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Word Problems Worksheet 1 – This 6 problem algebra worksheet will help you practice creating and solving systems of equations to represent real-life situations. You will use the “ elimination ” method to eliminate variables from standard form equations. Word Problems Worksheet 1 RTF Word Problems Worksheet 1 PDF View Answers

Word Problems Worksheet 3 – This 6 problem algebra worksheet will help you practice creating and solving systems of equations to represent real-life situations. Most of the problems involve money, so make sure you’re ready for some decimals. Word Problems Worksheet 3 RTF Word Problems Worksheet 3 PDF View Answers

Word Problems Worksheet 4 – This 6 problem algebra worksheet will help you practice creating and solving systems of equations to represent real-life situations. Most of the problems involve money, and a few distractors are introduced. Word Problems Worksheet 4 RTF Word Problems Worksheet 4 PDF View Answers

Word Problems Worksheet 5 – This 8 problem algebra worksheet features more abstract word problems like “ The sum of x and y is 42.  the difference of x and y is 13.  Find x and y .” There are a few negative integers, so be careful! Word Problems Worksheet 5 RTF Word Problems Worksheet 5 PDF View Answers

Word Problems Worksheet 6 – This 8 problem algebra worksheet features more abstract word problems like “ The sum of  twice a number, x, and twice another number, y, is 118.  The value of y is one less than twice the value of x.  Find x and y. ” One of the problems even has an infinite number of solutions! Word Problems Worksheet 6 RTF Word Problems Worksheet 6 PDF View Answers

These free  systems of equations   worksheets  will help you practice solving real-life systems of equations using the “ elimination ” method.  You will  need to create and solve a system of equations to represent each situation.  The exercises can also be solved using other algebraic methods if you choose.

This is a progressive series that starts simple with problems involving buying movie tickets and collecting for fundraisers.  Eventually the negative numbers , decimals , the  distributive property  and “ the opposite of x ” come into play.

Each worksheet will help students master Common Core skills in the Algebra strand.  They are great for ambitious students in pre-algebra or algebra classes.

These free  elmination  worksheets are printable and available in a variety of formats.  Each sheet includes an example to help you get started.  Of course, answer keys are provided as well.

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There are multiple problems with your worksheets. You have word problems which are unanswerable or answers which make no sense such as fractions of people. Your answer sheets are frequently incorrect. For a quick example: System of Equations- Word Problems #1 – KEY. Question #1 states that there are 9 tickets purchased, the Answer shows a=6 and b=2. 6+2 is 8 not 9. Question 5 on that same worksheet is unanswerable. I’ve found problems on worksheet 1, 2, 5 so far. How can you claim to have the best materials with so many clear problems?

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How to Solve Systems of Equations Word Problems? (+FREE Worksheet!)

Learn how to create and solve systems of equations word problems by the elimination method.

How to Solve Systems of Equations Word Problems? (+FREE Worksheet!)

Related Topics

  • How to Solve One-Step Equations
  • How to Solve One-Step Inequalities
  • How to Solve Multi-Step Inequalities
  • How to Solve Systems of Equations
  • How to Graph Single–Variable Inequalities

Step by step guide to solve systems of equations word Problems

  • Find the key information in the word problem that can help you define the variables.
  • Define two variables: \(x\) and \(y\)
  • Write two equations.
  • Use the elimination method for solving systems of equations.
  • Check the solution by substituting the ordered pair into the original equations.

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Systems of equations word problems – example 1:.

Tickets to a movie cost \($8\) for adults and \($5\) for students. A group of friends purchased \(20\) tickets for \($115.00\). How many adult tickets did they buy?

Let \(x\) be the number of adult tickets and \(y\) be the number of student tickets. There are \(20\) tickets. Then: \(x+y=20\). The cost of adults’ tickets is \($8\) and for students ticket is \($5\), and the total cost is \($115\). So, \(8x+5y=115\). Now, we have a system of equations: \(\begin{cases}x+y=20 \\ 8x+5y=115\end{cases}\) Multiply the first equation by \(-5\) and add to the second equation: \(-5(x+y= 20)=- \ 5x-5y=- \ 100\) \(8x+5y+(-5x-5y)=115-100→3x=15→x=5→5+y=20→y=15\). There are \(5\) adult tickets and \(15\) student tickets.

