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Adding fractions Subtracting fractions Dividing fractions Multiplying fractions Rational numbersInverse operations

Reciprocal mathHere you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.

Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.

**Equations with fractions** involve solving equations where the unknown variable is part of the numerator and/or denominator of a fraction.

The numerator (top number) in a fraction is divided by the denominator (bottom number).

To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.

The inverse operation of addition is subtraction.

The inverse operation of subtraction is addition.

The inverse operation of multiplication is division.

The inverse operation of division is multiplication.

For example,

\begin{aligned} \cfrac{2x+3}{5} \, &= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}

How does this relate to 7th grade and 8th grade math?

**Grade 7: Expressions and Equations (7.EE.A.1)**Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.

**Grade 8: Expressions and Equations (8.EE.C.7)**

Solve linear equations in one variable.

**Grade 8: Expressions and Equations (8.EE.C.7b)**

Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

In order to solve equations with fractions:

**Identify the operations that are being applied to the unknown variable.****Apply the inverse operations, one at a time, to both sides of the equation**.**When you have the variable on one side, you have the final answer.****Check the answer by substituting the answer back into the original equation**.

Use this worksheet to check your 7th grade and 8th grade students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREEUse this worksheet to check your 7th grade and 8th grade students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREESolve for x \text{: } \cfrac{x}{5}=4 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, the x is divided by 5.

\cfrac{x}{5}

2**Apply the inverse operations, one at a time, to both sides of the equation.**

The inverse of “dividing by 5 ” is “multiplying by 5 ”.

You will multiply both sides of the equation by 5.

3**When you have the variable on one side, you have the final answer.**

The final answer is x=20.

4**Check the answer by substituting the answer back into the original equation.**

You can check the answer by substituting the answer back into the original equation.

\cfrac{20}{5}=20\div5=4

Solve for x \text{: } \cfrac{x}{3}=8 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, the x is divided by 3.

\cfrac{x}{3}

**Apply the inverse operations, one at a time, to both sides of the equation.**

The inverse of “dividing by 3 ” is “multiplying by 3 ”.

You will multiply both sides of the equation by 3.

**Write the final answer, checking that it is correct.**

The final answer is x=24.

You can check the answer by substituting the answer back into the original equation.

\cfrac{24}{3}=24\div3=8

Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).

\cfrac{x \, + \, 1}{2}

**Apply the inverse operations, one at a time, to both sides of the equation.**

First, clear the fraction by multiplying both sides of the equation by 2.

Then, subtract 1 from both sides.

**Write the final answer, checking that it is correct.**

The final answer is x=13.

You can check the answer by substituting the answer back into the original equation.

\cfrac{13 \, +1 \, }{2}=\cfrac{14}{2}=14\div2=7

Solve for x \text{: } \cfrac{x}{4}-2=3 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, x is divided by 4 and then 2 is subtracted.

\cfrac{x}{4}-2

**Apply the inverse operations, one at a time, to both sides of the equation.**

First, add 2 to both sides of the equation.

Then, multiply both sides of the equation by 4.

**Write the final answer, checking that it is correct.**

The final answer is x=20.

You can check the answer by substituting the answer back into the original equation.

\cfrac{20}{4}-2=20\div4-2=5-2=3

Solve for x \text{: } \cfrac{3x}{5}+1=7 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, x is multiplied by 3, then divided by 5 , and then 1 is added.

\cfrac{3x}{5}+1

**Apply the inverse operations, one at a time, to both sides of the equation.**

First, subtract 1 from both sides of the equation.

Then, multiply both sides of the equation by 5.

Finally, divide both sides by 3.

**Write the final answer, checking that it is correct.**

The final answer is x=10.

You can check the answer by substituting the answer back into the original equation.

\cfrac{3 \, \times \, 10}{5}+1=\cfrac{30}{5}+1=6+1=7

Solve for x \text{: } \cfrac{2x-1}{7}=3 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, x is multiplied by 2, then 1 is subtracted, and the last operation is divided by 7 (the denominator).

\cfrac{2x-1}{7}

**Apply the inverse operations, one at a time, to both sides of the equation.**

First, multiply both sides of the equation by 7.

