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).