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\mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
\mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
\mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
\mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
\mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
How do you solve word problems?
To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
How do you identify word problems in math?
Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
Is there a calculator that can solve word problems?
Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
What is an age problem?
An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,...

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Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

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1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable $x .$ Twice the variable is $2 x .$ Adding $2 x$ to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is $3 x .$ We need to subtract $3 x$ from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by $x .8$ less than $x$ mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is $10-8=2$.

d) Subtract $5 p^{2}-7 p+2$ from $3 p^{2}+4 p$ and simplify.

Solution: We need to calculate $3 p^{2}+4 p$ minus $5 p^{2}-7 p+2:$

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by $x$ dimes and $y$ quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of $10 x$ cents. Each quarter is worth 25 cents, so that this gives a total of $25 y$ cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for $x$ :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is $13 .$ Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for $x$. First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives $x=\frac{18}{2}=9$.

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add $x$ :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is $x=\frac{9}{3}=3$

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting $x$ gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: $x=3$.

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as $x+9,$ while subtracting 7 from three times the number is written as $3 x-7$. We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for $x$ by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract $x:$

\[16 = 2x.\nonumber\]

After dividing by $2,$ we obtain the answer $x=8$

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by $5 \%$. How much does the loaf of bread cost now, when its price was $\$ 2.40$ last year?

Solution: We calculate the price increase as $5 \% \cdot \$ 2.40 .$ We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore $\$ 2.52$.

b) To complete a job, three workers get paid at a rate of $\$ 12$ per hour. If the total pay for the job was $\$ 180,$ then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by $x$. Then the total price is calculated as the price per hour $(\$ 12)$ times the number of workers times the number of hours $(3) .$ We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is $4 \times 60$ feet $=240$ feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by $x .$ If 4 blocks weigh $28,$ then a block weighs $x=\frac{28}{4}=7$

How many blocks weigh $70 ?$ Well, we only need to find $\frac{70}{7}=10 .$ So, the answer is $10 .$

Note You can solve this problem by setting up and solving the fractional equation $\frac{28}{4}=\frac{70}{x}$. Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by $x$. Then the length is $2 x+3$. The perimeter is 20 in on one hand and $2($length$)+2($width$)$ on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that $C=2 \pi r$ where $C$ is the circumference and $r$ is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by $2 \pi$ gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let $x$ equal the side of the triangle. Then the perimeter is, on the one hand, $60,$ and on other hand $3 x .$ So $3 x=60$ and dividing both sides of the equation by 3 gives $x=20$ meters.

h) If a gardener has $\$ 600$ to spend on a fence which costs $\$ 10$ per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is $P=2 L+2 W$. Let $x$ be the width of the rectangle. Then the length is $2 x .$ The perimeter is $P=2(2 x)+2 x=6 x$. The largest perimeter is $\$ 600 /(\$ 10 / f t)=60$ ft. So $60=6 x$ and dividing both sides by 6 gives $x=60 / 6=10$. So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let $b$ be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is $\frac{1}{2}(3.2+b) \cdot 4=$ $6.4+2 b .$ So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

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Sum of Five Consecutive Integers

Free Sum of Five Consecutive Integers Calculator - Finds five consecutive integers, if applicable, who have a sum equal to a number. Sum of 5 consecutive integers

Sum of Four Consecutive Integers

Free Sum of Four Consecutive Integers Calculator - Finds four consecutive integers, if applicable, who have a sum equal to a number. Sum of 4 consecutive integers

Sum of the First (n) Numbers

Free Sum of the First (n) Numbers Calculator - Determines the sum of the first (n) * Whole Numbers * Natural Numbers * Even Numbers * Odd Numbers * Square Numbers * Cube Numbers * Fourth Power Numbers

Sum of Three Consecutive Integers

Free Sum of Three Consecutive Integers Calculator - Finds three consecutive integers, if applicable, who have a sum equal to a number. Sum of 3 consecutive integers

Free Sun Shadow Calculator - This solves for various components and scenarios of the sun shadow problem

Unit Savings

Free Unit Savings Calculator - A discount and savings word problem using 2 people and full prices versus discount prices.

Work Word Problems

Free Work Word Problems Calculator - Given Person or Object A doing a job in (r) units of time and Person or Object B doing a job in (s) units of time, this calculates how long it would take if they combined to do the job.

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Solving Linear Equations Calculator

Instructions: Use this linear equation calculator to solve any linear equation you want, of one or more variables . Please type in the linear equation you want to solve.

This Linear Equation Calculator

This linear equation calculator will allow you solve linear equations you provide, showing all the steps. For example, you may be interested in solving something like '1/3 x +1/4 y = 1/6', which is a linear equation with two variables, x and y.

