Solve by Substitution x+y=4 , 2x+3y=0

$x + y = 4$ , $2 x + 3 y = 0$

Step 1

Subtract $y$ from both sides of the equation.

$x = 4 - y$

$2 x + 3 y = 0$

Step 2

Replace all occurrences of $x$ with $4 - y$ in each equation.

Replace all occurrences of $x$ in $2 x + 3 y = 0$ with $4 - y$ .

$2 (4 - y) + 3 y = 0$

$x = 4 - y$

Simplify the left side.

Simplify $2 (4 - y) + 3 y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 4 + 2 (- y) + 3 y = 0$

$x = 4 - y$

Multiply $2$ by $4$ .

$8 + 2 (- y) + 3 y = 0$

$x = 4 - y$

Multiply $- 1$ by $2$ .

$8 - 2 y + 3 y = 0$

$x = 4 - y$

$8 - 2 y + 3 y = 0$

$x = 4 - y$

Add $- 2 y$ and $3 y$ .

$8 + y = 0$

$x = 4 - y$

$8 + y = 0$

$x = 4 - y$

$8 + y = 0$

$x = 4 - y$

$8 + y = 0$

$x = 4 - y$

Step 3

Subtract $8$ from both sides of the equation.

$y = - 8$

$x = 4 - y$

Step 4

Replace all occurrences of $y$ with $- 8$ in each equation.

Replace all occurrences of $y$ in $x = 4 - y$ with $- 8$ .

$x = 4 - (- 8)$

$y = - 8$

Simplify the right side.

Simplify $4 - (- 8)$ .

Multiply $- 1$ by $- 8$ .

$x = 4 + 8$

$y = - 8$

Add $4$ and $8$ .

$x = 12$

$y = - 8$

$x = 12$

$y = - 8$

$x = 12$

$y = - 8$

$x = 12$

$y = - 8$

Step 5

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

$(12, - 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(12, - 8)$

Equation Form:

$x = 12, y = - 8$

Step 7

Solve by Substitution x+y=4 , 2x+3y=0

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