Question

How do you solve the system `y=x+7`, `y=(x+3)^2-8`?

Answer 1

There are two solutions x=1, y=8 and x=-6, y=1

Explanation:

Replace y in the second equation with the right hand side of the irst equation to give `x+7=(x+3)^2-8`. Expand the square on the right hand side `x+7=x^2+6x+9-8`. Subtract `x+7` from both sides and tidy up `x^2+5x-6=0`. Factorise `(x+6)(x-1)=0`. This gives the two solutions `x=1` and `x=-6`. Using the first equation, if `x=1` then `y=8` and if `x=-6` then `y=1`.