What is the answer to solvefor x,x^0y=xy+2x+2y+3 ?

The answer to solvefor x,x^0y=xy+2x+2y+3 is x=-(y+3)/(2+y);\quad y\ne-2

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solvefor x,x^0y=xy+2x+2y+3

solve for

x, x^{0} y = x y + 2 x + 2 y + 3

x = - \frac{y + 3}{2 + y}; y \neq = - 2

Solution steps

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

Apply rule

a^{0} = 1, a \neq = 0

x^{0} = 1

1 \cdot y = x y + 2 x + 2 y + 3

Multiply:

1 \cdot y = y

y = x y + 2 x + 2 y + 3

Move

y

to the right side

0 = x y + y + 2 x + 3

Move

0

to the right side

= x y + y + 2 x + 3

Subtract

x y + 2 x

from both sides

- (x y + 2 x) = x y + y + 2 x + 3 - (x y + 2 x)

Simplify

- x y - 2 x = y + 3

Factor

- x y - 2 x : - x (y + 2)

- x (y + 2) = y + 3

Divide both sides by

- (y + 2); y \neq = - 2

x = - \frac{y + 3}{2 + y}; y \neq = - 2

- 29 x + 5 a = - 29

(n - 15) (c + 3) = 20

f = (a^{2} + 16)^{2} - (a^{2} - 16)^{2}

7 x + 2 x - 15 + 69 = - 3 x + 10 x - 54

x, 3 x - 2 y + 3 z = 8

What is the answer to solvefor x,x^0y=xy+2x+2y+3 ?
The answer to solvefor x,x^0y=xy+2x+2y+3 is x=-(y+3)/(2+y);\quad y\ne-2