Question

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?

Answer 1

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

Answer 2

3

Answer 3

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

Answer 4

it is actually 11.

Answer 5

\[a - b = 3\]
\[a ^{2} - b ^{2} = 33\]

Rearrange eqaution 1:
b = 3-a

Substitute into equation 2:

\[a ^{2} - (3-a) ^{2} = 33\]
Solve for a...

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

Yes, you way is better.

Answer 10

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