Exercises for Solving Systems of Equations Word Problems

  • A farmhouse shelters \(10\) animals, some are pigs and some are ducks. Altogether there are \(36\) legs. How many of each animal are there?
  • A class of \(195\) students went on a field trip. They took vehicles, some cars, and some buses. Find the number of cars and the number of buses they took if each car holds \(5\) students and each bus holds \(45\) students.
  • The difference of two numbers is \(6\). Their sum is \(14\). Find the numbers.
  • The sum of the digits of a certain two–digit number is \(7\). Reversing is increasing the number by \(9\). What is the number?
  • The difference of two numbers is \(18\). Their sum is \(66\). Find the numbers.

Download Systems of Equations Word Problems Worksheet

  • There are \(8\) pigs and \(2\) ducks.
  • There are \(3\) cars and \(4\) buses.
  • \(10\) and \(4\).
  • \(24\) and \(42\).

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by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

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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

  • Highlight the important information in the problem that will help write two equations.
  • Define your variables
  • Write two equations
  • Use one of the methods for solving systems of equations to solve.
  • Check your answers by substituting your ordered pair into the original equations.
  • Answer the questions in the real world problems. Always write your answer in complete sentences!

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:

  • hot dogs cost $1.50
  • Sodas cost $0.50
  • Made a total of $78.50
  • Sold 87 hot dogs and sodas combined

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.

Solving a systems using substitution

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?

  • 3 soft tacos + 3 burritos cost $11.25
  • 4 soft tacos + 2 burritos cost $10.00

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.

Solving Systems Using Combinations

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:

solving systems of equations by elimination word problems

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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5.2: Solve Systems of Equations by Substitution

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Learning Objectives

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

  • Solve a system of equations by substitution
  • Solve applications of systems of equations by substitution

Before you get started, take this readiness quiz.

  • Simplify −5(3−x). If you missed this problem, review Example 1.10.43 .
  • Simplify 4−2(n+5). If you missed this problem, review Example 1.10.49 .
  • Solve for y. 8y−8=32−2y If you missed this problem, review Example 2.3.22 .
  • Solve for x. 3x−9y=−3 If you missed this problem, review Example 2.6.22 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

\(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example \(\PageIndex{1}\).

Example \(\PageIndex{1}\): How to Solve a System of Equations by Substitution

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.

Try It \(\PageIndex{2}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.\)

Try It \(\PageIndex{3}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.\)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

  • Solve one of the equations for either variable.
  • Substitute the expression from Step 1 into the other equation.
  • Solve the resulting equation.
  • Substitute the solution in Step 3 into one of the original equations to find the other variable.
  • Write the solution as an ordered pair.
  • Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example \(\PageIndex{4}\).

Example \(\PageIndex{4}\)

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

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It \(\PageIndex{5}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.\)

Try It \(\PageIndex{6}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.\)

(−1,−3)

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example \(\PageIndex{7}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It \(\PageIndex{8}\)

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

(1,−2)

Try It \(\PageIndex{9}\)

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

In Example \(\PageIndex{7}\) it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example \(\PageIndex{10}\) it will be easier to solve for x .

Example \(\PageIndex{10}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)

We will solve the first equation for \(x\) and then substitute the expression into the second equation.

Try It \(\PageIndex{11}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.\)

(−2,−3)

Try It \(\PageIndex{12}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.\)

When both equations are already solved for the same variable, it is easy to substitute!

Example \(\PageIndex{13}\)

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

Since both equations are solved for y , we can substitute one into the other.

Try It \(\PageIndex{14}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.\)

Try It \(\PageIndex{15}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.\)

Be very careful with the signs in the next example.

Example \(\PageIndex{16}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+2 y=4} \\ {6 x-y=8}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for y .

Try It \(\PageIndex{17}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)

\((2,\frac{3}{2})\)

Try It \(\PageIndex{18}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.\)

\((−\frac{1}{2},−2)\)

In Example , it will take a little more work to solve one equation for x or y .

