What are the solutions to the equation solvefor y,x^2+y^2+18y+81=18x+2xy ?

The solutions to the equation solvefor y,x^2+y^2+18y+81=18x+2xy are y=-9+x

Find the zeros of solvefor y,x^2+y^2+18y+81=18x+2xy

The zeros of solvefor y,x^2+y^2+18y+81=18x+2xy are y=-9+x

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solve for

y, x^{2} + y^{2} + 18 y + 81 = 18 x + 2 x y

y = - 9 + x

Solution steps

x^{2} + y^{2} + 18 y + 81 = 18 x + 2 x y

Subtract

18 x + 2 x y

from both sides

x^{2} + y^{2} + 18 y + 81 - (18 x + 2 x y) = 18 x + 2 x y - (18 x + 2 x y)

Simplify

y^{2} + (18 - 2 x) y + x^{2} + 81 - 18 x = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( 18 - 2 x ) \pm ( 18 - 2 x ) ^{2} - 4 \cdot 1 \cdot ( x ^{2} + 81 - 18 x )}{2 \cdot 1}

(18 - 2 x)^{2} - 4 \cdot 1 \cdot (x^{2} + 81 - 18 x) = 0

y_{1, 2} = \frac{- ( 18 - 2 x ) \pm 0}{2 \cdot 1}

y = \frac{- ( 18 - 2 x )}{2 \cdot 1}

Simplify

\frac{- ( 18 - 2 x )}{2 \cdot 1} : - 9 + x

y = - 9 + x

The solution to the quadratic equation is:

y = - 9 + x

- 4 x^{2} - 7 x + 36 = 0

4 x^{2} + 23 x = - 28

2 x^{2} + 3 x + 0.5 = 0

2 (x - 7)^{2} - 31 = 11

0 = 50 + 20 \cdot t - 4.9 \cdot t^{2}

What are the solutions to the equation solvefor y,x^2+y^2+18y+81=18x+2xy ?
The solutions to the equation solvefor y,x^2+y^2+18y+81=18x+2xy are y=-9+x
Find the zeros of solvefor y,x^2+y^2+18y+81=18x+2xy
The zeros of solvefor y,x^2+y^2+18y+81=18x+2xy are y=-9+x