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

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

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

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

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solve for

x, (x + y)^{2} = 2 x + 2 y

x = - y + 2, x = - y

Solution steps

(x + y)^{2} = 2 x + 2 y

Expand

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

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

Move

2 y

to the left side

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

Move

2 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 2 y - 2 ) \pm ( 2 y - 2 ) ^{2} - 4 \cdot 1 \cdot ( y ^{2} - 2 y )}{2 \cdot 1}

(2 y - 2)^{2} - 4 \cdot 1 \cdot (y^{2} - 2 y) = 2

x_{1, 2} = \frac{- ( 2 y - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 2 y - 2 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( 2 y - 2 ) - 2}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = - y + 2, x = - y

v^{2} - 9 v + 18 = 0

8 (- x^{2} - 1) = - 24

7 p^{2} = 35 p + 168

9 x^{2} - 48 x + 54 = 0

4.905 t^{2} - 90.27 t - 150 = 0

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