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Common Core: 8th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 8th Grade Math

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All Common Core: 8th Grade Math Resources

7 Diagnostic Tests 75 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

$\frac{x+17}{3}=5$

Possible Answers:

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we need to multiply each side by $3\textup:$

$3\left ( \frac{x+17}{3}\right )=5(3)$

x+17=15

Next, we need to subtract from each side:

$\frac{\begin{array}[b]{r}x+17=15\\ -17-17\end{array}}{\\\\x=-2}$

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Example Question #1 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

$\frac{x+28}{3x}=5$

Possible Answers:

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we need to multiply each side by $3x\textup:$

$3x\left ( \frac{x+28}{3x}\right )=5(3x)$

x+28=15x

Next, we need to combine like terms, so we subtract from both sides:

$\frac{\begin{array}[b]{r}x+28=15x\\ -x\ \ \ \ \ \ \ \ \ \ -x\end{array}}{\\\\28=14x}$

Finally, we can divide by both sides:

$\frac{28}{14}=\frac{14x}{14}$

2=x

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Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

$\frac{-2x+5}{6x}=3$

Possible Answers:

$\frac{1}{5}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we need to multiply each side by $6x\textup:$

$6x\left ( \frac{-2x+5}{6x}\right )=3(6x)$

-2x+5=18x

Next, we need to combine like terms, so we add to both sides:

$\frac{\begin{array}[b]{r}-2x+5=18x\\ +2x\ \ \ \ \ \ \ +2x\end{array}}{\\\\5=20x}$

Finally, we can divide by both sides:

$\frac{5}{20}=\frac{20x}{20}$

$x=\frac{5}{20}=\frac{1}{4}$

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Example Question #12 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

$\frac{-4x+10}{8x}=2$

Possible Answers:

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we need to multiply each side by $8x\textup:$

$8x\left ( \frac{-4x+10}{8x}\right )=2(8x)$

-4x+10=16x

Next, we need to combine like terms, so we add to both sides:

$\frac{\begin{array}[b]{r}-4x+10=16x\\ +4x \ \ \ \ \ \ \ \ +4x\end{array}}{\\\\10=20x}$

Finally, we can divide by both sides:

$\frac{10}{20}=\frac{20x}{20}$

$x=\frac{10}{20}=\frac{1}{2}$

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Example Question #141 : Expressions & Equations

Solve for $x\textup:$

$\frac{x-7}{2x}=5+20$

Possible Answers:

$\frac{1}{7}$

$-\frac{1}{7}$

Correct answer:

$-\frac{1}{7}$

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we need to multiply each side by $2x\textup:$

$2x\left ( \frac{x-7}{2x}\right )=2x(5+20)$

x-7=10x+40x

x-7=50x

Next, we need to combine like terms, so we subtract from both sides:

$\frac{\begin{array}[b]{r}x-7=50x\\ -x \ \ \ \ \ \ \ \ -x\end{array}}{\\\\-7=49x}$

Finally, we can divide by both sides:

$\frac{-7}{49}=\frac{49x}{49}$

$x=-\frac{7}{49}=-\frac{1}{7}$

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Example Question #14 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

14x+9=5x-36

Possible Answers:

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, we want to combine like terms. Let's start by moving the values to one side:

$\frac{\begin{array}[b]{r}14x+9=5x-36\\ -5x\ \ \ \ \ \ \ -5x\ \ \ \ \ \ \ \end{array}}{\\\\9x+9=-36}$

Next, we can subtract from both sides:

$\frac{\begin{array}[b]{r}9x+9=-36\\ -9 -9\ \end{array}}{\\\\9x=45}$

Finally, we divide from both sides:

$\frac{9x}{9}=\frac{45}{9}$

x=5

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Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

12=3(x+6)

Possible Answers:

$\frac{2}{3}$

$\frac{1}{3}$

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, the first thing we want to do is distribute the :

12=3(x+6)

12=3x+18

Next, we can subtract from both sides:

$\frac{\begin{array}[b]{r}12=3x+18\\ -18\ \ \ \ \ \ -18\ \end{array}}{\\\\-6=3x}$

Finally, we divide from both sides:

$\frac{-6}{3}=\frac{3x}{3}$

x=-2

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Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

32=4(x+40)

Possible Answers:

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, the first thing we want to do is distribute the :

32=4(x+40)

32=4x+160

Next, we can subtract 160 from both sides:

$\frac{\begin{array}[b]{r}32=4x+160\\ -160\ \ \ \ \ \ \ \ \ \ \-160 \end{array}}{\\\\-128=4x}$

Finally, we divide from both sides:

$\frac{-128}{4}=\frac{4x}{4}$

x=-32

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Example Question #182 : Grade 8

Solve for $x\textup:$

4=2(x+16)

Possible Answers:

$-\frac{1}{7}$

$\frac{1}{7}$

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, the first thing we want to do is distribute the :

4=2(x+16)

4=2x+32

Next, we can subtract from both sides:

$\frac{\begin{array}[b]{r}4=2x+32\\ -32\ \ \ \ \ \ \ -32 \end{array}}{\\\\-28=2x}$

Finally, we divide from both sides:

$\frac{-28}{2}=\frac{2x}{2}$

x=-14

Report an Error

Example Question #12 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Solve for $x\textup:$

100=5(2x+4)

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{8}$

Correct answer:

Explanation:

In order to solve for , we need to isolate the to one side of the equation.

For this problem, the first thing we want to do is distribute the :

100=5(2x+4)

100=10x+20

Next, we can subtract from both sides:

$\frac{\begin{array}[b]{r}100=10x+20\\ -20\ \ \ \ \ \ \ \ -20 \end{array}}{\\\\80=10x}$

Finally, we divide from both sides:

$\frac{80}{10}=\frac{10x}{10}$

x=8

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All Common Core: 8th Grade Math Resources

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Common Core: 8th Grade Math : Expressions & Equations

Study concepts, example questions & explanations for Common Core: 8th Grade Math

All Common Core: 8th Grade Math Resources

Example Questions

Example Question #5 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #1 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #12 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #141 : Expressions & Equations

Example Question #14 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #11 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

Example Question #182 : Grade 8

Example Question #12 : Solve Linear Equations With Rational Number Coefficients: Ccss.Math.Content.8.Ee.C.7b

All Common Core: 8th Grade Math Resources

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