array}\)

\(\left( \frac { 7 } { 9 } , \frac { 1 } { 3 } \right)\)

Exercise \(\PageIndex{3}\)

Solve by elimination: \(\left\{ \begin{array} { l } { 2 x + 5 y = 5 } \\ { 3 x + 2 y = - 9 } \end{array} \right.\).

\((-5, 3)\)

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At this point, we explore what happens when solving dependent and inconsistent systems using the elimination method.

Example \(\PageIndex{7}\):

Solve by elimination: \(\left\{ \begin{array} { c } { 3 x - y = 7 } \\ { 6 x - 2 y = 14 } \end{array} \right.\).

To eliminate the variable \(x\), we could multiply the first equation by \(−2\).

0e355e8f73169140ce519a886e768fbe.png

Now adding the equations we have

\(\begin{aligned} -6 x + \color{red}{2 y} & \color{Black}{=} -14 \\ \pm \quad \color{black}{6x}- \color{red}{2 y} & \color{Black}{=} 14 \\ \hline 0 & = 0\quad\color{Cerulean}{True} \end{aligned}\)

A true statement indicates that this is a dependent system. The lines coincide, and we need \(y\) in terms of \(x\) to present the solution set in the form \((x, mx + b)\). Choose one of the original equations and solve for \(y\). Since the equations are equivalent, it does not matter which one we choose.

\(\begin{aligned} 3 x - y & = 7 \\ - y & = - 3 x + 7 \\ \color{Cerulean}{- 1}\color{Black}{ (} - y ) & = \color{Cerulean}{- 1}\color{Black}{ (} - 3 x + 7 ) \\ y & = 3 x - 7 \end{aligned}\)

\(( x , 3 x - 7 )\)

Exercise \(\PageIndex{4}\)

Solve by elimination: \(\left\{ \begin{array} { l } { 3 x + 15 y = - 15 } \\ { 2 x + 10 y = 30 } \end{array} \right.\).

No solution, \(\varnothing\)

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Given a linear system where the equations have fractional coefficients, it is usually best to clear the fractions before beginning the elimination method.

Example \(\PageIndex{8}\):

Solve: \(\left\{ \begin{array} { l } { - \frac { 1 } { 10 } x + \frac { 1 } { 2 } y = \frac { 4 } { 5 } } \\ { \frac { 1 } { 7 } x + \frac { 1 } { 3 } y = - \frac { 2 } { 21 } } \end{array} \right.\).

Recall that we can clear fractions by multiplying both sides of an equation by the least common multiple of the denominators (LCD). Take care to distribute and then simplify.

This results in an equivalent system where the equations have integer coefficients,

\(\left\{ \begin{array} { l l } { -\frac{1}{10} x + \frac{1}{2}y = \frac{4}{5} } & { \stackrel { x 10 } { \Rightarrow } } \\ { \frac{1}{7} x + \frac{1}{3} y = -\frac{2}{21} } & { \stackrel { x21 } { \Rightarrow } } \end{array} \right. \left\{ \begin{array} { c } { - x +5y = 8 } \\ { 3 x + 7 y = -2 } \end{array} \right.\)

Solve using the elimination method.

f199eea33ed4bb7832221e93c546f85f.png

Figure \(\PageIndex{7}\)

\(\begin{aligned} \color{red}{-3 x} + \color{black}{15 y} & \color{Black}{=} 24 \\ \pm \quad \color{red}{3x}+ \color{black}{7 y} & \color{Black}{=} -2 \\ \hline 22y & = 22 \\ y&=1 \end{aligned}\)

\(\begin{aligned} 3 x + 7 y & = - 2 \\ 3 x + 7 ( \color{OliveGreen}{1}\color{Black}{ )} & = - 2 \\ 3 x + 7 & = - 2 \\ 3 x & = - 9 \\ x & = - 3 \end{aligned}\)

We can use a similar technique to clear decimals before solving.

Exercise \(\PageIndex{5}\)

Solve using elimination: \(\left\{ \begin{array} { l } { \frac { 1 } { 3 } x - \frac { 2 } { 3 } y = 3 } \\ { \frac { 1 } { 3 } x - \frac { 1 } { 2 } y = \frac { 8 } { 3 } } \end{array} \right.\).

