+1348 Find one ordered pair $(x,y)$ of real numbers such that $x + y = 10$ and $x^3 + y^3 = 162 + x^2 + y^2.$
+24 x + y = 10
x^3 + y^3 = 162 + x^2 + y^2
x^3 + y^3 factors into (x+y)(x^2+xy+y^2)
we know x+y is 10 from the first equation so we plug that in
so now we have:
10(x^2+xy+y^2) = 162+x^2+y^2
expanding gives us
10x^2+10xy+10y^2 = 162+x^2+y^2
combine like terms
9x^2 + 9y^2 + 10xy - 162 = 0
divdie both sides by 9
x^2 + y^2 + 9/10xy - 18 = 0
Since x+y = 10, (x+y)^2 = 100 and x^2+y^2 = 100 - 2xy
plugging that in gives
100-2xy+9/10xy-18=0
combine like terms
82-11/10xy=0
11/10xy = 82
xy = 820/11
now knowing the sum and product of x and y, we can create quadratic satisfying these properties using vieta's and solve
x+y = -b/a
xy = c/a
let a = 1 for convenience
we get
n^2 -10n + 820/11 = 0
solving gives
x and y = 5 + i(sqrt(545/11)) and 5 - i(sqrt(545/11))
+10 If we factor the left side of the second equation, we get the equation:
\((x+y)(x^2+xy+y^2)=162+x^2+y^2\)
Since the first equation gives x+y=10, the second equation now becomes:
\(10(x^2+xy+y^2)=162+x^2+y^2 \)
\(10x^2+10xy+10y^2= 162+x^2+y^2\)
\(9x^2+9y^2+10xy=162\)
Now substitute the first equation into the second
\(9(10-y)^2+9y^2+10(10-y)y=162\)
\(9(100-20y+y^2)+9y^2+100y-10y^2=162\)
\(900-180y+9y^2+9y^2+100y-10y^2=162\)
\(8y^2-80y+900=162\)
\(8y^2-80y+738=0\)
However, putting this into the quadratic formula, we find that the discriminant is negative, so therefore there are no real solutions.
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