Solve 8^2+(6-y)^2=y^2

$$8 ^ { 2 } + ( 6 - y ) ^ { 2 } = y ^ { 2 }$$

Solve for y

$$y = \frac{25}{3} = 8\frac{1}{3} \approx 8.333333333$$

Steps for Solving Linear Equation

Calculate 8 to the power of 2 and get 64.

$$64+\left(6-y\right)^{2}=y^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$64+36-12y+y^{2}=y^{2}$$

Add 64 and 36 to get 100.

$$100-12y+y^{2}=y^{2}$$

Subtract y^{2} from both sides.

$$100-12y+y^{2}-y^{2}=0$$

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

$$100-12y=0$$

Subtract 100 from both sides. Anything subtracted from zero gives its negation.

$$-12y=-100$$

Divide both sides by -12.

$$y=\frac{-100}{-12}$$

Reduce the fraction \frac{-100}{-12} to lowest terms by extracting and canceling out -4.

$$y=\frac{25}{3}$$