\((5, -2)\)

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Summary of the Methods for Solving Linear Systems

We have reviewed three methods for solving linear systems of two equations with two variables. Each method is valid and can produce the same correct result. In this section, we summarize the strengths and weaknesses of each method.

The graphing method is useful for understanding what a system of equations is and what the solutions must look like. When the equations of a system are graphed on the same set of axes, we can see that the solution is the point where the graphs intersect. The graphing is made easy when the equations are in slope-intercept form. For example,

  • \(\left\{ \begin{array} { l } { y = 5 x + 15 } \\ { y = - 5 x + 5 } \end{array} \right.\)

343553149db59973f7cf3eba035c0c1a.png

The simultaneous solution \((−1, 10)\) corresponds to the point of intersection. One drawback of this method is that it is very inaccurate. When the coordinates of the solution are not integers, the method is practically unusable. If we have a choice, we typically avoid this method in favor of the more accurate algebraic techniques.

The substitution method, on the other hand, is a completely algebraic method. It requires you to solve for one of the variables and substitute the result into the other equation. The resulting equation has one variable for which you can solve. This method is particularly useful when there is a variable within the system with coefficient of \(1\). For example,

\(\left\{ \begin{array} { l } { 10 x + y = 20 } \\ { 7 x + 5 y = 14 } \end{array} \right. \color{Cerulean}{Choose\: the\: substitution\: method.} \quad\)

In this case, it is easy to solve for \(y\) in the first equation and then substitute the result into the other equation. One drawback of this method is that it often leads to equivalent equations with fractional coefficients, which are tedious to work with. If there is not a coefficient of \(1\), then it usually is best to choose the elimination method.

The elimination method is a completely algebraic method which makes use of the addition property of equations. We multiply one or both of the equations to obtain equivalent equations where one of the variables is eliminated if we add them together. For example,

\(\left\{ \begin{array} { l } { 2 x - 3 y = 9 } \\ { 5 x - 8 y = - 16 } \end{array} \color{Cerulean}\:\:{Choose\: the\:elimination\: method.}\right. \quad\)

To eliminate the terms involving \(x\), we would multiply both sides of the first equation by \(5\) and both sides of the second equation by \(−2\). This results in an equivalent system where the variable \(x\) is eliminated when we add the equations together. Of course, there are other combinations of numbers that achieve the same result. We could even choose to eliminate the variable \(y\). No matter which variable is eliminated first, the solution will be the same. Note that the substitution method, in this case, would require tedious calculations with fractional coefficients. One weakness of the elimination method, as we will see later in our study of algebra, is that it does not always work for nonlinear systems.

Key Takeaways

  • The substitution method requires that we solve for one of the variables and then substitute the result into the other equation. After performing the substitution step, the resulting equation has one variable and can be solved using the techniques learned up to this point.
  • The elimination method is another completely algebraic method for solving a system of equations. Multiply one or both of the equations in a system by certain numbers to obtain an equivalent system where at least one variable in both equations have opposite coefficients. Adding these equivalent equations together eliminates that variable, and the resulting equation has one variable for which you can solve.
  • It is a good practice to first rewrite the equations in standard form before beginning the elimination method.
  • Solutions to systems of two linear equations with two variables, if they exist, are ordered pairs \((x, y)\).
  • If the process of solving a system of equations leads to a false statement, then the system is inconsistent and there is no solution, \(Ø\).
  • If the process of solving a system of equations leads to an identity, then the system is dependent and there are infinitely many solutions that can be expressed using the form \((x, mx + b)\).

Exercise \(\PageIndex{6}\)

Solve by substitution.

