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Find all ordered pairs x, y of real numbers such that x+y=10 and x^3 + y^3 = 300 + x^2 + y^2.
For example, to enter the solutions (2, 4) and (-3, 9), you would enter "(2,4),(-3,9)" (without the quotation marks).

 Sep 8, 2024
 #1
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\(x+y=10 \)

\(x^3 + y^3 = 300 + x^2 + y^2\)

\(x^3+y^3=(x + y)(x^2 − xy + y^2)\)

\(10(x^2-xy+y^2)=300+x^2+y^2\)

\(9x^2-10xy+9y^2=300\)

\(9(x+y)^2-28xy=300\)

\(28xy=600\)

\(xy=\frac{150}{7}\)

 

\(x+y=10,xy=\frac{150}{7}\)

 

I think you can solve on your own now. If you are having trouble, leave a note and ill do the rest.

 Sep 8, 2024

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