<p>Hi, is there an alternative solution to problem 7, pg 680?</p>
<p>x^2 + y^2 = 7</p>
<p>What does x equal? </p>
<p>a. 3
b. 4
c. 5
d. 6
e. 7</p>
<p>All I did was just plug in numbers on the x based on the number choice starting from seven. I'm just curious if there is a more quicker way to do that rather than plugging in numbers.</p>
<p>in order to solve it wouldnt you need a system of 2 equations? if not i solved it and got x=-sqrt(7-y^2),x=sqrt(7-y^2)</p>
<p>The problem you posted is wrong. You can't solve it (very confident).</p>
<p>The problem is x^2-y^2=7. Difference of perfect squares.
(x+y)(x-y)=7 </p>
<p>x and y are both positive integers and 7 is a prime number. x + y is obviously greater than x-y</p>
<p>Add these two equations
X+y =7
x-y =1 </p>
<p>2x = 8
x = 4</p>
<p>Definitely. If x^2 + y^2 = 7, then x can't be 3,4,5,6, or 7 because y^2 would have to be negative and that's not possible.</p>
<p>My bad, I accidentally put a + there. It's supposed to be x^2 - y^2..</p>
<ul>
<li>edit</li>
</ul>
<p>thanks duper for editing the problem</p>
<p>Duper:</p>
<p>How did you get:</p>
<p>X+y =7
x-y =1</p>
<p>Well since 7 is a prime number, the factors have to be 7 and 1. X+Y > X-Y; therefore
x+y =7
x-y=1</p>
<p>The solution wasn't that easy to find at first (personally), I probably would have plugged in answers if I were taking the real exam to save some time.</p>
<p>Duper:</p>
<p>Thanks. Awesome trick.</p>