Solve by Substitution x+12y=68 , x=8y-12

$x + 12 y = 68$ , $x = 8 y - 12$

Step 1

Replace all occurrences of $x$ with $8 y - 12$ in each equation.

Replace all occurrences of $x$ in $x + 12 y = 68$ with $8 y - 12$ .

$(8 y - 12) + 12 y = 68$

$x = 8 y - 12$

Simplify the left side.

Add $8 y$ and $12 y$ .

$20 y - 12 = 68$

$x = 8 y - 12$

$20 y - 12 = 68$

$x = 8 y - 12$

$20 y - 12 = 68$

$x = 8 y - 12$

Step 2

Solve for $y$ in $20 y - 12 = 68$ .

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

Add $12$ to both sides of the equation.

$20 y = 68 + 12$

$x = 8 y - 12$

Add $68$ and $12$ .

$20 y = 80$

$x = 8 y - 12$

$20 y = 80$

$x = 8 y - 12$

Divide each term in $20 y = 80$ by $20$ and simplify.

Divide each term in $20 y = 80$ by $20$ .

$\frac{20 y}{20} = \frac{80}{20}$

$x = 8 y - 12$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 y}{20} = \frac{80}{20}$

$x = 8 y - 12$

Divide $y$ by $1$ .

$y = \frac{80}{20}$

$x = 8 y - 12$

$y = \frac{80}{20}$

$x = 8 y - 12$

$y = \frac{80}{20}$

$x = 8 y - 12$

Simplify the right side.

Divide $80$ by $20$ .

$y = 4$

$x = 8 y - 12$

$y = 4$

$x = 8 y - 12$

$y = 4$

$x = 8 y - 12$

$y = 4$

$x = 8 y - 12$

Step 3

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

Replace all occurrences of $y$ in $x = 8 y - 12$ with $4$ .

$x = 8 (4) - 12$

$y = 4$

Simplify the right side.

Simplify $8 (4) - 12$ .

Multiply $8$ by $4$ .

$x = 32 - 12$

$y = 4$

Subtract $12$ from $32$ .

$x = 20$

$y = 4$

$x = 20$

$y = 4$

$x = 20$

$y = 4$

$x = 20$

$y = 4$

Step 4

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

$(20, 4)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(20, 4)$

Equation Form:

$x = 20, y = 4$

Step 6

Solve by Substitution x+12y=68 , x=8y-12

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