Next, add 1 to both sides.

Finally, divide both sides by 3.

**Write the final answer, checking that it is correct.**

The final answer is x=11.

You can check the answer by substituting the answer back into the original equation.

\cfrac{2 \, \times \, 11-1}{7}=\cfrac{22-1}{7}=\cfrac{21}{7}=3

Solve for x \text{: } \cfrac{24}{x}=6 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, x is the denominator. 24 is divided by x.

\cfrac{24}{x}

**Apply the inverse operations, one at a time, to both sides of the equation.**

You need to multiply both sides of the equation by x.

Then, you can divide both sides by 6.

**Write the final answer, checking that it is correct.**

The final answer is x=4.

You can check the answer by substituting the answer back into the original equation.

\cfrac{24}{4}=24\div4=6

Solve for x \text{: } \cfrac{18}{x}-6=3 .

**Identify the operations that are being applied to the unknown variable.**

The unknown is x.

Looking at the left hand side of the equation, x is the denominator. 18 is divided by x , and then 6 is subtracted.

\cfrac{18}{x}-6

**Apply the inverse operations, one at a time, to both sides of the equation.**

First, add 6 to both sides of the equation.

Then, multiply both sides of the equation by x.

Finally, divide both sides by 9.

**Write the final answer, checking that it is correct.**

The final answer is x=2.

You can check the answer by substituting the answer back into the original equation.

\cfrac{18}{2}-6=9-6=3

- When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.

- Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.

- Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.

- Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.

**The solution to an equation can be any type of number**

The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.

**The unknown of an equation can be on either side of the equation**

The unknown, represented by a letter, is often on the left hand side of the equations; however, it doesn’t have to be. It could also be on the right hand side of an equation.

**When multiplying both sides of an equation, multiply each and every term**

When multiplying each side of the equation by a number, it is a common mistake to forget to multiply every term.

For example,

Solve:

\cfrac{x}{2}+3=9

Here, the + 3 was not multiplied by 2, resulting in the incorrect answer.

This person has correctly multiplied each term by the denominator.

**Lowest common denominator (LCD)**

It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.

- Math equations
- Rearranging equations
- How to find the equation of a line
- Substitution
- Linear equations
- Writing linear equations
- Solving equations
- Identity math
- One step equations

1. Solve: \cfrac{x}{6}=3

x=9

x=36

x=12

x=18

You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.

The final answer is x = 18.

You can check the answer by substituting the answer back into the original equation.

\cfrac{18}{6}=18 \div 6=3

2. Solve: \cfrac{x \, + \, 4}{2}=7

x=18

x=10

x=26

x=30

First, clear the fraction by multiplying both sides of the equation by 2.

Then subtract 4 from both sides.

The final answer is x = 10.

You can check the answer by substituting the answer back into the original equation.

\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7

3. Solve: \cfrac{x}{8}-5=1

x=40

x=64

x=48

x=56

First, add 5 to both sides of the equation.

Then multiply both sides of the equation by 8.

The final answer is x = 48.

You can check the answer by substituting the answer back into the original equation.

\cfrac{48}{8}-5=48 \div 8-5=1

4. Solve: \cfrac{3x \, + \, 2}{4}=2

x=4

x=16

x=12

x=2

First, multiply both sides of the equation by 4.

Next, subtract 2 from both sides.

Finally, divide both sides by 3.

The final answer is x = 2.

You can check the answer by substituting the answer back into the original equation.

\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2

5. Solve: \cfrac{4x}{7}-2=6

x=11

x=14

x=7

x=10

First, add 2 to both sides of the equation.

Then multiply both sides of the equation by 7.

Finally, divide both sides by 4.

The final answer is x = 14.

You can check the answer by substituting the answer back into the original equation.

\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6

6. Solve: \cfrac{42}{x}=7

x=5

x=294

x=7

x=6

You need to multiply both sides of the equation by x.

Then you divide both sides by 7.

The final answer is x = 6.

You can check the answer by substituting the answer back into the original equation.

\cfrac{42}{6}=42 \div 6=7

Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.

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[FREE] Common Core Practice Tests (3rd to 8th Grade)

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