Solve the System of @WORD x+y=0 , 2x-y=5

$x + y = 0$ , $2 x - y = 5$

Step 1

Subtract $y$ from both sides of the equation.

$x = - y$

$2 x - y = 5$

Step 2

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

Replace all occurrences of $x$ in $2 x - y = 5$ with $- y$ .

$2 (- y) - y = 5$

$x = - y$

Simplify the left side.

Simplify $2 (- y) - y$ .

Multiply $- 1$ by $2$ .

$- 2 y - y = 5$

$x = - y$

Subtract $y$ from $- 2 y$ .

$- 3 y = 5$

$x = - y$

$- 3 y = 5$

$x = - y$

$- 3 y = 5$

$x = - y$

$- 3 y = 5$

$x = - y$

Step 3

Divide each term in $- 3 y = 5$ by $- 3$ and simplify.

Divide each term in $- 3 y = 5$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{5}{- 3}$

$x = - y$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{5}{- 3}$

$x = - y$

Divide $y$ by $1$ .

$y = \frac{5}{- 3}$

$x = - y$

$y = \frac{5}{- 3}$

$x = - y$

$y = \frac{5}{- 3}$

$x = - y$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{5}{3}$

$x = - y$

$y = - \frac{5}{3}$

$x = - y$

$y = - \frac{5}{3}$

$x = - y$

Step 4

Replace all occurrences of $y$ with $- \frac{5}{3}$ in each equation.

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

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

$y = - \frac{5}{3}$

Simplify the right side.

Multiply $- (- \frac{5}{3})$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{5}{3})$

$y = - \frac{5}{3}$

Multiply $\frac{5}{3}$ by $1$ .

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

Step 5

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

$(\frac{5}{3}, - \frac{5}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{5}{3}, - \frac{5}{3})$

Equation Form:

$x = \frac{5}{3}, y = - \frac{5}{3}$

Step 7

Solve the System of @WORD x+y=0 , 2x-y=5

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