Pre-Algebra Examples

$49 = \frac{1}{2} \cdot (h (4 + 3))$

Step 1

Rewrite the equation as $\frac{1}{2} \cdot (h (4 + 3)) = 49$ .

$\frac{1}{2} \cdot (h (4 + 3)) = 49$

Step 2

Multiply both sides of the equation by $\frac{1}{\frac{1}{2} (4 + 3)}$ .

$\frac{1}{\frac{1}{2} (4 + 3)} (\frac{1}{2} \cdot (h (4 + 3))) = \frac{1}{\frac{1}{2} (4 + 3)} \cdot 49$

Step 3

Simplify both sides of the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Cancel the common factor of $\frac{1}{2} \cdot (4 + 3)$ .

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Step 3.1.1.1

Factor $\frac{1}{2} \cdot (4 + 3)$ out of $\frac{1}{2} \cdot (h (4 + 3))$ .

$\frac{1}{\frac{1}{2} (4 + 3)} (\frac{1}{2} (4 + 3) (h)) = \frac{1}{\frac{1}{2} (4 + 3)} \cdot 49$

Step 3.1.1.2

Cancel the common factor.

$\frac{1}{\frac{1}{2} (4 + 3)} (\frac{1}{2} (4 + 3) h) = \frac{1}{\frac{1}{2} (4 + 3)} \cdot 49$

Step 3.1.1.3

Rewrite the expression.

$h = \frac{1}{\frac{1}{2} (4 + 3)} \cdot 49$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\frac{1}{\frac{1}{2} (4 + 3)} \cdot 49$ .

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Step 3.2.1.1

Add $4$ and $3$ .

$h = \frac{1}{\frac{1}{2} \cdot 7} \cdot 49$

Step 3.2.1.2

Combine $\frac{1}{2}$ and $7$ .

$h = \frac{1}{\frac{7}{2}} \cdot 49$

Step 3.2.1.3

Multiply the numerator by the reciprocal of the denominator.

$h = 1 (\frac{2}{7}) \cdot 49$

Step 3.2.1.4

Multiply $\frac{2}{7}$ by $1$ .

$h = \frac{2}{7} \cdot 49$

Step 3.2.1.5

Cancel the common factor of $7$ .

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Step 3.2.1.5.1

Factor $7$ out of $49$ .

$h = \frac{2}{7} \cdot (7 (7))$

Step 3.2.1.5.2

Cancel the common factor.

$h = \frac{2}{7} \cdot (7 \cdot 7)$

Step 3.2.1.5.3

Rewrite the expression.

$h = 2 \cdot 7$

Step 3.2.1.6

Multiply $2$ by $7$ .

$h = 14$