Solve by Graphing 6x-2y=8 4x+y=-1

$6 x - 2 y = 8$ $4 x + y = - 1$

Step 1

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

$- 2 y = 8 - 6 x$

$4 x + y = - 1$

Step 2

Divide each term in $- 2 y = 8 - 6 x$ by $- 2$ and simplify.

Divide each term in $- 2 y = 8 - 6 x$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{8}{- 2} + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{8}{- 2} + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

Divide $y$ by $1$ .

$y = \frac{8}{- 2} + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

$y = \frac{8}{- 2} + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

$y = \frac{8}{- 2} + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

Simplify the right side.

Simplify each term.

Divide $8$ by $- 2$ .

$y = - 4 + \frac{- 6 x}{- 2}$

$4 x + y = - 1$

Cancel the common factor of $- 6$ and $- 2$ .

Factor $- 2$ out of $- 6 x$ .

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

$4 x + y = - 1$

Cancel the common factors.

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

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

$4 x + y = - 1$

Cancel the common factor.

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

$4 x + y = - 1$

Rewrite the expression.

$y = - 4 + \frac{3 x}{1}$

$4 x + y = - 1$

Divide $3 x$ by $1$ .

$y = - 4 + 3 x$

$4 x + y = - 1$

$y = - 4 + 3 x$

$4 x + y = - 1$

$y = - 4 + 3 x$

$4 x + y = - 1$

$y = - 4 + 3 x$

$4 x + y = - 1$

$y = - 4 + 3 x$

$4 x + y = - 1$

$y = - 4 + 3 x$

$4 x + y = - 1$

Step 3

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

$y = - 1 - 4 x$

$y = - 4 + 3 x$

Step 4

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

$(\frac{3}{7}, - \frac{19}{7})$

Step 5

Solve by Graphing 6x-2y=8 4x+y=-1

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