What are the solutions to the equation solvefor x,(6x-x^2y)=(x^2-2x) ?

The solutions to the equation solvefor x,(6x-x^2y)=(x^2-2x) are x=0,x= 8/(1+y);\quad y\ne-1

Find the zeros of solvefor x,(6x-x^2y)=(x^2-2x)

The zeros of solvefor x,(6x-x^2y)=(x^2-2x) are x=0,x= 8/(1+y);\quad y\ne-1

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solve for

x, (6 x - x^{2} y) = (x^{2} - 2 x)

x = 0, x = \frac{8}{1 + y}; y \neq = - 1

Solution steps

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

Refine

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

Move

2 x

to the left side

- x^{2} y + 8 x = x^{2}

Move

x^{2}

to the left side

- x^{2} y + 8 x - x^{2} = 0

Factor

- x^{2} y + 8 x - x^{2} : - x (x + y x - 8)

- x (x + y x - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + y x - 8 = 0

Solve

x + y x - 8 = 0 : x = \frac{8}{1 + y}; y \neq = - 1

The solutions to the quadratic equation are:

x = 0, x = \frac{8}{1 + y}; y \neq = - 1

1 2^{2} = x \times x + 7

(x - 2 - 5) (x - 2 + 5) = 0

15 x^{2} - 4 x = 0

15 x^{2} + 7 x - 12 = 3 x^{2} + x - 12

40 (- 3 x^{2} - 4) = 0

What are the solutions to the equation solvefor x,(6x-x^2y)=(x^2-2x) ?
The solutions to the equation solvefor x,(6x-x^2y)=(x^2-2x) are x=0,x= 8/(1+y);\quad y\ne-1
Find the zeros of solvefor x,(6x-x^2y)=(x^2-2x)
The zeros of solvefor x,(6x-x^2y)=(x^2-2x) are x=0,x= 8/(1+y);\quad y\ne-1