Solve by Substitution x+y=12 , x-y=4

$x + y = 12$ , $x - y = 4$

Step 1

Subtract $y$ from both sides of the equation.

$x = 12 - y$

$x - y = 4$

Step 2

Replace all occurrences of $x$ with $12 - y$ in each equation.

Replace all occurrences of $x$ in $x - y = 4$ with $12 - y$ .

$(12 - y) - y = 4$

$x = 12 - y$

Simplify the left side.

Subtract $y$ from $- y$ .

$12 - 2 y = 4$

$x = 12 - y$

$12 - 2 y = 4$

$x = 12 - y$

$12 - 2 y = 4$

$x = 12 - y$

Step 3

Solve for $y$ in $12 - 2 y = 4$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$- 2 y = 4 - 12$

$x = 12 - y$

Subtract $12$ from $4$ .

$- 2 y = - 8$

$x = 12 - y$

$- 2 y = - 8$

$x = 12 - y$

Divide each term in $- 2 y = - 8$ by $- 2$ and simplify.

Divide each term in $- 2 y = - 8$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2}$

$x = 12 - y$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2}$

$x = 12 - y$

Divide $y$ by $1$ .

$y = \frac{- 8}{- 2}$

$x = 12 - y$

$y = \frac{- 8}{- 2}$

$x = 12 - y$

$y = \frac{- 8}{- 2}$

$x = 12 - y$

Simplify the right side.

Divide $- 8$ by $- 2$ .

$y = 4$

$x = 12 - y$

$y = 4$

$x = 12 - y$

$y = 4$

$x = 12 - y$

$y = 4$

$x = 12 - y$

Step 4

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

Replace all occurrences of $y$ in $x = 12 - y$ with $4$ .

$x = 12 - (4)$

$y = 4$

Simplify the right side.

Simplify $12 - (4)$ .

Multiply $- 1$ by $4$ .

$x = 12 - 4$

$y = 4$

Subtract $4$ from $12$ .

$x = 8$

$y = 4$

$x = 8$

$y = 4$

$x = 8$

$y = 4$

$x = 8$

$y = 4$

Step 5

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

$(8, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(8, 4)$

Equation Form:

$x = 8, y = 4$

Step 7

Solve by Substitution x+y=12 , x-y=4

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