Real world linear equations worksheet answer key

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Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. You can customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. The worksheets suit pre-algebra and algebra 1 courses (grades 6-9).

You can choose from SEVEN basic types of equations, ranging from simple to complex, explained below (such as one-step equations, variable on both sides, or having to use the distributive property). Customize the worksheets using the generator below.

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Ready-made worksheets

See also

Worksheets for simplifying expressions

Worksheets for evaluating expressions with variables

Worksheets for writing expressions with variables from verbal expressions

Worksheets for linear inequalities

Key to Algebra Workbooks

Key to Algebra offers a unique, proven way to introduce algebra to your students. New concepts are explained in simple language, and examples are easy to follow. Word problems relate algebra to familiar situations, helping students to understand abstract concepts. Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. Students begin their study of algebra in Books 1-4 using only integers. Books 5-7 introduce rational numbers and expressions. Books 8-10 extend coverage to the real number system.

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Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem:

● Read the problem carefully and note what is given and what is required and what is given.

● Denote the unknown by the variables as x, y, …….

● Translate the problem to the language of mathematics or mathematical statements.

● Form the linear equation in one variable using the conditions given in the problems.

● Solve the equation for the unknown.

● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.

Solution:

Then the other number = x + 9

Let the number be x.

Sum of two numbers = 25

According to question, x + x + 9 = 25

⇒ 2x + 9 = 25

⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)

⇒ 2x = 16

⇒ 2x/2 = 16/2 (divide by 2 on both the sides)

⇒ x = 8

Therefore, x + 9 = 8 + 9 = 17

Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?

Solution:

Let the common ratio be x.

Let the common ratio be x.

Their difference = 48

According to the question,

7x - 3x = 48

⇒ 4x = 48

⇒ x = 48/4

⇒ x = 12

Therefore, 7x = 7 × 12 = 84

          3x = 3 × 12 = 36

Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.

Solution:

Let the breadth of the rectangle be x,

Then the length of the rectangle = 2x

Perimeter of the rectangle = 72

Therefore, according to the question

2(x + 2x) = 72

⇒ 2 × 3x = 72

⇒ 6x = 72

⇒ x = 72/6

⇒ x = 12

We know, length of the rectangle = 2x

                      = 2 × 12 = 24

Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution:

Let Ron’s present age be x.

Then Aaron’s present age = x - 5

After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.

According to the question;

Ron will be twice as old as Aaron.

Therefore, x + 4 = 2(x - 5 + 4)

⇒ x + 4 = 2(x - 1)

⇒ x + 4 = 2x - 2

⇒ x + 4 = 2x - 2

⇒ x - 2x = -2 - 4

⇒ -x = -6

⇒ x = 6

Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1

Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.

Solution:

Let one part of the number be x

Then the other part of the number = x + 10

The ratio of the two numbers is 5 : 3

Therefore, (x + 10)/x = 5/3

⇒ 3(x + 10) = 5x

⇒ 3x + 30 = 5x

⇒ 30 = 5x - 3x

⇒ 30 = 2x

⇒ x = 30/2

⇒ x = 15

Therefore, x + 10 = 15 + 10 = 25

Therefore, the number = 25 + 15 = 40 

The two parts are 15 and 25.

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.

Solution:

Let Robert’s age be x years.

Then Robert’s father’s age = 4x

After 5 years, Robert’s age = x + 5

Father’s age = 4x + 5

According to the question,

4x + 5 = 3(x + 5)

⇒ 4x + 5 = 3x + 15

⇒ 4x - 3x = 15 - 5

⇒ x = 10

⇒ 4x = 4 × 10 = 40

Robert’s present age is 10 years and that of his father’s age = 40 years.

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.

Solution:

Let the first multiple of 5 be x.

Then the other multiple of 5 will be x + 5 and their sum = 55

Therefore, x + x + 5 = 55

⇒ 2x + 5 = 55

⇒ 2x = 55 - 5

⇒ 2x = 50

⇒ x = 50/2

⇒ x = 25

Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30

Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.

Solution:

Let the angle be x.

Complement of x = 90 - x

Given their difference = 12°

Therefore, (90 - x) - x = 12°

⇒ 90 - 2x = 12

⇒ -2x = 12 - 90

⇒ -2x = -78

⇒ 2x/2 = 78/2

⇒ x = 39

Therefore, 90 - x = 90 - 39 = 51

Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.

Solution:

The table cost $ 40 more than the chair.

Let us assume the cost of the chair to be x.

Then the cost of the table = $ 40 + x

The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)

Total cost of 2 tables and 3 chairs = $705

Therefore, 2(40 + x) + 3x = 705

80 + 2x + 3x = 705

80 + 5x = 705

5x = 705 - 80

5x = 625/5

x = 125 and 40 + x = 40 + 125 = 165

Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?

Solution:

Let the number be x, then 3/5 ᵗʰ of the number = 3x/5

Also, 1/2 of the number = x/2

According to the question,

3/5 ᵗʰ of the number is 4 more than 1/2 of the number.

⇒ 3x/5 - x/2 = 4

⇒ (6x - 5x)/10 = 4

⇒ x/10 = 4

⇒ x = 40

The required number is 40.

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

● Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

● Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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