Solve Linear Equations with Rational Number Coefficients: CCSS.Math.Content.8.EE.C.7b

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8th Grade Math › Solve Linear Equations with Rational Number Coefficients: CCSS.Math.Content.8.EE.C.7b

Questions 1 - 10

1

Solve for $x\textup:$

$\frac{1}{3}$

$\frac{2}{3}$

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 :

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:

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

2

Solve for $x\textup:$

$\frac{x+24}{7}=4$

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 $7\textup:$

$7\left ( \frac{x+24}{7}\right )=4(7)$

Next, we need to subtract from each side:

$\frac{\begin{array}[b]{r}x+24=28\ -24-24\end{array}}{\\x=4}$

3

Solve for $x\textup:$

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:

$\frac{\begin{array}[b]{r}3x+6=2x+23\ -2x\ \ \ \ \ -2x\ \ \ \ \ \ \ \end{array}}{\\x+6=23}$

Next, we can subtract from both sides:

$\frac{\begin{array}[b]{r}x+6=23\ -6 -6\ \end{array}}{\\x=17}$

4

Solve for $x\textup:$

$\frac{1}{2}$

$-\frac{1}{8}$

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 :

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:

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

5

Solve for :

Explanation

First. combine like terms to get

.

Then, add and subtract from both sides to separate the terms.

This gives you .

Finally, divide both sides by to get a solution of .

6

Solve for $x\textup:$

$\frac{1}{7}$

$-\frac{1}{7}$

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 :

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:

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

7

Solve for $x\textup:$

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

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 $3\textup:$

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

Next, we need to subtract from each side:

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

8

Solve for $x\textup:$

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

$\frac{1}{4}$

$\frac{1}{5}$

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 $6x\textup:$

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

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:

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

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

9

Solve for $x\textup:$

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 :

Next, we can subtract from both sides:

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

Finally, we divide from both sides:

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

10

Solve for $x\textup:$

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:

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

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