Question 1091985
<br>The solution already posted has an arithmetic error (or maybe just a typographical error) somewhere along the way.  Maybe you saw it....<br>
Step 1: If you expand everything in both equations, then you will have an x^2 and y^2 term in both equations; so when you compare the two equations by subtraction, the squared terms will cancel out, leaving you with a linear equation in x and y.<br>
Step 2: You can solve that linear equation for one of the variables and substitute into either of the original equations to solve for one variable; since you will be solving a quadratic equation, there will probably be two solutions.<br>
Step 3: substituting each of those values into one of the original equations will give you the corresponding values of the other variable.<br>
<br>
Step 1: Get a linear equation from the two given equations by expanding, then subtracting one equation from the other...<br>
{{{(x^2+2x+1)-(y^2-2y+1) = 20}}}
{{{x^2+2x-y^2+2y = 20}}}  (1)<br>
{{{x^2 - (y^2+4y+4) = 24}}}
{{{x^2-y^2-4y-4 = 24}}}
{{{x^2-y^2-4y = 28}}}  (2)<br>
{{{2x+6y = -8}}}   (by subtracting (2) from (1))
{{{x+3y = -4}}}  (3)<br>
Step 2: Solve (3) for x and substitute in the second original equation; solve the resulting quadratic equation for two values of y.<br>
{{{x = -3y-4}}}  (4)
{{{(-3y-4)^2 - (y+2)^2 = 24}}}
{{{9y^2+24y+16-y^2-4y-4 = 24}}}
{{{8y^2+20y-12 = 0}}}
{{{2y^2+5y-3 = 0}}}
{{{(2y-1)(y+3) = 0}}}<br>
The two values for y are 1/2 and -3.<br>
Step 3: Find the corresponding x value for each y value, using (4).<br>
{{{x = -3(1/2)-4 = -11/2}}}
{{{x = -3(-3)-4 = 5}}}<br>
The points of intersection of the two graphs (the pair of (x,y) values that satisfy both equations) are (-11/2,1/2) and (5,-3).