Solve the System of Equations 2x+y=0 5x+3y=2

$2 x + y = 0$ $5 x + 3 y = 2$

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

$y = - 2 x$

$5 x + 3 y = 2$

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

Replace all occurrences of $y$ in $5 x + 3 y = 2$ with $- 2 x$ .

$5 x + 3 (- 2 x) = 2$

$y = - 2 x$

Simplify the left side.

Simplify $5 x + 3 (- 2 x)$ .

Multiply $- 2$ by $3$ .

$5 x - 6 x = 2$

$y = - 2 x$

Subtract $6 x$ from $5 x$ .

$- x = 2$

$y = - 2 x$

$- x = 2$

$y = - 2 x$

$- x = 2$

$y = - 2 x$

$- x = 2$

$y = - 2 x$

Divide each term in $- x = 2$ by $- 1$ and simplify.

Divide each term in $- x = 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{2}{- 1}$

$y = - 2 x$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{2}{- 1}$

$y = - 2 x$

Divide $x$ by $1$ .

$x = \frac{2}{- 1}$

$y = - 2 x$

$x = \frac{2}{- 1}$

$y = - 2 x$

Simplify the right side.

Divide $2$ by $- 1$ .

$x = - 2$

$y = - 2 x$

$x = - 2$

$y = - 2 x$

$x = - 2$

$y = - 2 x$

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

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

$y = - 2 \cdot - 2$

$x = - 2$

Simplify the right side.

Multiply $- 2$ by $- 2$ .

$y = 4$

$x = - 2$

$y = 4$

$x = - 2$

$y = 4$

$x = - 2$

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

$(- 2, 4)$

The result can be shown in multiple forms.

Point Form:

$(- 2, 4)$

Equation Form:

$x = - 2, y = 4$

Solve the System of Equations 2x+y=0 5x+3y=2

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