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Write & Solve Linear Equations Worksheets

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Printable “Algebra” worksheets: Write Equations Using Symbols Write Linear & Nonlinear Expressions Write & Solve Linear Equations

Examples, solutions, videos, and worksheets to help Grade 7 and Grade 8 learn how to use of the properties of equality to solve linear equations having rational coefficients.

How to use of the properties of equality to solve linear equations?

Solving linear equations with rational coefficients using the properties of equality, involves isolating the variable on one side of the equation.

Here are the steps:

Understand the Properties of Equality The properties of equality allow you to perform the same operation on both sides of an equation without changing the equation’s solution. The key properties are: Addition Property of Equality: If a = b, then a+c = b+c. Subtraction Property of Equality: If a = b, then a−c = b−c. Multiplication Property of Equality: If a = b, then ac = bc. Division Property of Equality: If a = b and c ≠ 0 then a/c = b/c

Solving Linear Equations with Rational Coefficients Step 1: Eliminate Fractions (Optional but Helpful) You can start by eliminating fractions by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions involved. This step is not mandatory, but it simplifies the equation. Step 2: Isolate the Variable Use the addition or subtraction property of equality to move all terms containing the variable to one side of the equation and constants to the other side. Step 3: Solve for the Variable Use the multiplication or division property of equality to isolate the variable. Step 4: Check Your Solution Substitute the solution back into the original equation to ensure it satisfies the equation.

• Keep Equations Balanced: Whatever you do to one side of the equation, you must do to the other side.
• Simplify Fractions: Simplify fractions whenever possible to make calculations easier.
• Check the Solution: Always substitute your solution back into the original equation to verify that it works.

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Curriculum  /  Math  /  6th Grade  /  Unit 6: Equations and Inequalities  /  Lesson 6

Equations and Inequalities

Lesson 6 of 14

Criteria for Success

Tips for teachers, anchor problems, problem set, target task, additional practice.

Solve percent problems using equations.

Common Core Standards

Core standards.

The core standards covered in this lesson

Expressions and Equations

6.EE.B.7 — Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.

Ratios and Proportional Relationships

6.RP.A.3.C — Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.

Foundational Standards

The foundational standards covered in this lesson

6.RP.A.1 — Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." "For every vote candidate A received, candidate C received nearly three votes."

6.RP.A.2 — Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. Expectations for unit rates in this grade are limited to non-complex fractions. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid $75 for 15 hamburgers, which is a rate of $5 per hamburger."

The essential concepts students need to demonstrate or understand to achieve the lesson objective

• Determine, through repeated reasoning, an equation to represent the relationship between percent, whole, and part:   $$percent\times{whole}=part$$   (MP.8).
• Write an equation to represent a percent situation when given a part and a percent.
• Write and solve equations to find the whole, given the part and percent.

Suggestions for teachers to help them teach this lesson

In Unit 2, students solved percent problems by reasoning about diagrams, double number lines, and tables. Now having learned about equations in the form  $${px=q}$$ , students revisit percent problems to see how they can be modeled and solved efficiently using an equation (MP.4).

Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.

Problems designed to teach key points of the lesson and guiding questions to help draw out student understanding

25-30 minutes

• 30% of 120?

In general, if you’re given a percent and a whole, how can you find the part? Write this as an equation.

Guiding Questions

30% of what number is 12?

Solve this problem first by drawing a diagram. Then write and solve an equation to verify your solution. 

For each situation below, write and solve an equation to answer the question.

a.   There are 6 liters of water in a bucket, which is 20% of the maximum number of liters the bucket can hold. What is the maximum number of liters the bucket can hold?

b.   A softball team won 18 games, which was 60% of the games they played this season. How many games did the softball team play this season?

A set of suggested resources or problem types that teachers can turn into a problem set

15-20 minutes

Give your students more opportunities to practice the skills in this lesson with a downloadable problem set aligned to the daily objective.

A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved

5-10 minutes

Paula is saving money to buy a tablet. So far, she has saved $54, which is 45% of what she needs to buy the tablet. 

Write and solve an equation to find the price of the tablet. 

Student Response

An example response to the Target Task at the level of detail expected of the students.

The following resources include problems and activities aligned to the objective of the lesson that can be used for additional practice or to create your own problem set.

• Challenge: In a choir, there are 28 female singers which is 40% of the choir. How many female singers would have to be added to the group so exactly 50% of the choir were females?
• EngageNY Mathematics Grade 6 Mathematics > Module 1 — Lessons 27–29 (Revisit these lessons from Unit 2 and have students write equations to solve.)
• Open Up Resources Grade 6 Unit 6 Practice Problems — Lesson 7 #1–3

Topic A: Reasoning About and Solving Equations

Represent equations in the form  $${ x+p=q }$$ and  $${px=q}$$ using tape diagrams and balances.

