Equation system

Nov 2007
329
55
Solve the following system of equation for \(\displaystyle x,y\in\mathbb{R}\):
\(\displaystyle x^3+4y=y^3+16x\)
\(\displaystyle \frac {1+y^2}{1+x^2}=5\)
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Solve the following system of equation for \(\displaystyle x,y\in\mathbb{R}\):
\(\displaystyle x^3+4y=y^3+16x\)
\(\displaystyle \frac {1+y^2}{1+x^2}=5\)
\(\displaystyle x^3-16x = y^3-4y\)

\(\displaystyle x(x^2-16) = y(y^2-4)
\)

\(\displaystyle \frac {1+y^2}{1+x^2}=5\)

\(\displaystyle 1+y^2 = 5+5x^2\)

\(\displaystyle y^2 = 4+5x^2\)


\(\displaystyle x(x^2-16) = y(5x^2)\)

\(\displaystyle y = \frac{x^2-16}{5x}\)


\(\displaystyle \frac{(x^2-16)^2}{25x^2} = 4 + 5x^2\)

\(\displaystyle x^4 - 32x^2 + 256 = 100x^2 + 125x^4\)

\(\displaystyle 0 = 124x^4 + 132x^2 - 256\)

\(\displaystyle 0 = 31x^4 + 33x^2 - 64\)

\(\displaystyle 0 = (x^2-1)(31x^2 + 64)\)

\(\displaystyle 0 = (x-1)(x+1)(31x^2+64)\)

can you finish?
 
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