Solve (2x+y-8)^2=(2x+y)^2+quad4y^4

$$( 2 x + y - 8 ) ^ { 2 } = ( 2 x + y ) ^ { 2 } + \quad 4 y ^ { 4 }$$

Solve for x

$$x=-\frac{y^{4}}{8}-\frac{y}{2}+2$$

Steps for Solving Linear Equation

Square 2x+y-8.

$$4x^{2}+4xy-32x+y^{2}-16y+64=\left(2x+y\right)^{2}+4y^{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+y\right)^{2}.

$$4x^{2}+4xy-32x+y^{2}-16y+64=4x^{2}+4xy+y^{2}+4y^{4}$$

Subtract 4x^{2} from both sides.

$$4x^{2}+4xy-32x+y^{2}-16y+64-4x^{2}=4xy+y^{2}+4y^{4}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4xy-32x+y^{2}-16y+64=4xy+y^{2}+4y^{4}$$

Subtract 4xy from both sides.

$$4xy-32x+y^{2}-16y+64-4xy=y^{2}+4y^{4}$$

Combine 4xy and -4xy to get 0.

$$-32x+y^{2}-16y+64=y^{2}+4y^{4}$$

Subtract y^{2} from both sides.

$$-32x-16y+64=y^{2}+4y^{4}-y^{2}$$

Combine y^{2} and -y^{2} to get 0.

$$-32x-16y+64=4y^{4}$$

Add 16y to both sides.

$$-32x+64=4y^{4}+16y$$

Subtract 64 from both sides.

$$-32x=4y^{4}+16y-64$$

Divide both sides by -32.

$$\frac{-32x}{-32}=\frac{4y^{4}+16y-64}{-32}$$

Dividing by -32 undoes the multiplication by -32.

$$x=\frac{4y^{4}+16y-64}{-32}$$

Divide 4y^{4}+16y-64 by -32.

$$x=-\frac{y^{4}}{8}-\frac{y}{2}+2$$