6.EE.B.6 6.EE.B.7

Define and identify solutions to equations.

Write equations for real-world situations.

Solve one-step equations with addition and subtraction.

Solve one-step equations with multiplication and division.

6.EE.B.7 6.RP.A.3.C

Solve multi-part equations leading to the form  $${x+p=q }$$  and $${px=q}$$ .

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Topic B: Reasoning About and Solving Inequalities

Define and identify solutions to inequalities.

6.EE.B.5 6.EE.B.8

Write and graph inequalities for real-world conditions. (Part 1)

Write and graph inequalities for real-world conditions. (Part 2)

Solve one-step inequalities.

6.EE.B.6 6.EE.B.8

Topic C: Representing and Analyzing Quantitative Relationships

Write equations for and graph ratio situations. Define independent and dependent variables.

6.EE.C.9 6.RP.A.3.A

Represent the relationship between two quantities in graphs, equations, and tables. (Part 1)

Represent the relationship between two quantities in graphs, equations, and tables. (Part 2)

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Word Problems on Linear Equations

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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Class vi math, class 6 linear equations, introduction to linear equation, solving a linear equation by trial and error method, rules to follow for linear equation, rule of transposition.

Linear Equation Test

Linear Equation Worksheet

Answer Sheet

An equation in which the highest power of the variables involved is 1, is known as linear equation. In other words, an equation of the form of ax + b = c, where a, b, c are constants, a ≠ 0 and 'x' is the variable, is called a linear equation in the variable 'x'.

The equal sign in an equation divides it into two sides. Both the sides are known as left hand side (LHS) and right hand side (RHS). A number which when substituted for the variable in an equation, makes LHS = RHS and is known as solution of the equation.

In this method, generally we make a guess of the solution of the equation. We try several values of the variable and find LHS and RHS. When LHS = RHS for a particular value of the variable, we say that it is a solution of the equation. Let’s see some examples.

• We can add the same number to both the sides of the equation.
• We can subtract same number from both sides of the equation.
• We can multiply non-zero number to both the sides of an equation.
• We can divide non-zero number to both the sides of an equation.

Let's solve some linear equations by using above rules.

Example 1. x + 1 = 10. Find the value of 'x'. Solution x + 1 = 10 Add −1 to both side of the equation. x + 1 − 1 = 10 − 1 x = 9

Example 2. y − 2 = 22, find the value of 'y'. Solution. y − 2 = 22 Add 2 to both side of the equation y − 2 + 2 = 22 + 2 y = 24

Example 3. 4 × a = 12, find the value of 'a'. Solution. 4 × a = 12 Divide 4 to both side of the equation 4 x a ⁄ 4 = 12 ⁄ 4 a = 3

Example 4. n ÷ 5 = 3, find the value of 'n'. Solution. n ÷ 5 = 3 Multiply 5 to the both side of the equation n ⁄ 5 × 5 = 3 × 5 n = 15

If term transposed from one side of the equation to the other side, then it's sign will change. Let's see some examples.

Example 1. p + 4 = 12. Find the value of p. Solution. p + 4 = 12 Transposing +4 from LHS to RHS by changing it's sign p = 12 - 4 p = 8

Example 2. 2n = 9 − n. Find the value of n. Solution. 2n = 9 − n Transposing −n from RHS to LHS by changing it's sign 2n + n = 9 3n = 9 Dividing 3 both sides of LHS and RHS, we shall get the value of n. 3n ⁄ 3 = 9 ⁄ 3 n = 3

Example 3. m ⁄ 5 − 4 = 7. Find the value of m. Solution. m ⁄ 5 − 4 = 7 Transposing -4 from LHS to RHS by changing it's sign m ⁄ 5 = 7 + 4 m ⁄ 5 = 11 Multiplying 5 to both LHS and RHS m x 5 ⁄ 5 = 11 x 5 m = 55

Example 4. 3(q − 2) = 2(q + 1) − 3. Find the value of q. Solution. 3(q − 2) = 2(q + 1) − 3 First we should remove the brackets. 3q − 6 = 2q + 2 − 3 3q − 2q = 2 − 3 + 6 q = 5

Class-6 Linear Equation Test

Linear Equation Test - 1

Linear Equation Test - 2

Class-6 Linear Equation Worksheet

Linear Equation Worksheet - 1

Linear Equation Worksheet - 2

Linear Equation Worksheet - 3

Linear-Equation-Answer Download the pdf

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