Solve 4^2+left(6-xright)^2=x^2

$${ 4 }^{ 2 } + { \left(6-x \right) }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x = \frac{13}{3} = 4\frac{1}{3} \approx 4.333333333$$

Steps for Solving Linear Equation

Calculate 4 to the power of 2 and get 16.

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

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

$$16+36-12x+x^{2}=x^{2}$$

Add 16 and 36 to get 52.

$$52-12x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$52-12x+x^{2}-x^{2}=0$$

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

$$52-12x=0$$

Subtract 52 from both sides. Anything subtracted from zero gives its negation.

$$-12x=-52$$

Divide both sides by -12.

$$x=\frac{-52}{-12}$$

Reduce the fraction \frac{-52}{-12} to lowest terms by extracting and canceling out -4.

$$x=\frac{13}{3}$$