system of equations word problem how to

Writing Systems of Linear Equations from Word Problems

Some word problems require the use of systems of linear equations . Here are clues to know when a word problem requires you to write a system of linear equations:

(i) There are two different quantities involved: for instance, the number of adults and the number of children, the number of large boxes and the number of small boxes, etc. (ii) There is a value associated with each quantity: for instance, the price of an adult ticket or a children's ticket, or the number of items in a large box as opposed to a small box.

Such problems often require you to write two different linear equations in two variables. Typically, one equation will relate the number of quantities (people or boxes) and the other equation will relate the values (price of tickets or number of items in the boxes).

Here are some steps to follow:

1. Understand the problem.

Understand all the words used in stating the problem. Understand what you are asked to find. Familiarize the problem situation.

2. Translate the problem to an equation.

Assign a variable (or variables) to represent the unknown. Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

Use substitution , elimination or graphing method to solve the problem.

The cost of admission to a popular music concert was $ 162 for 12 children and 3 adults. The admission was $ 122 for 8 children and 3 adults in another music concert. How much was the admission for each child and adult?

1 . Understand the problem:

The admission cost for 12 children and 3 adults was $ 162 . The admission cost for 8 children and 3 adults was $ 122 .

2 . Translate the problem to an equation.

Let x represent the admission cost for each child. Let y represent the admission cost for each adult. The admission cost for 12 children plus 3 adults is equal to $ 162 . That is, 12 x + 3 y = 162 . The admission cost for 8 children plus 3 adults is equal to $122. That is, 8 x + 3 y = 122 .

3 . Carry out the plan and solve the problem.

Subtract the second equation from the first. 12 x + 3 y = 162 8 x + 3 y = 122 _ 4 x = 40 x = 10 Substitute 10 for x in 8 x + 3 y = 122 . 8 ( 10 ) + 3 y = 122 80 + 3 y = 122 3 y = 42 y = 14 Therefore, the cost of admission for each child is $ 10 and each adult is $ 14 .

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Unit 6: Lesson 3

Systems of linear equations word problems | Lesson

What are systems of linear equations word problems, and how frequently do they appear on the test.

How do I solve systems of linear equations word problems?

Systems of equations examples, how do i write systems of linear equations.

Let's look at an example!

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Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1. Let's start by identifying the important information:

2. Define your variables.

Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50 (Equation related to cost)

x + y = 87 (Equation related to the number sold)

4. Solve!

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6. Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7. Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25 (Equation representing your lunch)

4x + 2y = 10 (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

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Systems of Equations Word Problems Lesson

How to Solve Word Problems with Linear Systems

Six-step method for applications of linear systems.

Skills Check:

Solve each word problem.

Jason and Mallorie are selling baked goods for a school fundraiser. Students can buy boxes of assorted donuts and boxes of cupcakes. Jason sold 3 boxes of donuts and 14 boxes of cupcakes for a total of $178. Mallorie sold 14 boxes of donuts and 6 boxes of cupcakes for a total of $178. What is the cost each of one box of assorted donuts and one box of cupcakes?

Please choose the best answer.

A boat traveled 119 miles each way downstream and back. The trip downstream took 7 hours. The trip back took 17 hours. What is the speed of the boat in still water? What is the speed of the current?

The senior class at Malcolm High School and the junior class at Ridge Point High School both planned trips to the city park. The senior class at Malcolm High School rented and filled 13 vans and 10 buses with 369 students. Ridge Point High School rented and filled 18 vans and 16 buses with 554 students. Each van carried the same number of students; additionally, each bus carried the same number of students. Find the number of students in each van and in each bus.

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Word Problems – System Of Equations

After you finish this lesson, view all of our Pre-Algebra lessons and practice problems. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

Solving Word Problems

To solve a word problem using a system of equations , it is important to; – Identify what we don’t know – Declare variables – Use sentences to create equations

An example on how to do this:

Mary and Jose each bought plants from the same store. Mary spent $188 on 7 cherry trees and 11 rose bushes. Jose spent $236 on 13 cherry trees and 11 rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

$x$

The y-values cancel each other out, so now you are left with only x-values and real numbers.

$-6x = -48$

Then, you plug in your x-value into an original equation in order to find the y-value.

$7(8) + 11y = 188$

Cost of a cherry tree: $8 Cost of a rose bush: $12

Examples of Word Problems – System Of Equations

$7$

Let’s solve this by following steps.

1. What we don’t know:

Cost of a single coffee

Cost of single cupcake

2. Declare variables :

$x$

3. Use sentences to create equations.

$3x+y=7$

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

$y=7-3x$

$28-10x=8$

$x = 2$

2. Declare variables:

$v$

High School A rented and filled 8 vans and 8 buses with 240 students.

$8v+8b=240$

High School B rented and filled 4 vans and 1 bus with 54 students.

$4v+b=54$

Video-Lesson Transcript

To solve a word problem using system of equations, it is important to: 1. Identify what we don’t know 2. Declare variables. 3. Use sentences to create equations.

Let’s have an example:

$\$188$

Let’s solve this by following steps above.

1. What we don’t know: cost of a cherry tree cost of a rose bush

$= \$188$

Now, we have a system of equations

$7x + 11y = 188$

We can solve this by process of substitution, elimination or fraction.

$-1$

Here we’ll have a negated equation

$-13x - 11y = -236$

Let’s do the process of elimination now

$+ {-13x} + 11y = 236$

We’ll have

$6x = -48$

Now, we have

$x = 8$

Remember, our declared variable?

