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MissPacGirl

  • 2 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?

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  1. MissPacGirl
    • 2 years ago
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    x+y=3 (x+y)(x=y)=33 x-y=|a|

  2. nenadmatematika
    • 2 years ago
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    3

  3. MissPacGirl
    • 2 years ago
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    oops, I meant (x-y) not (x=y)

  4. MissPacGirl
    • 2 years ago
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    it is actually 11.

  5. adnanchowdhury
    • 2 years ago
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    \[a - b = 3\] \[a ^{2} - b ^{2} = 33\] Rearrange eqaution 1: b = 3-a Substitute into equation 2: \[a ^{2} - (3-a) ^{2} = 33\] Solve for a...

  6. MissPacGirl
    • 2 years ago
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    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. satellite73
    • 2 years ago
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    \[x+y=3\] \[x^2-y^2=33\] \[(x+y)(x-y)=33\] \[x-y=\frac{33}{x+y}\] \[x-y=\frac{33}{3}=11\]

  8. MissPacGirl
    • 2 years ago
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    I did it a bit different, but that's okay! :D

  9. adnanchowdhury
    • 2 years ago
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    Yes, you way is better.

  10. nenadmatematika
    • 2 years ago
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    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

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