## anonymous 4 years ago The sum of two numbers is 3. The difference of the squares of the numbers is 33. What is the absolute value of the difference of the two numbers?

1. anonymous

x+y=3 (x+y)(x=y)=33 x-y=|a|

3

3. anonymous

oops, I meant (x-y) not (x=y)

4. anonymous

it is actually 11.

5. anonymous

$a - b = 3$ $a ^{2} - b ^{2} = 33$ Rearrange eqaution 1: b = 3-a Substitute into equation 2: $a ^{2} - (3-a) ^{2} = 33$ Solve for a...

6. anonymous

I substituted so ir is 3(x-y)=33 divided 3 from both sides. x-y=33/3 x-y=11 absolute value of 11 is 11.

7. anonymous

$x+y=3$ $x^2-y^2=33$ $(x+y)(x-y)=33$ $x-y=\frac{33}{x+y}$ $x-y=\frac{33}{3}=11$

8. anonymous

I did it a bit different, but that's okay! :D

9. anonymous

Yes, you way is better.