  • \(\left\{ \begin{array} { l } { y = - 5 x + 1 } \\ { 4 x - 3 y = - 41 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x = 2 y - 3 } \\ { x + 3 y = - 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = x } \\ { 2 x + 3 y = 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = \frac { 1 } { 2 } x + \frac { 1 } { 3 } } \\ { x - 6 y = 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 4 x + 1 } \\ { - 4 x + y = 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - 3 x + 5 } \\ { 3 x + y = 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 2 x + 3 } \\ { 2 x - y = - 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = \frac { 2 } { 3 } x - 1 } \\ { 6 x - 9 y = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - 2 } \\ { - 2 x - y = - 6 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - \frac { 1 } { 5 } x + 3 } \\ { 7 x - 5 y = 9 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + y = 1 } \\ { 3 x - 5 y = 19 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - y = 3 } \\ { - 2 x + 3 y = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x + y = 2 } \\ { 3 x - 2 y = 17 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - 3 y = - 11 } \\ { 3 x + 5 y = - 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 2 y = - 3 } \\ { 3 x - 4 y = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x - y = 12 } \\ { 9 x - y = 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 2 y = - 6 } \\ { - 4 x - 8 y = 24 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 3 y = - 6 } \\ { - 2 x - 6 y = - 12 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 3 x + y = - 4 } \\ { 6 x - 2 y = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - 5 y = - 10 } \\ { 2 x - 10 y = - 20 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - y = 9 } \\ { 4 x + 3 y = - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - y = 5 } \\ { 4 x + 2 y = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - 5 y = 1 } \\ { 4 x + 10 y = 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 7 y = - 3 } \\ { 6 x + 14 y = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 10 x - y = 3 } \\ { - 5 x + \frac { 1 } { 2 } y = 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - \frac { 1 } { 3 } x + \frac { 1 } { 6 } y = \frac { 2 } { 3 } } \\ { \frac { 1 } { 2 } x - \frac { 1 } { 3 } y = - \frac { 3 } { 2 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 3 } x + \frac { 2 } { 3 } y = 1 } \\ { \frac { 1 } { 4 } x - \frac { 1 } { 3 } y = - \frac { 1 } { 12 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 7 } x - y = \frac { 1 } { 2 } } \\ { \frac { 1 } { 4 } x + \frac { 1 } { 2 } y = 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - \frac { 3 } { 5 } x + \frac { 2 } { 5 } y = \frac { 1 } { 2 } } \\ { \frac { 1 } { 3 } x - \frac { 1 } { 12 } y = - \frac { 1 } { 3 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x = \frac { 2 } { 3 } y } \\ { x - \frac { 2 } { 3 } y = 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - \frac { 1 } { 2 } x + \frac { 1 } { 2 } y = \frac { 5 } { 8 } } \\ { \frac { 1 } { 4 } x + \frac { 1 } { 2 } y = \frac { 1 } { 4 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - y = 0 } \\ { - x + 2 y = 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 3 x } \\ { 2 x - 3 y = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 3 x + 4 y = 20 } \\ { 2 x + 8 y = 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x - 3 y = - 1 } \\ { 3 x + 2 y = 7 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 3 x + 7 y = 2 } \\ { 2 x + 7 y = 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x = 5 } \\ { x = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 4 } \\ { 5 y = 20 } \end{array} \right.\)

1. \((-2,11)\)

3. \((2,2)\)

5. \(\varnothing\)

7. \(( x , 2 x + 3 )\)

9. \(( 4 , - 2 )\)

11. \(( 3 , - 2 )\)

13. \(( 3 , - 4 )\)

15. \(\left( - \frac { 8 } { 5 } , - \frac { 7 } { 10 } \right)\)

17. \(\left( x , - \frac { 1 } { 2 } x - 3 \right)\)

19. \(\varnothing\)

21. \(( 2 , - 3 )\)

23. \(\left( \frac { 1 } { 2 } , 0 \right)\)

25. \(\varnothing\)

27. \(( 1,1 )\)

29. \(\left( - \frac { 11 } { 10 } , - \frac { 2 } { 5 } \right)\)

31. \(\left( - \frac { 1 } { 2 } , \frac { 3 } { 4 } \right)\)

33. \((0,0)\)

35. \((1, 2)\)

37. \(\varnothing\)

Exercise \(\PageIndex{7}\)

Solve by elimination.