Once you specified a valid linear equation you want to solve, then you can click on "Calculate" and you will be provided with corresponding steps needed to arrive to the solution.

Solving linear equation is the easiest among the broader task of solving polynomial equations , which can be a lot harder, especially for polynomial with a higher degree.

What is a Linear Equation

A linear equation is a math equation in which you have that both sides of the equation are linear expressions. A linear expression is the sum or subtraction of constants or constant multiplied by a variable.

For example, '2x + 3y = 1' is a linear equation , but '2x = cos(x)' is not. It is important to distinguish between a linear expression and a linear equation.

Following the same example, '2x + 3y' is a linear expression, but it is not a linear equation, because there is no equality involved. In order to have a linear equation, YOU NEED to have an equality sign in it.

Linear Equations Formula

A linear equation formula will depend on the number of variables we use. For example, the general linear equation formula for one variable x is:

Some will argue that there is no need to have a constant on the left side, and they would write:

Now, the general linear equation formula for two variables x and y is:

In general, the general linear equation formula for $n$ variables is:

Notice that we put a "+" in general, but the constants $a_1$, ..., $a_n$ can also be negative.

How to Solve Linear Equations

Step 1: Make sure you are dealing with an actual linear equation. Then, identify how many variables are involved in the equation
Step 2: If you have only one variable, say x, you can solve for x, manipulating the terms of the equation, putting x on one side and then solving for x. Solving for x in this case is expected to lead to a numeric solution
Step 3: If you have more than one variable, then you choose one variable, say x, and then solve for x , in terms of the other variables. Here you don't get a numeric solution, but instead, you get x (or whatever variable you chose) in terms of other variables

Observe that we are dealing here with one linear equation. You can use this system of equations calculator if you are dealing with multiple linear equations.

Having an equation calculator with steps can prove to be extremely useful, as it is sometimes hard to find the correct strategy to use for certain equations. Of course linear equations are simple, but we can find that solving polynomial equations , or solving trigonometric equations , for example, can be tremendously laborious and challenging.

How do you find the linear equation?

Linear equations appear naturally in algebra problems and all sorts of algebra equations. Linear functions are extremely common both in Algebra and Calculus and will appear literally EVERYWHERE.

You can, for example, use the slope intercept form or the point slope form to calculate a linear function. Usually, you will work the linear equations in standard form , which the way we presented before:

Normally, we don't work with n generic variables, we work with two or three variables, which would look like:

respectively.

Advantages of working with Linear Equations

Step 1: Linear equations are simple! They are easy to calculate and easy to interpret
Step 2: There are no tricks needed to solve a linear equation: pass the terms with to one side, group them and simplify
Step 3: Linear equations are very common, and have a clear graphical interpretation

Naturally, if we could choose, we would always work with linear equation, but unfortunately reality is not that generous, as it the case that very frequently we will need to deal with equations more difficult than linear equations.

How do you know if a function is linear?

Fractions are one of the corner stones of algebra and of any general algebraic expression to calculate . Fractions are simple operands, but which can be compounded into more complicated terms using operations like sum, multiplication, etc., and then using functions we can construct even more advanced expressions.

The center of all algebraic calculator starts with the power of basic numbers of fractions.

Example: Solving linear equations one variable

Solve the following: $\frac{1}{3} x + \frac{5}{4} = \frac{5}{6}$

We need to solve the following given linear equation:

The linear equation has only one variable, which is $x$, so the objective is to solve for it.

Putting $x$ on the left hand side and the constant on the right hand side we get

Now, solving for $x$, by dividing both sides of the equation by $\frac{1}{3}$, the following is obtained

and simplifying we finally get the following

Therefore, the solving for $x$ for given linear equation leads to $x=-\frac{5}{4}$. This concludes the solution calculation.

Other useful equation calculators

Using an equation solver can completely come in handy, especially when dealing with difficult equations. The case of linear equations is truly reduced to a class of simple equations to solve, and you will find equations that will be a lot more challenging.

Coming next in terms of difficulty you will find the polynomial equations , for which you can use a methodology which ensures you have the best chance of finding as many solutions as possible, but you are not guaranteed to find all of them at times. This polynomial calculator will guarantee you get as many solutions as possible.

Then you have the even more complicated of non-polynomial non-linear equations, for which you need to come up with an astute approach usually, if you want to come closer to the solution. Trigonometric equations are notorious for being difficult and reliant on a precise substitution.

