Solve 6^2+(16-x)^2=x^2

$$6 ^ { 2 } + ( 16 - x ) ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x = \frac{73}{8} = 9\frac{1}{8} = 9.125$$

Steps for Solving Linear Equation

Calculate 6 to the power of 2 and get 36.

$$36+\left(16-x\right)^{2}=x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16-x\right)^{2}.

$$36+256-32x+x^{2}=x^{2}$$

Add 36 and 256 to get 292.

$$292-32x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$292-32x+x^{2}-x^{2}=0$$

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

$$292-32x=0$$

Subtract 292 from both sides. Anything subtracted from zero gives its negation.

$$-32x=-292$$

Divide both sides by -32.

$$x=\frac{-292}{-32}$$

Reduce the fraction \frac{-292}{-32} to lowest terms by extracting and canceling out -4.

$$x=\frac{73}{8}$$