  • \(\left\{ \begin{array} { l } { 6 x + y = 3 } \\ { 3 x - y = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + y = 3 } \\ { 2 x - y = 9 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x - y = - 6 } \\ { 5 x + y = - 18 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 3 y = 5 } \\ { - x - 2 y = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - x + 4 y = 4 } \\ { x - y = - 7 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - x + y = 2 } \\ { x - y = - 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - y = - 2 } \\ { 6 x + 4 y = 2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x + 2 y = - 3 } \\ { 10 x - y = 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 2 x + 14 y = 28 } \\ { x - 7 y = 21 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 2 x + y = 4 } \\ { 12 x - 6 y = - 24 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 8 y = 3 } \\ { 3 x + 12 y = 6 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - 3 y = 15 } \\ { 4 x + 10 y = 14 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 4 x + 3 y = - 10 } \\ { 3 x - 9 y = 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 4 x - 5 y = - 3 } \\ { 8 x + 3 y = - 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 2 x + 7 y = 56 } \\ { 4 x - 2 y = - 112 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 9 x - 15 y = - 15 } \\ { 3 x + 5 y = - 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 6 x - 7 y = 4 } \\ { 2 x + 6 y = - 7 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 4 x + 2 y = 4 } \\ { - 5 x - 3 y = - 7 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 7 x + 3 y = 9 } \\ { 2 x + 5 y = - 14 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 9 x - 3 y = 3 } \\ { 7 x + 2 y = - 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x - 3 y = - 7 } \\ { - 7 x + 6 y = 11 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x + 9 y = 8 } \\ { 3 x + 7 y = - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x + 2 y = 5 } \\ { 3 x + 3 y = - 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 3 x + 6 y = - 12 } \\ { 2 x - 4 y = 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 25 x + 15 y = - 1 } \\ { 15 x + 10 y = - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - 3 y = 2 } \\ { 18 x - 12 y = 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - 2 x - 3 } \\ { - 3 x - 2 y = 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 28 x + 6 y = 9 } \\ { 6 y = 4 x - 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 x - 3 y = 9 } \\ { 5 x - 8 y = - 16 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x - \frac { 1 } { 3 } y = \frac { 1 } { 6 } } \\ { \frac { 5 } { 2 } x + y = \frac { 7 } { 2 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 4 } x - \frac { 1 } { 9 } y = 1 } \\ { x + y = \frac { 3 } { 4 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x - \frac { 1 } { 4 } y = \frac { 1 } { 3 } } \\ { \frac { 1 } { 4 } x + \frac { 1 } { 2 } y = - \frac { 19 } { 6 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - \frac { 14 } { 3 } x + 2 y = 4 } \\ { - \frac { 1 } { 3 } x + \frac { 1 } { 7 } y = \frac { 4 } { 21 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 0.025 x + 0.1 y = 0.5 } \\ { 0.11 x + 0.04 y = - 0.2 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 1.3 x + 0.1 y = 0.35 } \\ { 0.5 x + y = - 2.75 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + y = 5 } \\ { 0.02 x + 0.03 y = 0.125 } \end{array} \right.\)

1. \(\left( \frac { 1 } { 3 } , 1 \right)\)

3. \((-4,2)\)

5. \((-8,-1)\)

7. \(\left( - \frac { 1 } { 3 } , 1 \right)\)

9. \(\varnothing\)

11. \(\left( 1 , \frac { 1 } { 4 } \right)\)

13. \((-1,-2)\)

15. \((-28,0)\)

17. \(\left( - \frac { 1 } { 2 } , - 1 \right)\)

19. \((1,2)\)

21. \((-1,-4)\)

23. \((-5,2)\)

25. \(\left( x , \frac { 1 } { 2 } x - 2 \right)\)

27. \(\left( - \frac { 3 } { 10 } , - \frac { 13 } { 15 } \right)\)

29. \(\left( \frac { 3 } { 4 } , - 2 \right)\)

31. \((120,77)\)

33. \(\left( 3 , - \frac { 9 } { 4 } \right)\)

35. \(\varnothing\)

37. \(( 0.5 , - 3 )\)

Exercise \(\PageIndex{8}\)

Solve using any method.

