Solve (48/5)^2+(x-36/5)^2=x^2

$$( \frac { 48 } { 5 } ) ^ { 2 } + ( x - \frac { 36 } { 5 } ) ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=10$$

Steps for Solving Linear Equation

Calculate \frac{48}{5} to the power of 2 and get \frac{2304}{25}.

$$\frac{2304}{25}+\left(x-\frac{36}{5}\right)^{2}=x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{36}{5}\right)^{2}.

$$\frac{2304}{25}+x^{2}-\frac{72}{5}x+\frac{1296}{25}=x^{2}$$

Add \frac{2304}{25} and \frac{1296}{25} to get 144.

$$144+x^{2}-\frac{72}{5}x=x^{2}$$

Subtract x^{2} from both sides.

$$144+x^{2}-\frac{72}{5}x-x^{2}=0$$

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

$$144-\frac{72}{5}x=0$$

Subtract 144 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{72}{5}x=-144$$

Multiply both sides by -\frac{5}{72}, the reciprocal of -\frac{72}{5}.

$$x=-144\left(-\frac{5}{72}\right)$$

Multiply -144 and -\frac{5}{72} to get 10.

$$x=10$$