Example \(\PageIndex{19}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-3 y=6} \\ {15 y-20 x=-30}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for x .

Try It \(\PageIndex{20}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.\)

infinitely many solutions

Try It \(\PageIndex{21}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.\)

Look back at the equations in Example \(\PageIndex{22}\). Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example \(\PageIndex{22}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)

The second equation is already solved for y , so we can substitute for y in the first equation.

Try It \(\PageIndex{23}\)

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

no solution

Try It \(\PageIndex{24}\)

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

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

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

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example \(\PageIndex{25}\)

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It \(\PageIndex{26}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

The numbers are 3 and 7.

Try It \(\PageIndex{27}\)

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

The numbers are 2 and −8.

In the Example \(\PageIndex{28}\), we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example \(\PageIndex{28}\)

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and width of the rectangle.

Try It \(\PageIndex{29}\)

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

The length is 12 and the width is 8.

Try It \(\PageIndex{30}\)

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

The length is 23 and the width is 6.

For Example \(\PageIndex{31}\) we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example \(\PageIndex{31}\)

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It \(\PageIndex{32}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 22 degrees and 68 degrees.

Try It \(\PageIndex{33}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 36 degrees and 54 degrees.

Example \(\PageIndex{34}\)

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It \(\PageIndex{35}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

There would need to be 160 policies sold to make the total pay the same.

Try It \(\PageIndex{36}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Kenneth would need to sell 1,000 suits.

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

  • Instructional Video-Solve Linear Systems by Substitution
  • Instructional Video-Solve by Substitution

Key Concepts

Elimination Calculator

gives you step-by-step help on solving systems by elimination.

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About Elimination

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WORD PROBLEMS USING ELIMINATION METHOD

Problem 1 :

Find the value of two numbers if their sum is 12 and their difference is 4.

Let x and y be the two numbers.

Given : T heir sum is 12.

x + y = 12 ----(1)

Given : T heir difference is 4.

x - y = 4 ----(1)

(1) + (2) :

Divide both sides by 2.

Substitute x = 8 into (1).

Subtract 12 from both sides.

Therefore, the numbers are 8 and 4.

Problem 2 :

Kevin bought pen and pencils in a total of 50. The cost of each pen is $2 and that of a pencil is $1.50. If he had paid a total of $85 for his purchase, find the cost of a pen and a pencil.

Let x and y be the costs of a pen and a pencil respectively.

Given : Kevin bought pen and pencils in a total of 50.

x + y = 50 ----(1)

Given : The cost of each pen is $2, a pencil is  $1.50 and Kevin paid a total of $85 for his purchase.

2x + 1.5y = 85 ----(2)

1.5(1) - (2) :

Divide both sides by 0.5.

Substitute x = 20 into (1).

20 + y = 50

Subtract 20 from both sides.

number of pens = 20

number of pencils = 30

Problem 3 :

The sum of two numbers is 7. The difference between 5 times the larger and 3 times the smaller is equal 11. Find the numbers. 

Let x and y be the two numbers such that x > y.

Given : The sum of two numbers is 7.

x + y = 7 ----(1)

Given : The difference between 5 times the larger and 3 times the smaller is equal 11.

5x - 3y = 11 ----(2)

3(1) + (2) :

Divide both sides by 8.

Substitute x = 4 into (1).

Subtract 3 from both sides.

Therefore, the two numbers are 4 and 3.

Problem 4 :

Chase and Sara went to the candy store. Chase bought 5 pieces of fudge and 3 pieces of bubble gum for a total of $5.70. Sara bought 2 pieces of fudge and 10 pieces of bubble gum for a total of $3.60. Find the cost of 1 piece of fudge and 1 piece of bubble gum?

Let f and g be the costs of 1 piece of fudge and 1 piece of bubble.

From the given information,

5f + 3b = 5.7 ----(1)

2f + 10b = 3.6 ----(2)

2(1) - 5(2) :

Divide both sides by -44.

Substitute x = 0.15 into (2).

