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

Card 0 of 76

Question

Solve for .

Answer

Subtract x from both sides of the second equation.

Divide both sides by to get $y=\frac{18-x}{5}$ .

Plug in y to the other equation.

$5x+15= 10*\frac{18-x}{5}$

Divide 10 by 5 to eliminate the fraction, yielding .

Distribute the 2 to get .

Add to each side, and subtract 15 from each side to get .

Divide both sides by 7 to get $\frac{x}{7}=\frac{21}{7}$ , which simplifies to .

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Question

Solve for :

Answer

can be simplified to become

Then, you can further simplify by adding 5 and to both sides to get .

Then, you can divide both sides by 5 to get .

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Question

Solve for :

Answer

First, you must multiply the left side of the equation using the distributive property.

This gives you .

Next, subtract from both sides to get .

Then, divide both sides by to get .

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Question

Solve for :

Answer

Combine like terms on the left side of the equation:

Use the distributive property to simplify the right side of the equation:

Next, move the 's to one side and the integers to the other side:

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Question

Solve for :

Answer

To solve for , you must first combine the 's on the right side of the equation. This will give you $\ 6x-1=9x+8$ .

Then, subtract and from both sides of the equation to get .

Finally, divide both sides by to get the solution .

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Question

Solve for :

Answer

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 .

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Question

Solve for :

Answer

First, combine like terms within the equation to get

.

Then, add and subtract from both sides to get

.

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

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Question

Solve for $x\textup:$

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

Answer

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:

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

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Question

Solve for $x\textup:$

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

Answer

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:

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

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Question

Solve for $x\textup:$

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

Answer

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 $3x\textup:$

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

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:

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

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Question

Solve for $x\textup:$

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

Answer

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:

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

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

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Question

Solve for $x\textup:$

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

Answer

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 $8x\textup:$

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

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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Question

Solve for $x\textup:$

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

Answer

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 $2x\textup:$

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

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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Question

Solve for $x\textup:$

Answer

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:

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

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Question

Solve for $x\textup:$

Answer

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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Question

Solve for $x\textup:$

Answer

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:

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

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Question

Solve for $x\textup:$

Answer

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:

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

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Question

Solve for $x\textup:$

Answer

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:

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

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Question

Solve for $x\textup:$

Answer

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:

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

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Question

Solve for .

Answer

Subtract x from both sides of the second equation.

Divide both sides by to get $y=\frac{18-x}{5}$ .

Plug in y to the other equation.

$5x+15= 10*\frac{18-x}{5}$

Divide 10 by 5 to eliminate the fraction, yielding .

Distribute the 2 to get .

Add to each side, and subtract 15 from each side to get .

Divide both sides by 7 to get $\frac{x}{7}=\frac{21}{7}$ , which simplifies to .

Compare your answer with the correct one above

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