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Tiger Algebra Calculator

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SAT Mathematics : Solving Linear Equations in Word Problems

Study concepts, example questions & explanations for sat mathematics, all sat mathematics resources, example questions, example question #31 : solving word problems.

Jackie plans to buy one video game and a number of hardcover books. The video game costs $40 and each book costs $30. If she must spend at least $120 in order to get free shipping, what is the minimum number of books she must buy in order to get free shipping?

Since you know that Jackie will purchase exactly one video game for $40, you can set up your equation (or inequality) here as:

$solve linear equations word problems calculator$

Now you can subtract 40 from both sides:

$solve linear equations word problems calculator$

And when you divide both sides by 30 you'll see that you have a number between 2 and 3. Note that you do not have to do the full decimal calculation, because you cannot buy part of a book! So since she can't buy 2.67 books, the minimum number she can purchase is 3.

Example Question #32 : Solving Word Problems

On the first day of the week, a bakery had an inventory of 450 loaves of bread. It bakes 210 loaves of bread and sells 240 loaves of bread each day that it is open, and then closes for a baking day when it runs out of loaves. How many days can it be open before it must close for a baking day?

If the bakery bakes 210 loaves of bread and sells 240 loaves of bread, then that means that, total, it loses 30 loaves of bread per day. Since you know that it starts with 450 loaves of bread, you can use this information to write a linear equation relating the number of loaves of bread left with how many days it has been since the bakery has closed for a baking day. It should look like:

B = 450 - 30d

Where B represents the number of loaves left and d represents the number of days since the bakery’s last baking day.

In order for the bakery to need to close for a baking day, the number of loaves left must equal 0. If you substitute in B = 0, you get: 0 = 450−30d.

Now you can solve: add 30d to both sides to get:

And then divide both sides by 30 to get d = 15

Example Question #1 : Solving Linear Equations In Word Problems

Katharine currently has $10,000 saved for a down payment on purchasing her first house, which she can do when her savings has reached $50,000. Each month she earns $6,500 but incurs $4,000 in expenses. If her earnings and expenses remain constant, how many months will it take until she has reached her savings goal?

To turn this problem into an equation, you can start by putting Katharine's goal of $50,000 on one side of the equation, and then arranging all of the inputs to that goal (ways she earns money) and impediments (ways she loses money) on the other. If you use m to represent the number of months, an initial equation would look like:

$50,000 = $10,000 + $6,500m - $4,000m

Which, qualitatively, is that her goal of $50,000 is equal to the $10,000 she has plus the $6,500 she earns each month, minus the $4,000 she loses each month.

Now you can combine like terms to simplify the equation:

$40,000 = $2,500m

And then divide 40000 by 2500 (which simplifies to 400 divided by 25) to solve:

m = 16 months

Example Question #34 : Solving Word Problems

$solve linear equations word problems calculator$

Then you'll need to find a common denominator of 6, and rewrite what you have to reflect that common denominator:

$solve linear equations word problems calculator$

You can then combine like terms on the right-hand side of the equation:

$solve linear equations word problems calculator$

And then multiply both sides by 6 to finish the problem:

$solve linear equations word problems calculator$

Example Question #35 : Solving Word Problems

A wood beam 48 inches in length is cut into two smaller beams. If one of the new beams is 20 inches longer than the other, what is the length of the longer beam?

$solve linear equations word problems calculator$

Example Question #36 : Solving Word Problems

$solve linear equations word problems calculator$

While it is tempting to set up and solve an equation for this problem, the algebra for this problem may be difficult to translate quickly and accurately. Because you are dealing with relatively simple numbers that add up to a total, you should see that this should be an easy problem to backsolve.

$solve linear equations word problems calculator$

Alternatively, you could set up the algebra:

$solve linear equations word problems calculator$

Example Question #37 : Solving Word Problems

Eight times one-third of a number is one greater than five times one half of that number. What is that number?

$solve linear equations word problems calculator$

This problem can be solved algebraically but is likely solved more quickly by simply back solving. Your algebra would set up as:

$solve linear equations word problems calculator$

So start with 6. Eight times one-third of 6 is 8(2) = 16. Is that one greater than five times one half of 6? It is. Five times one half of 6 is 5(3) = 15. The relationship works, so 6 is correct.

Example Question #38 : Solving Word Problems

$solve linear equations word problems calculator$

Importantly, note that if 60 hybrids are sold, that means that 60 of the total of 480 cars are sold. That affects both the number of hybrids AND the number of total cars, which is now reduced to 420. So your new proportion of hybrids is the 60 hybrids out of the 420 total cars:

$solve linear equations word problems calculator$

Example Question #39 : Solving Word Problems

Five years ago, Juliet was three times as old as Will. If Juliet is currently twice as old as Will, how old is Juliet?

