What is the answer to solvefor y,yx^2-4y-x^2=0 ?

The answer to solvefor y,yx^2-4y-x^2=0 is y=(x^2)/(x^2-4);\quad x\ne 2,x\ne-2

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solvefor y,yx^2-4y-x^2=0

solve for

y, y x^{2} - 4 y - x^{2} = 0

y = \frac{x ^{2}}{x ^{2} - 4}; x \neq = 2, x \neq = - 2

Solution steps

y x^{2} - 4 y - x^{2} = 0

Move

x^{2}

to the right side

y x^{2} - 4 y = x^{2}

Factor

y x^{2} - 4 y : y (x^{2} - 4)

y (x^{2} - 4) = x^{2}

Divide both sides by

x^{2} - 4; x \neq = 2, x \neq = - 2

y = \frac{x ^{2}}{x ^{2} - 4}; x \neq = 2, x \neq = - 2

n (x) = - x^{2} + 8 x - 9

(x + 8) (4 x + 1) = 0

y, x = a - 2 b y^{2}

x^{2} - 4 x - 3

22 q^{2} + 16 q - 5 = 6 q^{2}

What is the answer to solvefor y,yx^2-4y-x^2=0 ?
The answer to solvefor y,yx^2-4y-x^2=0 is y=(x^2)/(x^2-4);\quad x\ne 2,x\ne-2