Solve by Substitution 3x+5y=3 x+2y=0

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

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

$x = - 2 y$

$3 x + 5 y = 3$

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

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

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

$x = - 2 y$

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

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Multiply $- 2$ by $3$ .

$- 6 y + 5 y = 3$

$x = - 2 y$

Add $- 6 y$ and $5 y$ .

$- y = 3$

$x = - 2 y$

$- y = 3$

$x = - 2 y$

$- y = 3$

$x = - 2 y$

Multiply each term in $- y = 3$ by $- 1$

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

$(- y) \cdot - 1 = 3 \cdot - 1$

$x = - 2 y$

Multiply $(- y) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 y = 3 \cdot - 1$

$x = - 2 y$

Multiply $y$ by $1$ .

$y = 3 \cdot - 1$

$x = - 2 y$

$y = 3 \cdot - 1$

$x = - 2 y$

Multiply $3$ by $- 1$ .

$y = - 3$

$x = - 2 y$

$y = - 3$

$x = - 2 y$

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

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

$x = - 2 \cdot - 3$

$y = - 3$

Multiply $- 2$ by $- 3$ .

$x = 6$

$y = - 3$

$x = 6$

$y = - 3$

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

$(6, - 3)$

The result can be shown in multiple forms.

Point Form:

$(6, - 3)$

Equation Form:

$x = 6, y = - 3$

Solve by Substitution 3x+5y=3 x+2y=0

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