Solve by Graphing 4x-2y=8 , 2x=y+4

$4 x - 2 y = 8$ , $2 x = y + 4$

Step 1

Subtract $4 x$ from both sides of the equation.

$- 2 y = 8 - 4 x$

$2 x = y + 4$

Step 2

Divide each term in $- 2 y = 8 - 4 x$ by $- 2$ and simplify.

Divide each term in $- 2 y = 8 - 4 x$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{8}{- 2} + \frac{- 4 x}{- 2}$

$2 x = y + 4$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{8}{- 2} + \frac{- 4 x}{- 2}$

$2 x = y + 4$

Divide $y$ by $1$ .

$y = \frac{8}{- 2} + \frac{- 4 x}{- 2}$

$2 x = y + 4$

$y = \frac{8}{- 2} + \frac{- 4 x}{- 2}$

$2 x = y + 4$

$y = \frac{8}{- 2} + \frac{- 4 x}{- 2}$

$2 x = y + 4$

Simplify the right side.

Simplify each term.

Divide $8$ by $- 2$ .

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

$2 x = y + 4$

Cancel the common factor of $- 4$ and $- 2$ .

Factor $- 2$ out of $- 4 x$ .

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

$2 x = y + 4$

Cancel the common factors.

Factor $- 2$ out of $- 2$ .

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

$2 x = y + 4$

Cancel the common factor.

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

$2 x = y + 4$

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 4x-2y=8 , 2x=y+4

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