How do you solve the simultaneous equations ##x^2 + y^2=29## and ##y-x=3##?

Use the second equation to provide an expression for ##y## in terms of ##x## to substitute into the first equation to give a quadratic equation in ##x##.

First add ##x## to both sides of the second equation to get:

##y = x+3##

Then substitute this expression for ##y## into the first equation to get:

##29 = x^2+(x+3)^2 = 2x^2+6x+9##

Subtract ##29## from both ends to get:

##0 = 2x^2+6x-20##

Divide both sides by ##2## to get:

##0 = x^2+3x-10 = (x+5)(x-2)##

So ##x=2## or ##x=-5##

If ##x=2## then ##y = x+3 = 5##.

If ##x=-5## then ##y = x+3 = -2##

So the two solutions ##(x, y)## are ##(2, 5)## and ##(-5, -2)##