Solve the System of Equations 2x+y=-10 -y=2x+10

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

Step 1

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

$y = - 10 - 2 x$

$- y = 2 x + 10$

Step 2

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

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

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

$y = - 10 - 2 x$

Simplify the left side.

Simplify $- (- 10 - 2 x)$ .

Apply the distributive property.

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

$y = - 10 - 2 x$

Multiply.

Multiply $- 1$ by $- 10$ .

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

$y = - 10 - 2 x$

Multiply $- 2$ by $- 1$ .

$10 + 2 x = 2 x + 10$

$y = - 10 - 2 x$

$10 + 2 x = 2 x + 10$

$y = - 10 - 2 x$

$10 + 2 x = 2 x + 10$

$y = - 10 - 2 x$

$10 + 2 x = 2 x + 10$

$y = - 10 - 2 x$

$10 + 2 x = 2 x + 10$

$y = - 10 - 2 x$

Step 3

Solve for $x$ in $10 + 2 x = 2 x + 10$ .

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

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

$10 + 2 x - 2 x = 10$

$y = - 10 - 2 x$

Combine the opposite terms in $10 + 2 x - 2 x$ .

Subtract $2 x$ from $2 x$ .

$10 + 0 = 10$

$y = - 10 - 2 x$

Add $10$ and $0$ .

$10 = 10$

$y = - 10 - 2 x$

$10 = 10$

$y = - 10 - 2 x$

$10 = 10$

$y = - 10 - 2 x$

Since $10 = 10$ , the equation will always be true.

Always true

$y = - 10 - 2 x$

Always true

$y = - 10 - 2 x$

Step 4

Remove any equations from the system that are always true.

$y = - 10 - 2 x$

Step 5

Solve the System of Equations 2x+y=-10 -y=2x+10

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