Here's the question you clicked on:
MissPacGirl
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?
x+y=3 (x+y)(x=y)=33 x-y=|a|
oops, I meant (x-y) not (x=y)
\[a - b = 3\] \[a ^{2} - b ^{2} = 33\] Rearrange eqaution 1: b = 3-a Substitute into equation 2: \[a ^{2} - (3-a) ^{2} = 33\] Solve for a...
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.
\[x+y=3\] \[x^2-y^2=33\] \[(x+y)(x-y)=33\] \[x-y=\frac{33}{x+y}\] \[x-y=\frac{33}{3}=11\]
I did it a bit different, but that's okay! :D
Yes, you way is better.
just a second 7+(-4)=3 and 7^2-4^2=33 so the numbers are 7 and -4...oh no I'm sorry it's the absolute value of the DIFFERENCE...:D sorry :D you're right :D