 2(0.15) + 10b = 3.6

0.3 + 10b = 3.6

Subtract 0.3 from both sides.

Divide both sides by 10.

cost of 1 piece of fudge = $0.15

cost of 1  piece of bubble gum = $0.33

Problem 5 :

Daily earnings of Oliver and Henry are in the ratio 3 : 4 and their expenditures are in the ratio 5 : 7. If each saves $50 per day, find the daily earnings of each.

From the earnings ratio 3 : 4,

earnings of Oliver = 3x

earnings of Henry = 4x

From the expenditures ratio 5 : 7,

expenditure of Oliver = 5y

expenditure of Henry = 7y

The relationship between income, expenditure and savings :

Income - Expenditure = Savings

Then, we have

3x - 5y = 50 ----(1)

4x - 7y = 50 ----(2)

7(1) - 5(2) :

earnings of Oliver = 3(100) = $300

earnings of Henry = 4(100) = $400

Problem 6 :

The sum of the present ages of a father and his son is 45 years. 10 years hence, the difference between their ages is 25 years. Find the present age of the father and son.

Let f and s be the present ages of father and son.

Given : The sum of the present ages 45.

f + s = 45 ----(1)

Given : 10 years hence, the difference between the ages is 25.

(f + 10) - (s + 10) = 25

f + 10 - s - 10 = 25

f - s = 25 ----(2)

Substitute f = 35 into (1).

 35 + s = 45

Subtract 35 from both sides.

present age of the father = 35 years

present age of the son = 10 years

Problem 7 :

In a three digit number, the middle digit is zero and sum of the other two digits is 9. If 99 is added to it, the digits are reversed. Find the three digit number.

Let x 0 y  be the required three digit number.

x + y = 9 ----(1)

Given : When 99  is added to it, the digits are reversed.

x 0 y + 99 = y 0 x

100(x) + 10(0) + 1(y) + 99 = 100(y) + 10(0) + 1(x)

100x + 0 + y + 99 = 100y + 0 + x

100x + y + 99 = 100y + x

99x - 99y = -99

Divide both sides by 99.

x - y = -1 ----(2)

Substitute x  = 4 into (1).

Subtract 4 from both sides.

x 0 y = 405

Therefore, the three digit number is 405.

Problem 8 :

Two numbers are such that twice the greater number exceeds twice the sm aller one by 18. One-third of the smaller number and one-fifth of the greater number are together 21. Find the two numbers.

Let x  and  y  be the two numbers such that x > y .

Given : Twice the greater number exceeds twice the smaller one by 18.

2x - 2y = 18

x - y = 9 ----(1)

Given : One-third of the smaller number and one-fifth of the greater number are together 21.

5y  + 3x = 315

3x + 5y = 315 ----(2)

5(1) + (2) :

Substitute x  = 45 into (1).

Subtract 45 from both sides.

Therefore, the two numbers are 45 and 36.

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  2. PDF Systems of Equations Word Problems

    Systems of Equations Word Problems Date_____ Period____ 1) Find the value of two numbers if their sum is 12 and their difference is 4. 4 and 8 2) The difference of two numbers is 3. Their sum is 13. Find the numbers. 5 and 8 3) Flying to Kampala with a tailwind a plane averaged 158 km/h. On the return trip the plane only

  3. Word Problems

    Word Problems Worksheet 1 - This 6 problem algebra worksheet will help you practice creating and solving systems of equations to represent real-life situations. You will use the " elimination " method to eliminate variables from standard form equations. Word Problems Worksheet 1 RTF. Word Problems Worksheet 1 PDF. View Answers.

  4. Systems of equations with elimination (practice)

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    Systems of equations word problems. Google Classroom. Microsoft Teams. You might need: Calculator. Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h . If doubled, Malcolm's maximum speed would be 80 km/h more than Ravi's maximum speed. What were Malcolm's and Ravi's maximum speeds?

  6. PDF Systems of Equations Word Problems

    Systems of Equations Word Problems. 2) Flying with the wind a plane went 183 km/h. Flying into the same wind the plane only went 141 km/h. Find the speed of the plane in still air and the speed of the wind. 3) Castel and Gabriella are selling pies for a school fundraiser. Customers can buy apple pies and lemon meringue pies.