  • \(\left\{ \begin{array} { l } { 6 x = 12 y + 7 } \\ { 6 x + 24 y + 5 = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 2 x - 3 } \\ { 3 x + y = 12 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { x + 3 y = - 5 } \\ { y = \frac { 1 } { 3 } x + 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 1 } \\ { x = - 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = \frac { 1 } { 2 } } \\ { x + 9 = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = x } \\ { - x + y = 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = 5 x } \\ { y = - 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - \frac { 3 } { 2 } x + 1 } \\ { - 2 y + 2 = 3 x } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 7 y = - 2 x - 1 } \\ { 7 x = 2 y + 23 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 x + 9 y - 14 = 0 } \\ { 3 x + 2 y - 5 = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - \frac { 5 } { 16 } x + 10 } \\ { y = \frac { 5 } { 16 } x - 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - \frac { 6 } { 5 } x + 12 } \\ { x = 6 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 ( x - 3 ) + y = 0 } \\ { 3 ( 2 x + y - 1 ) = 15 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 - 2 ( x - y ) = - 3 } \\ { 4 x - 3 ( y + 1 ) = 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 2 ( x + 1 ) = 3 ( 2 y - 1 ) - 21 } \\ { 3 ( x + 2 ) = 1 - ( 3 y - 2 ) } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { x } { 2 } - \frac { y } { 3 } = - 7 } \\ { \frac { x } { 3 } - \frac { y } { 2 } = - 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - \frac { 1 } { 7 } x + y = - \frac { 2 } { 3 } } \\ { - \frac { 1 } { 14 } x + \frac { 1 } { 2 } y = \frac { 1 } { 3 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { x } { 4 } - \frac { y } { 2 } = \frac { 3 } { 4 } } \\ { \frac { x } { 3 } + \frac { y } { 6 } = \frac { 1 } { 6 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { y = - \frac { 5 } { 3 } x + \frac { 1 } { 2 } } \\ { \frac { 1 } { 3 } x + \frac { 1 } { 5 } y = \frac { 1 } { 10 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { \frac { 1 } { 15 } x - \frac { 1 } { 12 } y = \frac { 1 } { 3 } } \\ { - \frac { 3 } { 10 } x + \frac { 3 } { 8 } y = - \frac { 3 } { 2 } } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 0.2 x - 0.05 y = 0.43 } \\ { 0.3 x + 0.1 y = - 0.3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 0.1 x + 0.3 y = 0.3 } \\ { 0.05 x - 0.5 y = - 0.63 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 0.15 x - 0.25 y = - 0.3 } \\ { - 0.75 x + 1.25 y = - 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 0.15 x + 1.25 y = 0.4 } \\ { - 0.03 x + 0.25 y = 0.08 } \end{array} \right.\)

1. \(\left( \frac { 1 } { 2 } , - \frac { 1 } { 3 } \right)\)

3. \(\left( - 10 , \frac { 5 } { 3 } \right)\)

5. \(\left( - 9 , \frac { 1 } { 2 } \right)\)

7. \(( - 2 , - 10 )\)

9. \(( 3 , - 1 )\)

11. \(( 32,0 )\)

13. \(( x , - 2 x + 6 )\)

15. \(( - 4,3 )\)

17. \(\varnothing\)

19. \(\left( x - \frac { 5 } { 3 } x + \frac { 1 } { 2 } \right)\)

21. \(( 0.8 , - 5.4 )\)

23. \(\varnothing\)

Exercise \(\PageIndex{9}\)

  • Explain to a beginning algebra student how to choose a method for solving a system of two linear equations. Also, explain what solutions look like and why.
  • Make up your own linear system with two variables and solve it using all three methods. Explain which method was preferable in your exercise.

1. Answer may vary

11 A means of solving a linear system by solving for one of the variables and substituting the result into the other equation.

12 Once a value is found for a variable, substitute it back into one of the original equations, or its equivalent, to determine the corresponding value of the other variable.

13 A means of solving a system by adding equivalent equations in such a way as to eliminate a variable.

14 Often used when referring to the elimination method for solving systems.

15 If \(A, B, C\), and \(D\) are algebraic expressions, where \(A = B\) and \(C = D\), then \(A + C = B + D\).

how to solve algebraic equations with 2 variables

How Do You Solve Two Equations with Two Variables?

Trying to solve two equations each with the same two unknown variables? Take one of the equations and solve it for one of the variables. Then plug that into the other equation and solve for the variable. Plug that value into either equation to get the value for the other variable. This tutorial will take you through this process of substitution step-by-step!

  • substitution
  • solve by substitution
  • apply distributive property
  • linear equation
  • 2 equations 2 unknowns

Background Tutorials

Evaluating expressions.

What is a Variable?

What is a Variable?

You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!

Using the Distributive Property

How Do You Use the Distributive Property to Simplify an Expression?

