Solve by Graphing 6x-3y=12 , 2x=y+4

$6 x - 3 y = 12$ , $2 x = y + 4$

Step 1

Subtract $6 x$ from both sides of the equation.

$- 3 y = 12 - 6 x$

$2 x = y + 4$

Step 2

Divide each term in $- 3 y = 12 - 6 x$ by $- 3$ and simplify.

Divide each term in $- 3 y = 12 - 6 x$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{12}{- 3} + \frac{- 6 x}{- 3}$

$2 x = y + 4$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{12}{- 3} + \frac{- 6 x}{- 3}$

$2 x = y + 4$

Divide $y$ by $1$ .

$y = \frac{12}{- 3} + \frac{- 6 x}{- 3}$

$2 x = y + 4$

$y = \frac{12}{- 3} + \frac{- 6 x}{- 3}$

$2 x = y + 4$

$y = \frac{12}{- 3} + \frac{- 6 x}{- 3}$

$2 x = y + 4$

Simplify the right side.

Simplify each term.

Divide $12$ by $- 3$ .

$y = - 4 + \frac{- 6 x}{- 3}$

$2 x = y + 4$

Cancel the common factor of $- 6$ and $- 3$ .

Factor $- 3$ out of $- 6 x$ .

$y = - 4 + \frac{- 3 (2 x)}{- 3}$

$2 x = y + 4$

Cancel the common factors.

Factor $- 3$ out of $- 3$ .

$y = - 4 + \frac{- 3 (2 x)}{- 3 \cdot 1}$

$2 x = y + 4$

Cancel the common factor.

$y = - 4 + \frac{- 3 (2 x)}{- 3 \cdot 1}$

$2 x = y + 4$

Rewrite the expression.

$y = - 4 + \frac{2 x}{1}$

$2 x = y + 4$

Divide $2 x$ by $1$ .

$y = - 4 + 2 x$

$2 x = y + 4$

$y = - 4 + 2 x$

$2 x = y + 4$

$y = - 4 + 2 x$

$2 x = y + 4$

$y = - 4 + 2 x$

$2 x = y + 4$

$y = - 4 + 2 x$

$2 x = y + 4$

$y = - 4 + 2 x$

$2 x = y + 4$

Step 3

Rewrite the equation as $y + 4 = 2 x$ .

$y + 4 = 2 x$

$y = - 4 + 2 x$

Step 4

Subtract $4$ from both sides of the equation.

$y = 2 x - 4$

$y = - 4 + 2 x$

Step 5

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

Always true

Step 6

Solve by Graphing 6x-3y=12 , 2x=y+4

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