Solve by Substitution 2x=y-10 2x+7=2y

$2 x = y - 10$ $2 x + 7 = 2 y$

Step 1

Divide each term in $2 x = y - 10$ by $2$ and simplify.

Divide each term in $2 x = y - 10$ by $2$ .

$\frac{2 x}{2} = \frac{y}{2} + \frac{- 10}{2}$

$2 x + 7 = 2 y$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{y}{2} + \frac{- 10}{2}$

$2 x + 7 = 2 y$

Divide $x$ by $1$ .

$x = \frac{y}{2} + \frac{- 10}{2}$

$2 x + 7 = 2 y$

$x = \frac{y}{2} + \frac{- 10}{2}$

$2 x + 7 = 2 y$

$x = \frac{y}{2} + \frac{- 10}{2}$

$2 x + 7 = 2 y$

Simplify the right side.

Divide $- 10$ by $2$ .

$x = \frac{y}{2} - 5$

$2 x + 7 = 2 y$

$x = \frac{y}{2} - 5$

$2 x + 7 = 2 y$

$x = \frac{y}{2} - 5$

$2 x + 7 = 2 y$

Step 2

Replace all occurrences of $x$ with $\frac{y}{2} - 5$ in each equation.

Replace all occurrences of $x$ in $2 x + 7 = 2 y$ with $\frac{y}{2} - 5$ .

$2 (\frac{y}{2} - 5) + 7 = 2 y$

$x = \frac{y}{2} - 5$

Simplify the left side.

Simplify $2 (\frac{y}{2} - 5) + 7$ .

Simplify each term.

Apply the distributive property.

$2 (\frac{y}{2}) + 2 \cdot - 5 + 7 = 2 y$

$x = \frac{y}{2} - 5$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 (\frac{y}{2}) + 2 \cdot - 5 + 7 = 2 y$

$x = \frac{y}{2} - 5$

Rewrite the expression.

$y + 2 \cdot - 5 + 7 = 2 y$

$x = \frac{y}{2} - 5$

$y + 2 \cdot - 5 + 7 = 2 y$

$x = \frac{y}{2} - 5$

Multiply $2$ by $- 5$ .

$y - 10 + 7 = 2 y$

$x = \frac{y}{2} - 5$

$y - 10 + 7 = 2 y$

$x = \frac{y}{2} - 5$

Add $- 10$ and $7$ .

$y - 3 = 2 y$

$x = \frac{y}{2} - 5$

$y - 3 = 2 y$

$x = \frac{y}{2} - 5$

$y - 3 = 2 y$

$x = \frac{y}{2} - 5$

$y - 3 = 2 y$

$x = \frac{y}{2} - 5$

Step 3

Solve for $y$ in $y - 3 = 2 y$ .

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

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

$y - 3 - 2 y = 0$

$x = \frac{y}{2} - 5$

Subtract $2 y$ from $y$ .

$- y - 3 = 0$

$x = \frac{y}{2} - 5$

$- y - 3 = 0$

$x = \frac{y}{2} - 5$

Add $3$ to both sides of the equation.

$- y = 3$

$x = \frac{y}{2} - 5$

Divide each term in $- y = 3$ by $- 1$ and simplify.

Divide each term in $- y = 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{3}{- 1}$

$x = \frac{y}{2} - 5$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{3}{- 1}$

$x = \frac{y}{2} - 5$

Divide $y$ by $1$ .

$y = \frac{3}{- 1}$

$x = \frac{y}{2} - 5$

$y = \frac{3}{- 1}$

$x = \frac{y}{2} - 5$

Simplify the right side.

Divide $3$ by $- 1$ .

$y = - 3$

$x = \frac{y}{2} - 5$

$y = - 3$

$x = \frac{y}{2} - 5$

$y = - 3$

$x = \frac{y}{2} - 5$

$y = - 3$

$x = \frac{y}{2} - 5$

Step 4

Replace all occurrences of $y$ with $- 3$ in each equation.

Replace all occurrences of $y$ in $x = \frac{y}{2} - 5$ with $- 3$ .

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

$y = - 3$

Simplify the right side.

Simplify $\frac{- 3}{2} - 5$ .

Move the negative in front of the fraction.

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

$y = - 3$

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

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

$y = - 3$

Combine $- 5$ and $\frac{2}{2}$ .

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

$y = - 3$

Combine the numerators over the common denominator.

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

$y = - 3$

Simplify the numerator.

Multiply $- 5$ by $2$ .

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

$y = - 3$

Subtract $10$ from $- 3$ .

$x = \frac{- 13}{2}$

$y = - 3$

$x = \frac{- 13}{2}$

$y = - 3$

Move the negative in front of the fraction.

$x = - \frac{13}{2}$

$y = - 3$

$x = - \frac{13}{2}$

$y = - 3$

$x = - \frac{13}{2}$

$y = - 3$

$x = - \frac{13}{2}$

$y = - 3$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- \frac{13}{2}, - 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{13}{2}, - 3)$

Equation Form:

$x = - \frac{13}{2}, y = - 3$

Step 7

Solve by Substitution 2x=y-10 2x+7=2y

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