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

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

Step 1

Add $2 y$ to both sides of the equation.

$x = 10 + 2 y$

$x + 2 y = - 2$

Step 2

Replace all occurrences of $x$ with $10 + 2 y$ in each equation.

Replace all occurrences of $x$ in $x + 2 y = - 2$ with $10 + 2 y$ .

$(10 + 2 y) + 2 y = - 2$

$x = 10 + 2 y$

Simplify the left side.

Add $2 y$ and $2 y$ .

$10 + 4 y = - 2$

$x = 10 + 2 y$

$10 + 4 y = - 2$

$x = 10 + 2 y$

$10 + 4 y = - 2$

$x = 10 + 2 y$

Step 3

Solve for $y$ in $10 + 4 y = - 2$ .

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

Subtract $10$ from both sides of the equation.

$4 y = - 2 - 10$

$x = 10 + 2 y$

Subtract $10$ from $- 2$ .

$4 y = - 12$

$x = 10 + 2 y$

$4 y = - 12$

$x = 10 + 2 y$

Divide each term in $4 y = - 12$ by $4$ and simplify.

Divide each term in $4 y = - 12$ by $4$ .

$\frac{4 y}{4} = \frac{- 12}{4}$

$x = 10 + 2 y$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{- 12}{4}$

$x = 10 + 2 y$

Divide $y$ by $1$ .

$y = \frac{- 12}{4}$

$x = 10 + 2 y$

$y = \frac{- 12}{4}$

$x = 10 + 2 y$

$y = \frac{- 12}{4}$

$x = 10 + 2 y$

Simplify the right side.

Divide $- 12$ by $4$ .

$y = - 3$

$x = 10 + 2 y$

$y = - 3$

$x = 10 + 2 y$

$y = - 3$

$x = 10 + 2 y$

$y = - 3$

$x = 10 + 2 y$

Step 4

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

Replace all occurrences of $y$ in $x = 10 + 2 y$ with $- 3$ .

$x = 10 + 2 (- 3)$

$y = - 3$

Simplify the right side.

Simplify $10 + 2 (- 3)$ .

Multiply $2$ by $- 3$ .

$x = 10 - 6$

$y = - 3$

Subtract $6$ from $10$ .

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

Step 5

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

$(4, - 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 3)$

Equation Form:

$x = 4, y = - 3$

Step 7

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

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