Pre-Algebra Examples

$\frac{1}{2} r + 3 (\frac{3}{4} r - 2) = \frac{1}{4} r + 8$

Step 1

Simplify $\frac{1}{2} r + 3 (\frac{3}{4} r - 2)$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

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

$\frac{r}{2} + 3 (\frac{3}{4} r - 2) = \frac{1}{4} r + 8$

Step 1.1.2

Combine $\frac{3}{4}$ and $r$ .

$\frac{r}{2} + 3 (\frac{3 r}{4} - 2) = \frac{1}{4} r + 8$

Step 1.1.3

Apply the distributive property.

$\frac{r}{2} + 3 \frac{3 r}{4} + 3 \cdot - 2 = \frac{1}{4} r + 8$

Step 1.1.4

Multiply $3 \frac{3 r}{4}$ .

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Step 1.1.4.1

Combine $3$ and $\frac{3 r}{4}$ .

$\frac{r}{2} + \frac{3 (3 r)}{4} + 3 \cdot - 2 = \frac{1}{4} r + 8$

Step 1.1.4.2

Multiply $3$ by $3$ .

$\frac{r}{2} + \frac{9 r}{4} + 3 \cdot - 2 = \frac{1}{4} r + 8$

Step 1.1.5

Multiply $3$ by $- 2$ .

$\frac{r}{2} + \frac{9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.2

To write $\frac{r}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{r}{2} \cdot \frac{2}{2} + \frac{9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{r}{2}$ by $\frac{2}{2}$ .

$\frac{r \cdot 2}{2 \cdot 2} + \frac{9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.3.2

Multiply $2$ by $2$ .

$\frac{r \cdot 2}{4} + \frac{9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.4

Combine the numerators over the common denominator.

$\frac{r \cdot 2 + 9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.5

Add $r \cdot 2$ and $9 r$ .

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Step 1.5.1

Reorder $r$ and $2$ .

$\frac{2 \cdot r + 9 r}{4} - 6 = \frac{1}{4} r + 8$

Step 1.5.2

Add $2 \cdot r$ and $9 r$ .

$\frac{11 \cdot r}{4} - 6 = \frac{1}{4} r + 8$

$\frac{11 r}{4} - 6 = \frac{1}{4} r + 8$

Step 2

Combine $\frac{1}{4}$ and $r$ .

$\frac{11 r}{4} - 6 = \frac{r}{4} + 8$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$r = \frac{28}{5}$

Step 4