Pre-Algebra Examples

5 (2 k + 10) = 40

$5 (2 k + 10) = 40$

Step 1

Divide each term in

5 (2 k + 10) = 40

$5 (2 k + 10) = 40$ by

5

$5$ and simplify.

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Step 1.1

Divide each term in

5 (2 k + 10) = 40

$5 (2 k + 10) = 40$ by

5

$5$ .

\frac{5 (2 k + 10)}{5} = \frac{40}{5}

$\frac{5 (2 k + 10)}{5} = \frac{40}{5}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of

5

$5$ .

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Step 1.2.1.1

Cancel the common factor.

\frac{5 (2 k + 10)}{5} = \frac{40}{5}

$\frac{5 (2 k + 10)}{5} = \frac{40}{5}$

Step 1.2.1.2

Divide

2 k + 10

$2 k + 10$ by

1

$1$ .

2 k + 10 = \frac{40}{5}

$2 k + 10 = \frac{40}{5}$

2 k + 10 = \frac{40}{5}

$2 k + 10 = \frac{40}{5}$

2 k + 10 = \frac{40}{5}

$2 k + 10 = \frac{40}{5}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide

40

$40$ by

5

$5$ .

2 k + 10 = 8

$2 k + 10 = 8$

2 k + 10 = 8

$2 k + 10 = 8$

2 k + 10 = 8

$2 k + 10 = 8$

Step 2

Move all terms not containing

k

$k$ to the right side of the equation.

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Step 2.1

Subtract

10

$10$ from both sides of the equation.

2 k = 8 - 10

$2 k = 8 - 10$

Step 2.2

Subtract

10

$10$ from

8

$8$ .

2 k = - 2

$2 k = - 2$

2 k = - 2

$2 k = - 2$

Step 3

Divide each term in

2 k = - 2

$2 k = - 2$ by

2

$2$ and simplify.

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Step 3.1

Divide each term in

2 k = - 2

$2 k = - 2$ by

2

$2$ .

\frac{2 k}{2} = \frac{- 2}{2}

$\frac{2 k}{2} = \frac{- 2}{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of

2

$2$ .

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Step 3.2.1.1

Cancel the common factor.

\frac{2 k}{2} = \frac{- 2}{2}

$\frac{2 k}{2} = \frac{- 2}{2}$

Step 3.2.1.2

Divide

k

$k$ by

1

$1$ .

k = \frac{- 2}{2}

$k = \frac{- 2}{2}$

k = \frac{- 2}{2}

$k = \frac{- 2}{2}$

k = \frac{- 2}{2}

$k = \frac{- 2}{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide

- 2

$- 2$ by

2

$2$ .

k = - 1

$k = - 1$

k = - 1

$k = - 1$

k = - 1

$k = - 1$

5 (2 k + 10) = 40

$5 (2 k + 10) = 40$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$