Algebra Examples

$\frac{1}{6} (12 x + 48) = - \frac{1}{8} (8 x - 16) + 3 x$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- \frac{1}{8} \cdot (8 x - 16) + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2

Simplify $- \frac{1}{8} \cdot (8 x - 16) + 3 x$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$- \frac{1}{8} (8 x) - \frac{1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.2

Cancel the common factor of $8$ .

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Step 2.1.2.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$\frac{- 1}{8} (8 x) - \frac{1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.2.2

Factor $8$ out of $8 x$ .

$\frac{- 1}{8} (8 (x)) - \frac{1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.2.3

Cancel the common factor.

$\frac{- 1}{8} (8 x) - \frac{1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.2.4

Rewrite the expression.

$- 1 x - \frac{1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.3

Cancel the common factor of $8$ .

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Step 2.1.3.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$- 1 x + \frac{- 1}{8} \cdot - 16 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.3.2

Factor $8$ out of $- 16$ .

$- 1 x + \frac{- 1}{8} \cdot (8 (- 2)) + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.3.3

Cancel the common factor.

$- 1 x + \frac{- 1}{8} \cdot (8 \cdot - 2) + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.3.4

Rewrite the expression.

$- 1 x - 1 \cdot - 2 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.4

Multiply $- 1$ by $- 2$ .

$- 1 x + 2 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.1.5

Rewrite $- 1 x$ as $- x$ .

$- x + 2 + 3 x = \frac{1}{6} \cdot (12 x + 48)$

Step 2.2

Add $- x$ and $3 x$ .

$2 x + 2 = \frac{1}{6} \cdot (12 x + 48)$

Step 3

Simplify $\frac{1}{6} \cdot (12 x + 48)$ .

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Step 3.1

Apply the distributive property.

$2 x + 2 = \frac{1}{6} (12 x) + \frac{1}{6} \cdot 48$

Step 3.2

Cancel the common factor of $6$ .

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Step 3.2.1

Factor $6$ out of $12 x$ .

$2 x + 2 = \frac{1}{6} (6 (2 x)) + \frac{1}{6} \cdot 48$

Step 3.2.2

Cancel the common factor.

$2 x + 2 = \frac{1}{6} (6 (2 x)) + \frac{1}{6} \cdot 48$

Step 3.2.3

Rewrite the expression.

$2 x + 2 = 2 x + \frac{1}{6} \cdot 48$

Step 3.3

Cancel the common factor of $6$ .

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Step 3.3.1

Factor $6$ out of $48$ .

$2 x + 2 = 2 x + \frac{1}{6} \cdot (6 (8))$

Step 3.3.2

Cancel the common factor.

$2 x + 2 = 2 x + \frac{1}{6} \cdot (6 \cdot 8)$

Step 3.3.3

Rewrite the expression.

$2 x + 2 = 2 x + 8$

Step 4

Move all terms containing $x$ to the left side of the equation.

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Step 4.1

Subtract $2 x$ from both sides of the equation.

$2 x + 2 - 2 x = 8$

Step 4.2

Combine the opposite terms in $2 x + 2 - 2 x$ .

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Step 4.2.1

Subtract $2 x$ from $2 x$ .

$0 + 2 = 8$

Step 4.2.2

Add $0$ and $2$ .

$2 = 8$

Step 5

Since $2 \neq 8$ , there are no solutions.

No solution