How Do You Use the Distributive Property to Simplify an Expression?

In this tutorial you'll see how to apply the distributive property. Remember that this is important when you are trying to simplify an expression and get rid of parentheses!

Simplifying Expressions

How Do You Simplify an Expression?

How Do You Simplify an Expression?

Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial.

Solving Two-Step Equations

How Do You Solve a Two-Step Equation?

How Do You Solve a Two-Step Equation?

Solving an equation for a variable? Perform the order of operations in reverse! Check it out in this tutorial.

Solving Systems by Graphing

How Do You Check Your Answer When You Have Two Equations?

How Do You Check Your Answer When You Have Two Equations?

Imagine you have two equations with two variables that you're trying to solve for, and someone hands you the answer. How do you know that the answer is right? After watching this tutorial you'll see exactly what it takes to check that the answer you have is correct for BOTH equations!

Further Exploration

Solving systems using substitution.

How Do You Solve a Word Problem Using Two Equations?

How Do You Solve a Word Problem Using Two Equations?

Sometimes word problems describe a system of equations, two equations each with two unknowns. Solving word problems like this one aren't so bad if you know what to do. Check it out with this tutorial!

Solving Systems Using Elimination

What Are the Ways You Can Solve a System of Linear Equations?

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  • -x+3\gt 2x+1
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  • How do you solve algebraic expressions?
  • To solve an algebraic expression, simplify the expression by combining like terms, isolate the variable on one side of the equation by using inverse operations. Then, solve the equation by finding the value of the variable that makes the equation true.
  • What are the basics of algebra?
  • The basics of algebra are the commutative, associative, and distributive laws.
  • What are the 3 rules of algebra?
  • The basic rules of algebra are the commutative, associative, and distributive laws.
  • What is the golden rule of algebra?
  • The golden rule of algebra states Do unto one side of the equation what you do to others. Meaning, whatever operation is being used on one side of equation, the same will be used on the other side too.
  • What are the 5 basic laws of algebra?
  • The basic laws of algebra are the Commutative Law For Addition, Commutative Law For Multiplication, Associative Law For Addition, Associative Law For Multiplication, and the Distributive Law.

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  3. 4 Ways to Solve Systems of Algebraic Equations Containing Two Variables

    how to solve algebraic equations with 2 variables

  4. 4 Ways to Solve Systems of Algebraic Equations Containing Two Variables

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    how to solve algebraic equations with 2 variables

VIDEO

  1. How can solve algebraic equation ??

  2. Solving Equation for n: Mastering Mathematical Solution #mathematics #equations #solutions #mathhelp

  3. Algebraic Equations: Solving for X Step-by-Step. #shorts

  4. How to solve Algebraic Equations

  5. Can you solve this in 30 Seconds? Algebraic Equations

  6. Algebra 1: Linear Equations in One Variable

COMMENTS

  1. How to Solve Systems of Algebraic Equations Containing Two Variables

    Divide both sides of the equation to "solve for x." Once you have the x term (or whichever variable you are using) on one side of the equation, divide both sides of the equation to get the variable alone. [3] For example: 4x = 8 - 2y. (4x)/4 = (8/4) - (2y/4) x = 2 - ½y. 3. Plug this back into the other equation.

  2. Solutions to 2-variable equations (video)

    So divide both sides by 9. 9b = 7 7 divided by 9 (7/9) is 0.777777... So for simplicity I would leave it as a fraction, unless stated otherwise. Therefor, you get b = 7/9. Tada! For a more literal answer, the b in 9b - 5 = 2 is a letter, otherwise known as a variable.

  3. Completing solutions to 2-variable equations

    The reason both (-5,-8) and (5,8) are solutions to this equation is because they show up on the equation's line. Sal later says the (simplified) equation for the question is y=8x/5. To make this answer more clear, another equation, such as y=x-3, when x=-5, y is equal to -8, but when x equals 5, y equals 2. Hope this helps.

  4. Solving equations & inequalities

    Unit test. Level up on all the skills in this unit and collect up to 1100 Mastery points! There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions.

  5. Linear Equations in Two Variables

    To solve a system of two linear equations in two variables using the substitution method, we have to use the steps given below: Step 1: Solve one of the equations for one variable. Step 2: Substitute this in the other equation to get an equation in terms of a single variable. Step 3: Solve it for the variable.