  7. Systems of equations with elimination: TV & DVD

    11 years ago. To solve a system of equations by elimination, you have to manipulate one or both of the equations so that when you add them together, one of the variables drops out. 3x + 2y = -17. 1x - 3y = 9. If you were to multiply both sides of the second equation by -3, you would get: 3x + 2y = -17. -3x + 9y = -27.

  8. PDF Systems of Equations Elimination

    Infinite number of solutions. 23) −14 =. 24) (2, 0) (−1, 1) Create your own worksheets like this one with Infinite Algebra 1. Free trial available at KutaSoftware.com. ©q R2h041222 cK7uitqaL ASPovfPthwEanrQed vLOLrCy.6 w AAVlXl9 wrxivgghCtUsC xrmeAsfeGrivpe9du.Q Q iMwaHdMeB GwSijtZht xIrnOfNiRnFiotLeH 1AAlSgheWb4r0aG X1K.J.

  9. How to Use Elimination to Solve a System of Equations: Word Problems

    A Step-by-step Guide to Using Elimination to Solve a System of Equations: Word Problems. Elimination, as the name suggests, focuses on eliminating one variable to simplify a system of equations. This method is particularly effective for linear equation systems where addition or subtraction can effectively reduce the system to one equation with ...

  10. How to Solve Systems of Equations Word Problems? (+FREE Worksheet!)

    Step by step guide to solve systems of equations word Problems . Find the key information in the word problem that can help you define the variables. Define two variables: \(x\) and \(y\) Write two equations. Use the elimination method for solving systems of equations. Check the solution by substituting the ordered pair into the original ...

  11. 5.3: Solve Systems of Equations by Elimination

    Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form and the coefficients of m are opposites. Add. {n + m = 39 n − m = 9 _ 2n = 48 Solve for n. n = 24 Substitute n=24 into one of the original n + m = 39 equations and solve form. 24 + m = 39 m = 15 Step 6.

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  13. Substitution Method Practice Problems With Answers

    Do you want to learn how to solve systems of equations using the substitution method? Check out this webpage for ten (10) practice problems with detailed answers and explanations. You will also find links to other related topics in intermediate algebra, such as rational inequalities, distance formula, graphing a line, and literal equations.

  14. Elimination method review (systems of linear equations)

    Example 1. We're asked to solve this system of equations: 2 y + 7 x = − 5 5 y − 7 x = 12. We notice that the first equation has a 7 x term and the second equation has a − 7 x term. These terms will cancel if we add the equations together—that is, we'll eliminate the x terms: 2 y + 7 x = − 5 + 5 y − 7 x = 12 7 y + 0 = 7.

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  17. Systems of Linear Equations Word Problems: Elimination

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  18. 5.2: Solve Systems of Equations by Substitution

    Example 5.2.19. Solve the system by substitution. {4x − 3y = 6 15y − 20x = − 30. Solution. We need to solve one equation for one variable. We will solve the first equation for x. Solve the first equation for x. Substitute 3 4y + 3 2 for x in the second equation. Replace the x with 3 4y + 3 2.

  19. Systems of equations

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  20. Elimination Calculator

    Enter your equations separated by a comma in the box, and press Calculate! Or click the example. About Elimination Use elimination when you are solving a system of equations and you can quickly eliminate one variable by adding or subtracting your equations together. You can use this Elimination Calculator to practice solving systems.

  21. Word Problems Using Elimination Method

    WORD PROBLEMS USING ELIMINATION METHOD. Problem 1 : Find the value of two numbers if their sum is 12 and their difference is 4. Solution : Let x and y be the two numbers. Given : Their sum is 12. x + y = 12 ---- (1) Given : Their difference is 4. x - y = 4 ---- (1)

  22. System of Equations Elimination Calculator

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  23. Systems of equations with substitution

    You might need: Calculator. Solve the system of equations. 5 x − 7 y = 58 y = − x + 2. x =. y =. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.