Question

what are the solutions to the system of equations
{-x+y=4
{y+12=x^2+x

Answer 1

Answer:

Step-by-step explanation:

$y=4+x$          ............................................ (1)

$y=x^2+x-12$  ............................................ (2)

now equate both (1) and (2)

$4+x=x^2+x-12$

$=>x^2=16$

$x=-4,4$

now substitute the values of x ,

when $x=-4$ , y becomes 0

when $x=4$  , y becomes 8

so, solution set:

$(-4,0)$ and $(4,8)$