  6. 4.1 Solve Systems of Linear Equations with Two Variables

    We will solve larger systems of equations later in this chapter. An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. { 2 x + y = 7 x − 2 y = 6. A linear equation in two variables, such as 2 x + y = 7, has an infinite number of solutions.

  7. 4.1: Solve Systems of Linear Equations with Two Variables

    The steps to use to solve a system of linear equations by graphing are shown here. SOLVE A SYSTEM OF LINEAR EQUATIONS BY GRAPHING. Graph the first equation. Graph the second equation on the same rectangular coordinate system. Determine whether the lines intersect, are parallel, or are the same line.

  8. 7.1 Systems of Linear Equations: Two Variables

    Shortly we will investigate methods of finding such a solution if it exists. 2(4) + (7) = 15 True 3(4) − (7) = 5 True. In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution.

  9. Introduction to solving an equation with variables on both sides

    Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve...

  10. 4.2: Solve Systems of Linear Equations with Two Variables

    The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we'll do with the elimination method, too, but we'll have a different way to get there.

  11. 5.7: Solve Systems of Linear Equations with Two Variables

    A solution of a system of two linear equations is represented by an ordered pair (x, y). To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. Example 5.7.1.

  12. Intro to equations with variables on both sides

    Learn how to solve the equation 2x + 3 = 5x - 2 with the variable on both sides. We start by visualizing the equation, then isolate the variable by performing the same operations on both sides. Finally, we solve the equation to find the value of the variable. Created by Sal Khan.

  13. Solving Algebra Equations With Two Variables

    Tutorial on solving algebra equations with two variables. Elimination method, x and y, Cartesian coordinate systemLike MyBookSucks on Facebookhttp://www.fac...

  14. Solving Equations

    In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do. Here are some things we can do: Add or Subtract the same value from both sides. Clear out any fractions by Multiplying every term by the bottom parts. Divide every term by the same nonzero value.

  15. Algebraic Equations

    A linear algebraic equation is one in which the degree of the polynomial is 1. The general form of a linear equation is given as a 1 x 1 +a 2 x 2 +...+a n x n = 0 where at least one coefficient is a non-zero number. These linear equations are used to represent and solve linear programming problems. Example: 3x + 5 = 5 is a linear equation in ...

  16. Solving systems of equations in two variables

    We now wish to add the two equations but it will not result in either x or y being eliminated. Therefore we must multiply the second equation by 2 on both sides and get: $$2x-2y=8$$ $$2x+2y=2$$ Now we attempt to add our system of equations. We commence with the x-terms on the left, and the y-terms thereafter and finally with the numbers on the ...

  17. 4.2: Solving Linear Systems with Two Variables

    Explain to a beginning algebra student how to choose a method for solving a system of two linear equations. Also, explain what solutions look like and why. Make up your own linear system with two variables and solve it using all three methods. Explain which method was preferable in your exercise. Answer. 1. Answer may vary

  18. How Do You Solve Two Equations with Two Variables?

    Trying to solve two equations each with the same two unknown variables? Take one of the equations and solve it for one of the variables. Then plug that into the other equation and solve for the variable. Plug that value into either equation to get the value for the other variable. This tutorial will take you through this process of substitution ...

  19. Two-variable equations

    Pre-algebra 15 units · 179 skills. Unit 1 Factors and multiples. Unit 2 Patterns. Unit 3 Ratios and rates. Unit 4 Percentages. Unit 5 Exponents intro and order of operations. Unit 6 Variables & expressions. Unit 7 Equations & inequalities introduction. Unit 8 Percent & rational number word problems.

  20. Algebra Calculator

    To solve an algebraic expression, simplify the expression by combining like terms, isolate the variable on one side of the equation by using inverse operations. Then, solve the equation by finding the value of the variable that makes the equation true. What are the basics of algebra?

  21. Equation with variables on both sides: fractions

    AboutTranscript. To solve the equation (3/4)x + 2 = (3/8)x - 4, we first eliminate fractions by multiplying both sides by the least common multiple of the denominators. Then, we add or subtract terms from both sides of the equation to group the x-terms on one side and the constants on the other. Finally, we solve and check as normal.