Find the Points of Intersection y=4-x , 2y=8-2x

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

Step 1

Reorder $4$ and $- x$ .

$y = - x + 4$

$2 y = 8 - 2 x$

Step 2

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

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

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

$y = - x + 4$

Simplify the left side.

Simplify $2 (- x + 4)$ .

Apply the distributive property.

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

$y = - x + 4$

Multiply.

Multiply $- 1$ by $2$ .

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

$y = - x + 4$

Multiply $2$ by $4$ .

$- 2 x + 8 = 8 - 2 x$

$y = - x + 4$

$- 2 x + 8 = 8 - 2 x$

$y = - x + 4$

$- 2 x + 8 = 8 - 2 x$

$y = - x + 4$

$- 2 x + 8 = 8 - 2 x$

$y = - x + 4$

$- 2 x + 8 = 8 - 2 x$

$y = - x + 4$

Step 3

Solve for $x$ in $- 2 x + 8 = 8 - 2 x$ .

Move all terms containing $x$ to the left side of the equation.

Add $2 x$ to both sides of the equation.

$- 2 x + 8 + 2 x = 8$

$y = - x + 4$

Combine the opposite terms in $- 2 x + 8 + 2 x$ .

Add $- 2 x$ and $2 x$ .

$0 + 8 = 8$

$y = - x + 4$

Add $0$ and $8$ .

$8 = 8$

$y = - x + 4$

$8 = 8$

$y = - x + 4$

$8 = 8$

$y = - x + 4$

Since $8 = 8$ , the equation will always be true.

Always true

$y = - x + 4$

Always true

$y = - x + 4$

Step 4

Remove any equations from the system that are always true.

$y = - x + 4$

Step 5

Find the Points of Intersection y=4-x , 2y=8-2x

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