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anonymous

  • 5 years ago

2^x+2^y=10 and 4^x+4^y=68, solve x and y

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  1. anonymous
    • 5 years ago
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    let z=2^x and w=2^y

  2. anonymous
    • 5 years ago
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    the first equation becomes z+w=10 with that substitution

  3. anonymous
    • 5 years ago
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    now, replace 4 with 2^2, so the second equation becomes 2^2x+2^2y=68 now, 2^2x is that same as 2^x2, which is also the same as (2^x)^2...using a similar argument, for y 4^y is really (2^y)^2 but we called 2^x z and 2^y we called w, so we substitute in and get z^2+w^2=68

  4. anonymous
    • 5 years ago
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    now using the first equation z+w=10, solve for z and then substitute into the second equation. so z=10-w and the second equation is then (10-w)^2 +w^2=68 expand the (10-w)^2 with foil to get 100-20w+w^2+w^2=68 and then group like terms and move everything over

  5. anonymous
    • 5 years ago
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    2w^2-20w+32=0 dividing out by 2, w^2-10w+16=0 (w-8)(w-2)=0 w=2 and w=8 now, plug back in for w=2^y 2^y=2 and 2^y=8...so y=1 and y=3 plugging back in for y=1 gives x=3 and y=3 gives x=1 so the 2 solutions are (1,3) and (3,1)

  6. anonymous
    • 5 years ago
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    hoped that helped!

  7. anonymous
    • 5 years ago
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    thank you

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