Now we can say that

$= 8$

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System-of-Equations Word Problems

Exercises More Exercises

Many problems lend themselves to being solved with systems of linear equations. In "real life", these problems can be incredibly complex. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics.

In your studies, however, you will generally be faced with much simpler problems. What follows are some typical examples.

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System of Equations Word Problems

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

In the past , I would have set this up by picking a variable for one of the groups (say, " c " for "children") and then use "(total) less (what I've already accounted for)" (in this case, " 2200 − c ") for the other group. Using a system of equations, however, allows me to use two different variables for the two different unknowns.

number of adults: a

number of children: c

With these variables, I can create equations for the totals they've given me:

total number: a + c = 2200

total income: 4 a + 1.5 c = 5050

Now I can solve the system for the number of adults and the number of children. I will solve the first equation for one of the variables, and then substitute the result into the other equation:

a = 2200 − c

4(2200 − c ) + 1.5 c = 5050

8800 − 4 c + 1.5 c = 5050

8800 − 2.5 c = 5050

−2.5 c = −3750

Now I can back-solve for the value of the other variable:

a = 2200 − (1500) = 700

I now have values for both of my two variables. I can look back at my definitions for the variables in order to interpret these values within context. To answer the original question, there were:

1500 children and 700 adults.

You will probably start out with problems which, like the one above, seem very familiar. But you will then move on to more complicated problems.

The sum of the digits of a two-digit number is 7 . When the digits are reversed, the number is increased by 27 . Find the number.

The trick here is to work with the digits explicitly. I'll use " t " for the "tens" digit of the original number and " u " for the "units" (or "ones") digit. I then have:

The ten's digit stands for "ten times of this digit's value". Just as "26" is "10 times 2, plus 6 times 1", so also the two-digit number they've given me will be ten times the "tens" digit, plus one times the "units" digit. In other words:

original number: 10 t + 1 u

The new number has the values of the digits (represented by the variables) in reverse order. This gives me:

new number: 10 u + 1 t

And this new number is twenty-seven more than the original number. The keyword "is" means "equals", so I get:

(new number) is (old number) increased by (twenty-seven)

10 u + 1 t = (10 t + 1 u ) + 27

Now I have a system of equations that I can solve:

10 u + t = 10 t + u + 27

First I'll simplify the second equation:

9 u − 9 t = 27

u − t = 3

After reordering the variables in the first equation, I now have:

Adding down , I get:

Then t = 2 . Back-solving, this means that the original number was 25 and the new number (gotten by switching the digits) is 52 . Since 52 − 25 = 27 , this solution checks out.

The number is 25 .

How can I find a parabola's equation from just three points?

To find the parabola's equation, you'll plug the values from each of the three ( x , y ) pairs into y = ax 2 + bx + c . This will give you three equations in three unknowns, those unknowns being the coefficients, a , b , and c .

(If they've given you "nice" points like, say, all three intercepts, then life is pretty easy. You'd take the x -intercept values, convert them to factors of the quadratic, and then use the third intercept to find the value of any multiplier. Supposing the x -intercepts are at −3 and +2 ; then the factors would be x + 3 and x &minus 2 , which multiply to x 2 + x −6 . If you plug zero in for x , you'll get −6 . If the y -intercept is actually at y = −12 , then you know that you need to multiply through by 2 , so the equation is y = 2 x 2 + 2 x − 12 . Easy-peasy.

(But what does it look like when they give you not-nice points?)

Find the equation of the parabola that passes through the points (−1, 9) , (1, 5) , and (2, 12) .

Recalling that a parabola has a quadratic as its equation, I know that I am looking for an equation of the form ax 2 + bx + c = y . Also, I know that points are of the form ( x , y ) . Practically speaking, this mean that, in each of these points, they have given me values for x and y that make the quadratic equation true. Plugging the three points in the general equation for a quadratic, I get a system of three equations, where the variables stand for the unknown coefficients of that quadratic:

a (−1) 2 + b (−1) + c = 9

a (1) 2 + b (1) + c = 5

a (2) 2 + b (2) + c = 12

Simplifying the three equations, I get:

1 a − b + c = 9

1 a + b + c = 5

4 a + 2 b + c = 12

I won't display the solving of this problem, but the result is that a = 3, b = −2, and c = 4 , so the equation they're wanting is:

y = 3 x 2 − 2 x + 4

You may also see similar exercises referring to circles, using:

x 2 + y 2 + bx + cy + d = 0

...or other conics, though parabolas are the most common. Keep in mind that projectile problems (like shooting an arrow up in the air or dropping a penny from the roof of a tall building) are also parabola problems, using:

−( 1 / 2 ) gt 2 + v 0 t + h 0 = s

...where h 0 is the original height, v 0 is the initial velocity, s is the height at time t , usually measured in seconds, and g refers to gravity, being 9.8 if you're working in meters and 32 if you're working in feet).

All of these different permutations of the above example work the same way: Take the general equation for the curve, plug in the given points, and solve the resulting system of equations for the values of the coefficients. Warning: If you see an exercise of this sort in the homework, be advised that you may be expected to know the forms of the general equations (such as " ax 2 + bx + c = y " for parabolas) on the next text.

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Systems of Equations (Word Problems)

These lessons, with videos, examples and step-by-step solutions help Grade 8 students learn how to analyze and solve pairs of simultaneous linear equations.

A. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.

B. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 .

C. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair.

Common Core: 8.EE.8c

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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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