Solve by Substitution x+3y=12 5y=x+4

$x + 3 y = 12$ $5 y = x + 4$

Step 1

Subtract $3 y$ from both sides of the equation.

$x = 12 - 3 y$

$5 y = x + 4$

Step 2

Replace all occurrences of $x$ with $12 - 3 y$ in each equation.

Replace all occurrences of $x$ in $5 y = x + 4$ with $12 - 3 y$ .

$5 y = (12 - 3 y) + 4$

$x = 12 - 3 y$

Simplify the right side.

Add $12$ and $4$ .

$5 y = - 3 y + 16$

$x = 12 - 3 y$

$5 y = - 3 y + 16$

$x = 12 - 3 y$

$5 y = - 3 y + 16$

$x = 12 - 3 y$

Step 3

Solve for $y$ in $5 y = - 3 y + 16$ .

Move all terms containing $y$ to the left side of the equation.

Add $3 y$ to both sides of the equation.

$5 y + 3 y = 16$

$x = 12 - 3 y$

Add $5 y$ and $3 y$ .

$8 y = 16$

$x = 12 - 3 y$

$8 y = 16$

$x = 12 - 3 y$

Divide each term in $8 y = 16$ by $8$ and simplify.

Divide each term in $8 y = 16$ by $8$ .

$\frac{8 y}{8} = \frac{16}{8}$

$x = 12 - 3 y$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y}{8} = \frac{16}{8}$

$x = 12 - 3 y$

Divide $y$ by $1$ .

$y = \frac{16}{8}$

$x = 12 - 3 y$

$y = \frac{16}{8}$

$x = 12 - 3 y$

$y = \frac{16}{8}$

$x = 12 - 3 y$

Simplify the right side.

Divide $16$ by $8$ .

$y = 2$

$x = 12 - 3 y$

$y = 2$

$x = 12 - 3 y$

$y = 2$

$x = 12 - 3 y$

$y = 2$

$x = 12 - 3 y$

Step 4

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

Replace all occurrences of $y$ in $x = 12 - 3 y$ with $2$ .

$x = 12 - 3 \cdot 2$

$y = 2$

Simplify the right side.

Simplify $12 - 3 \cdot 2$ .

Multiply $- 3$ by $2$ .

$x = 12 - 6$

$y = 2$

Subtract $6$ from $12$ .

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

Step 5

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

$(6, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, 2)$

Equation Form:

$x = 6, y = 2$

Step 7

Solve by Substitution x+3y=12 5y=x+4

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