Find the point(s) of intersection of the graphs y=x^2+4x-21 and x+y=-27 using an algebraic method.

As, at the point of intersection, the values of x and y are the same in each graph (otherwise it wouldn't be a point of intersection!), we can approach the problem as solving a pair of simultaneous equations. Since the quadratic expression is already equal to y, we can substitute it into the linear equation, such that x+y=-27 becomes x+(x^2+4x-21)=-27. We can simplify this into the straightforward quadratic equation x^2+5x+6=0, and factorise into (x+2)(x+3)=0. (In order for the expression to equal zero, either x+2=0 or x+3=0, so) x=-2,-3. Substituting x=-2 into either of the initial graphs gives y=-25 (e.g. (-2)+y=-27 => y=-25), and substituting x=-3 into either of the initial graphs gives y=-24; therefore, the two points of intersection are (-2,-25) and (-3,-24).(When solving the simultaneous equations, we could have instead rearranged the linear equation to the form y=-27-x and then, since both (-27-x) and (x^2+4x-21) equal y, formed the quadratic equation x^2+4x-21=-x-27, which is equivalent to the same simplified quadratic equation as above: x^2+5x+6=0.)