$solve linear equations word problems calculator$

Example Question #40 : Solving Word Problems

On the same day, Devin bought a car for $40,000 that would then decrease in value by $1200 per year, and Anita purchased a piece of real estate for $6,000 that would then increase in value by $500 per year. After how many years will Anita’s land be worth as much as Devin’s car?

$solve linear equations word problems calculator$

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Solving Linear Equation Word Problems 9 Terrific Examples!

// Last Updated: January 20, 2020 - Watch Video //

What’s the first thing that comes to mind when you hear the phrase Word problems?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching linear word problems

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

For some, it’s a chance to solve a real-world example, so there’s a level of excitement and sense of wonder. For others, it’s groaning, and frustration on where to even begin.

Well, in this lesson we’re going to make Solving Linear Equation Word Problems manageable with easy to follow tricks and steps.

We already know how to solve all different types of equations. Yay!

And we also know how to translate algebraic expressions and equations. Double Yay!

Now it’s time to bring both of these together.

Finding the length of the missing side of a triangle

Solving equations and word problems Example

But, what about the tricks and steps?

Yes, there are some easy to follow steps that we are going to use to solve linear word problems.

Read the problem carefully and determine what is being asked.
Create a sidebar! Using different colors, symbols and diagrams and write an equation the relates all the information given.
Solve your equation and check your answer(s).

Now, these steps might not seem all that remarkable, but once you see them in action I guarantee that writing equations from word problems and solving them will become like second nature!

Again, the secret to success is your Sidebar. This is where you will write down all the information you’ve gleaned from the problem, and formulate a solution by writing an equation to model the situation, as Khan Academy accurately states.

Together we will walk through 9 examples in detail ranging from finding consecutive integers to finding hourly wages, profit and cost, distances for rectangles and triangles, and people’s ages.

Linear Word Problems (How-To) – Video

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Linear systems – word problems

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems . The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you accustomed to finding elements of linear systems inside of word problems. Example 1 : A farmhouse shelters 16 animals. Some of them are chickens and the others are cows. Altogether these animals have 60 legs. How many chickens and how many cows are in the farmhouse?

First, to make the calculations clearer, we will choose symbols to represent the number of cows and the number of chickens. Let us say that the chickens will be represented with x and the cows with y. Now, this task gave us enough information to make two equations. The first one is that the sum of the number of chickens (x) and the number of cows (y) is 16, since there are only 16 animals in the farmhouse. That equation should look like this: x + y = 16

The second piece of information we have is that the total number of legs in the farmhouse is 60. Since we know that cows have four legs each and chickens have two legs each, we have enough information to make another equation. This one will look like this: 2*x + 4*y = 60

Now we have a system of linear equations with two equations and two variables. The only thing left to do now is to solve the system. We will solve it here for you, but if you need to remind yourself how to do that step by step, read the article called Systems of linear equations . x = 16 – y 32 – 2y + 4y = 60 2y = 28 y = 14 x = 2

We can now see that there are two chickens and 14 cows in the farmhouse. The next example will be a bit harder,

Example 2: Rodney’s Kitchen Supplies makes and sells spoons and forks. It costs the store $2 to buy the supplies needed to make a fork, and $1 for the supplies needed to make a spoon. The store sells the forks for $4 and the spoons for 5$. Last month Rodney’s Kitchen Supplies spent $39 on supplies and sold the all of the forks and spoons that were made last month using those supplies for $93. How many forks and spoons did they make?

As we did in the first example, we will first designate symbols to available variables. So, the number of forks made will be represented with x and the number of spoons with y. Again, we have enough information to make two equations. The total cost of making a particular number of forks (x), which cost $2 to make each, and a particular number of spoons (y), which cost $1 to make each, is $39. So that will be our first equation and it will look like this: 2*x + y = 39 The other piece of information tells us that if we sell that number of forks (x) for $4 each and that number of spoons (y) for $5 each, we will make $93. And that will be our second equation: 4*x + 5*y = 93 This was the hard part. Now all we have to do is to solve this linear system to find how many spoons and how many forks did we make last month. y = 39 – 2x 4x + 5*(39 – 2x) = 93 4x + 195 – 10x = 93 -6x = 93 – 195 -6x = -102 |: (-6) x = 17 y = 5

We can see that last month the store made and sold 17 forks and five spoons.

Although they can seem complicated, mastery and understanding of linear systems and associated word problems will come with a bit of practice. With experience you will be able to recognize their elements and solve even complicated systems with ease. Feel free to use the math worksheets below to practice solving this type of linear systems.