Blue Book math question p. 475 #16

<p>I can't seem to find the correct way to do this one. </p>

<p>If x^2 - y^2 = 77 and x +y =1, what is the value of x?</p>

<p>It seems very simple and I tried doing solving it by making x=1-y and plugging that into the other equation, but I always seem to get a negative number.</p>

<p>x^2 - y^2 = (x - y)(x + y) = (x - y)(1) = x - y.</p>

<p>So x - y = 77.</p>

<p>So we have the system </p>

<p>x + y = 1
x - y = 77</p>

<p>Adding these two equations, we get 2x = 78, and x = 39.</p>

<p>it's pretty simple, you just need to plug in y to the original equation. i'm sorry for my bad english but i'll try to explain it:</p>

<p>x^2 - y^2 = 77
and x +y =1</p>

<p>you can see that x=1-y
when you plug it in in the first equation you get:<br>
(1-y)^2 - y^2 = 77</p>

<p>which leads to:
1-2y+y^2-y^2=77</p>

<p>cross the both y^2 because they have different sign, and you get</p>

<p>1-2y=77
2y=76
y=38</p>

<p>plug 38 in the first equation:
x^2-38^2=77
x^2=77+38^2
x=39</p>

<p>:)</p>

<p>This is the kind of problem that they'd definately have 78, or 2x, as an answer and make me lose the 800M haha</p>

<p>You could also graph the equation y=1-x and plus/minus (sqroot(x^2-77)) and find the points of intersection. No need for any work.</p>

<p>I don't have the book with me, but I think I remember this problem as having had different numbers:</p>

<p>Wasn't it x^2-y^2=77 and x+y = 11 ?</p>

<p>If you are doing it by algebra, it doesn't matter what the numbers were. But I really think I remember doing this just by playing around with numbers...x=9 and y=2 work. Like many SAT problems that are designed to look like they require algebra, this one was vulnerable to a little bit of hunting and playing.</p>