What are the solutions to the equation solvefor x,3x^2+y=3y^2+x ?

The solutions to the equation solvefor x,3x^2+y=3y^2+x are x=y,x=(-3y+1)/3

Find the zeros of solvefor x,3x^2+y=3y^2+x

The zeros of solvefor x,3x^2+y=3y^2+x are x=y,x=(-3y+1)/3

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solve for

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

x = y, x = \frac{- 3 y + 1}{3}

Solution steps

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

Move

x

to the left side

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

Move

3 y^{2}

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Factor

3 x^{2} - x + y - 3 y^{2} : (x - y) (3 x + 3 y - 1)

(x - y) (3 x + 3 y - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - y = 0 or 3 x + 3 y - 1 = 0

Solve

x - y = 0 : x = y

Solve

3 x + 3 y - 1 = 0 : x = \frac{- 3 y + 1}{3}

The solutions to the quadratic equation are:

x = y, x = \frac{- 3 y + 1}{3}

4 x^{2} + 32 x + 236 = - 8

(2 x - 3)^{2} = x + 2 (7 - x^{2})

x^{2} + 144 = 40 x

x^{2} + 11 x + 1

2 x^{2} = 3 x + 5

What are the solutions to the equation solvefor x,3x^2+y=3y^2+x ?
The solutions to the equation solvefor x,3x^2+y=3y^2+x are x=y,x=(-3y+1)/3
Find the zeros of solvefor x,3x^2+y=3y^2+x
The zeros of solvefor x,3x^2+y=3y^2+x are x=y,x=(-3y+1)/3