Solve by Graphing y=x^2+12x+30 , 8x-y=10

$y = x^{2} + 12 x + 30$ , $8 x - y = 10$

Step 1

Subtract $8 x$ from both sides of the equation.

$- y = 10 - 8 x$

$y = x^{2} + 12 x + 30$

Step 2

Divide each term in $- y = 10 - 8 x$ by $- 1$ and simplify.

Divide each term in $- y = 10 - 8 x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{10}{- 1} + \frac{- 8 x}{- 1}$

$y = x^{2} + 12 x + 30$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{10}{- 1} + \frac{- 8 x}{- 1}$

$y = x^{2} + 12 x + 30$

Divide $y$ by $1$ .

$y = \frac{10}{- 1} + \frac{- 8 x}{- 1}$

$y = x^{2} + 12 x + 30$

$y = \frac{10}{- 1} + \frac{- 8 x}{- 1}$

$y = x^{2} + 12 x + 30$

Simplify the right side.

Simplify each term.

Divide $10$ by $- 1$ .

$y = - 10 + \frac{- 8 x}{- 1}$

$y = x^{2} + 12 x + 30$

Move the negative one from the denominator of $\frac{- 8 x}{- 1}$ .

$y = - 10 - 1 \cdot (- 8 x)$

$y = x^{2} + 12 x + 30$

Rewrite $- 1 \cdot (- 8 x)$ as $- (- 8 x)$ .

$y = - 10 - (- 8 x)$

$y = x^{2} + 12 x + 30$

Multiply $- 8$ by $- 1$ .

$y = - 10 + 8 x$

$y = x^{2} + 12 x + 30$

$y = - 10 + 8 x$

$y = x^{2} + 12 x + 30$

$y = - 10 + 8 x$

$y = x^{2} + 12 x + 30$

$y = - 10 + 8 x$

$y = x^{2} + 12 x + 30$

Step 3

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

$y = - 26 + 48 i, x = - 2 + 6 i$

$y = - 26 - 48 i, x = - 2 - 6 i$

Step 4