Solve the System of Equations 2x-y=4 and y=-2x+8

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

Replace all occurrences of $y$ with $- 2 x + 8$ in each equation.

Replace all occurrences of $y$ in $2 x - y = 4$ with $- 2 x + 8$ .

$2 x - (- 2 x + 8) = 4$

$y = - 2 x + 8$

Simplify the left side.

Simplify $2 x - (- 2 x + 8)$ .

Simplify each term.

Apply the distributive property.

$2 x - (- 2 x) - 1 \cdot 8 = 4$

$y = - 2 x + 8$

Multiply $- 2$ by $- 1$ .

$2 x + 2 x - 1 \cdot 8 = 4$

$y = - 2 x + 8$

Multiply $- 1$ by $8$ .

$2 x + 2 x - 8 = 4$

$y = - 2 x + 8$

$2 x + 2 x - 8 = 4$

$y = - 2 x + 8$

Add $2 x$ and $2 x$ .

$4 x - 8 = 4$

$y = - 2 x + 8$

$4 x - 8 = 4$

$y = - 2 x + 8$

$4 x - 8 = 4$

$y = - 2 x + 8$

$4 x - 8 = 4$

$y = - 2 x + 8$

Solve for $x$ in $4 x - 8 = 4$ .

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

Add $8$ to both sides of the equation.

$4 x = 4 + 8$

$y = - 2 x + 8$

Add $4$ and $8$ .

$4 x = 12$

$y = - 2 x + 8$

$4 x = 12$

$y = - 2 x + 8$

Divide each term in $4 x = 12$ by $4$ and simplify.

Divide each term in $4 x = 12$ by $4$ .

$\frac{4 x}{4} = \frac{12}{4}$

$y = - 2 x + 8$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{12}{4}$

$y = - 2 x + 8$

Divide $x$ by $1$ .

$x = \frac{12}{4}$

$y = - 2 x + 8$

$x = \frac{12}{4}$

$y = - 2 x + 8$

$x = \frac{12}{4}$

$y = - 2 x + 8$

Simplify the right side.

Divide $12$ by $4$ .

$x = 3$

$y = - 2 x + 8$

$x = 3$

$y = - 2 x + 8$

$x = 3$

$y = - 2 x + 8$

$x = 3$

$y = - 2 x + 8$

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

Replace all occurrences of $x$ in $y = - 2 x + 8$ with $3$ .

$y = - 2 \cdot 3 + 8$

$x = 3$

Simplify the right side.

Simplify $- 2 \cdot 3 + 8$ .

Multiply $- 2$ by $3$ .

$y = - 6 + 8$

$x = 3$

Add $- 6$ and $8$ .

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

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

$(3, 2)$

The result can be shown in multiple forms.

Point Form:

$(3, 2)$

Equation Form:

$x = 3, y = 2$

Solve the System of Equations 2x-y=4 and y=-2x+8

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