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Adding fractions Subtracting fractions Dividing fractions Multiplying fractions Rational numbers Inverse 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 βbalancing methodβ 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?
In order to solve equations with fractions:
Use this worksheet to check your 7th grade and 8th grade studentsβ 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 studentsβ 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 .
The unknown is x.
Looking at the left hand side of the equation, the x is divided by 5.
\cfrac{x}{5}
2Apply the inverse operations, one at a time, to both sides of the equation.
The inverse of βdividing by 5 β is βmultiplying by 5 β.
You will multiply both sides of the equation by 5.
3When you have the variable on one side, you have the final answer.
The final answer is x=20.
4Check 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 βdividing by 3 β is βmultiplying by 3 β.
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
1. Solve: \cfrac{x}{6}=3
You will multiply both sides of the equation by 6, because the inverse of βdividing by 6 β is βmultiplying by 6 β.
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
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
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
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
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
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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