Question

Pls answer! I don’t know math at all

Answer 1

The solutions of the system of equations are (–2, 4) or (–5, 1).

Solution:

Given equations are:

$y=x+6$ – – – – (1)

$y=x^{2}+8 x+16$ – – – – (2)

Equate the values of y.

$x+6=x^{2}+8 x+16$

Subtract 6 from both sides of the equation.

$x=x^{2}+8 x+10$

Subtract x from both sides of the equation.

$0=x^{2}+7 x+10$

Switch the sides.

$x^{2}+7 x+10=0$

Factor this polynomial, we get

$(x+2)(x+5)=0$

x + 2 = 0 (or) x + 5 = 0

x = –2 (or) x = –5

Substitute x = –2 in equation (1),

y = –2 + 6

y = 4

Substitute x = –5 in equation (1),

y = –5 + 6

y = 1

The solutions of the system of equations are (–2, 4) or (–5, 1).