Solve for x (5x)/2-x=x/10-28/5

$\frac{5 x}{2} - x = \frac{x}{10} - \frac{28}{5}$

Step 1

Simplify $\frac{5 x}{2} - x$ .

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

$\frac{5 x}{2} - x \cdot \frac{2}{2} = \frac{x}{10} - \frac{28}{5}$

Simplify terms.

Combine $- x$ and $\frac{2}{2}$ .

$\frac{5 x}{2} + \frac{- x \cdot 2}{2} = \frac{x}{10} - \frac{28}{5}$

Combine the numerators over the common denominator.

$\frac{5 x - x \cdot 2}{2} = \frac{x}{10} - \frac{28}{5}$

Simplify the numerator.

Factor $x$ out of $5 x - x \cdot 2$ .

Factor $x$ out of $5 x$ .

$\frac{x \cdot 5 - x \cdot 2}{2} = \frac{x}{10} - \frac{28}{5}$

Factor $x$ out of $- x \cdot 2$ .

$\frac{x \cdot 5 + x (- 1 \cdot 2)}{2} = \frac{x}{10} - \frac{28}{5}$

Factor $x$ out of $x \cdot 5 + x (- 1 \cdot 2)$ .

$\frac{x (5 - 1 \cdot 2)}{2} = \frac{x}{10} - \frac{28}{5}$

Multiply $- 1$ by $2$ .

$\frac{x (5 - 2)}{2} = \frac{x}{10} - \frac{28}{5}$

Subtract $2$ from $5$ .

$\frac{x \cdot 3}{2} = \frac{x}{10} - \frac{28}{5}$

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

$\frac{3 x}{2} = \frac{x}{10} - \frac{28}{5}$

Step 2

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

Subtract $\frac{x}{10}$ from both sides of the equation.

$\frac{3 x}{2} - \frac{x}{10} = - \frac{28}{5}$

To write $\frac{3 x}{2}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{3 x}{2} \cdot \frac{5}{5} - \frac{x}{10} = - \frac{28}{5}$

Write each expression with a common denominator of $10$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 x}{2}$ by $\frac{5}{5}$ .

$\frac{3 x \cdot 5}{2 \cdot 5} - \frac{x}{10} = - \frac{28}{5}$

Multiply $2$ by $5$ .

$\frac{3 x \cdot 5}{10} - \frac{x}{10} = - \frac{28}{5}$

Combine the numerators over the common denominator.

$\frac{3 x \cdot 5 - x}{10} = - \frac{28}{5}$

Simplify the numerator.

Factor $x$ out of $3 x \cdot 5 - x$ .

Factor $x$ out of $3 x \cdot 5$ .

$\frac{x (3 \cdot 5) - x}{10} = - \frac{28}{5}$

Factor $x$ out of $- x$ .

$\frac{x (3 \cdot 5) + x \cdot - 1}{10} = - \frac{28}{5}$

Factor $x$ out of $x (3 \cdot 5) + x \cdot - 1$ .

$\frac{x (3 \cdot 5 - 1)}{10} = - \frac{28}{5}$

Multiply $3$ by $5$ .

$\frac{x (15 - 1)}{10} = - \frac{28}{5}$

Subtract $1$ from $15$ .

$\frac{x \cdot 14}{10} = - \frac{28}{5}$

Cancel the common factor of $14$ and $10$ .

Factor $2$ out of $x \cdot 14$ .

$\frac{2 (x \cdot 7)}{10} = - \frac{28}{5}$

Cancel the common factors.

Factor $2$ out of $10$ .

$\frac{2 (x \cdot 7)}{2 (5)} = - \frac{28}{5}$

Cancel the common factor.

$\frac{2 (x \cdot 7)}{2 \cdot 5} = - \frac{28}{5}$

Rewrite the expression.

$\frac{x \cdot 7}{5} = - \frac{28}{5}$

Move $7$ to the left of $x$ .

$\frac{7 x}{5} = - \frac{28}{5}$

Step 3

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

$7 x = - 28$

Step 4

Divide each term in $7 x = - 28$ by $7$ and simplify.

Divide each term in $7 x = - 28$ by $7$ .

$\frac{7 x}{7} = \frac{- 28}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 28}{7}$

Divide $x$ by $1$ .

$x = \frac{- 28}{7}$

Simplify the right side.

Divide $- 28$ by $7$ .

$x = - 4$

Solve for x (5x)/2-x=x/10-28/5

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