Solve for x 1/5((2-x)/2)-3/10(1-x)/2=17/20

$\frac{1}{5} (\frac{2 - x}{2}) - \frac{3}{10} \cdot \frac{1 - x}{2} = \frac{17}{20}$

Simplify $\frac{1}{5} \cdot \frac{2 - x}{2} - \frac{3}{10} \cdot \frac{1 - x}{2}$ .

Simplify each term.

Multiply $\frac{1}{5} \cdot \frac{2 - x}{2}$ .

Multiply $\frac{1}{5}$ by $\frac{2 - x}{2}$ .

$\frac{2 - x}{5 \cdot 2} - \frac{3}{10} \cdot \frac{1 - x}{2} = \frac{17}{20}$

Multiply $5$ by $2$ .

$\frac{2 - x}{10} - \frac{3}{10} \cdot \frac{1 - x}{2} = \frac{17}{20}$

Multiply $- \frac{3}{10} \cdot \frac{1 - x}{2}$ .

Multiply $\frac{1 - x}{2}$ by $\frac{3}{10}$ .

$\frac{2 - x}{10} - \frac{(1 - x) \cdot 3}{2 \cdot 10} = \frac{17}{20}$

Multiply $2$ by $10$ .

$\frac{2 - x}{10} - \frac{(1 - x) \cdot 3}{20} = \frac{17}{20}$

Move $3$ to the left of $1 - x$ .

$\frac{2 - x}{10} - \frac{3 (1 - x)}{20} = \frac{17}{20}$

To write $\frac{2 - x}{10}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{2 - x}{10} \cdot \frac{2}{2} - \frac{3 (1 - x)}{20} = \frac{17}{20}$

Write each expression with a common denominator of $20$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 - x}{10}$ by $\frac{2}{2}$ .

$\frac{(2 - x) \cdot 2}{10 \cdot 2} - \frac{3 (1 - x)}{20} = \frac{17}{20}$

Multiply $10$ by $2$ .

$\frac{(2 - x) \cdot 2}{20} - \frac{3 (1 - x)}{20} = \frac{17}{20}$

Combine the numerators over the common denominator.

$\frac{(2 - x) \cdot 2 - 3 (1 - x)}{20} = \frac{17}{20}$

Simplify the numerator.

Apply the distributive property.

$\frac{2 \cdot 2 - x \cdot 2 - 3 (1 - x)}{20} = \frac{17}{20}$

Multiply $2$ by $2$ .

$\frac{4 - x \cdot 2 - 3 (1 - x)}{20} = \frac{17}{20}$

Multiply $2$ by $- 1$ .

$\frac{4 - 2 x - 3 (1 - x)}{20} = \frac{17}{20}$

Apply the distributive property.

$\frac{4 - 2 x - 3 \cdot 1 - 3 (- x)}{20} = \frac{17}{20}$

Multiply $- 3$ by $1$ .

$\frac{4 - 2 x - 3 - 3 (- x)}{20} = \frac{17}{20}$

Multiply $- 1$ by $- 3$ .

$\frac{4 - 2 x - 3 + 3 x}{20} = \frac{17}{20}$

Subtract $3$ from $4$ .

$\frac{- 2 x + 1 + 3 x}{20} = \frac{17}{20}$

Add $- 2 x$ and $3 x$ .

$\frac{x + 1}{20} = \frac{17}{20}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x + 1 = 17$

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$x = 17 - 1$

Subtract $1$ from $17$ .

$x = 16$

Solve for x 1/5((2-x)/2)-3/10(1-x)/2=17/20

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