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MissPacGirl Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. MissPacGirl Group Title
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    x+y=3 (x+y)(x=y)=33 x-y=|a|

    • 2 years ago
  2. nenadmatematika Group Title
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    3

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

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

    • 2 years ago
  5. adnanchowdhury Group Title
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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...

    • 2 years ago
  6. MissPacGirl Group Title
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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.

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

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

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

    • 2 years ago
  10. nenadmatematika Group Title
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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

    • 2 years ago
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