Solve 9+{left(9+x/2-xright)}^2={left(9+x/2right)}^2

$$9+ { \left( \frac{ 9+x }{ 2 } -x \right) }^{ 2 } = { \left( \frac{ 9+x }{ 2 } \right) }^{ 2 }$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{9+x}{2}-x\right)^{2}.

$$9+\left(\frac{9+x}{2}\right)^{2}-2\times \frac{9+x}{2}x+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

To raise \frac{9+x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}-2\times \frac{9+x}{2}x+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

Express -2\times \frac{9+x}{2} as a single fraction.

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}+\frac{-2\left(9+x\right)}{2}x+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

Cancel out 2 and 2.

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}-\left(9+x\right)x+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

Use the distributive property to multiply -1 by 9+x.

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}+\left(-9-x\right)x+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

Use the distributive property to multiply -9-x by x.

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}-9x-x^{2}+x^{2}=\left(\frac{9+x}{2}\right)^{2}$$

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

$$9+\frac{\left(9+x\right)^{2}}{2^{2}}-9x=\left(\frac{9+x}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9-9x times \frac{2^{2}}{2^{2}}.

$$\frac{\left(9-9x\right)\times 2^{2}}{2^{2}}+\frac{\left(9+x\right)^{2}}{2^{2}}=\left(\frac{9+x}{2}\right)^{2}$$

Since \frac{\left(9-9x\right)\times 2^{2}}{2^{2}} and \frac{\left(9+x\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(9-9x\right)\times 2^{2}+\left(9+x\right)^{2}}{2^{2}}=\left(\frac{9+x}{2}\right)^{2}$$

Do the multiplications in \left(9-9x\right)\times 2^{2}+\left(9+x\right)^{2}.

$$\frac{36-36x+81+18x+x^{2}}{2^{2}}=\left(\frac{9+x}{2}\right)^{2}$$

Combine like terms in 36-36x+81+18x+x^{2}.

$$\frac{117-18x+x^{2}}{2^{2}}=\left(\frac{9+x}{2}\right)^{2}$$

To raise \frac{9+x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{117-18x+x^{2}}{2^{2}}=\frac{\left(9+x\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{117-18x+x^{2}}{4}=\frac{\left(9+x\right)^{2}}{2^{2}}$$

Divide each term of 117-18x+x^{2} by 4 to get \frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}.

$$\frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}=\frac{\left(9+x\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(9+x\right)^{2}.

$$\frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}=\frac{81+18x+x^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}=\frac{81+18x+x^{2}}{4}$$

Divide each term of 81+18x+x^{2} by 4 to get \frac{81}{4}+\frac{9}{2}x+\frac{1}{4}x^{2}.

$$\frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}=\frac{81}{4}+\frac{9}{2}x+\frac{1}{4}x^{2}$$

Subtract \frac{9}{2}x from both sides.

$$\frac{117}{4}-\frac{9}{2}x+\frac{1}{4}x^{2}-\frac{9}{2}x=\frac{81}{4}+\frac{1}{4}x^{2}$$

Combine -\frac{9}{2}x and -\frac{9}{2}x to get -9x.

$$\frac{117}{4}-9x+\frac{1}{4}x^{2}=\frac{81}{4}+\frac{1}{4}x^{2}$$

Subtract \frac{1}{4}x^{2} from both sides.

$$\frac{117}{4}-9x+\frac{1}{4}x^{2}-\frac{1}{4}x^{2}=\frac{81}{4}$$

Combine \frac{1}{4}x^{2} and -\frac{1}{4}x^{2} to get 0.

$$\frac{117}{4}-9x=\frac{81}{4}$$

Subtract \frac{117}{4} from both sides.

$$-9x=\frac{81}{4}-\frac{117}{4}$$

Subtract \frac{117}{4} from \frac{81}{4} to get -9.

$$-9x=-9$$

Divide both sides by -9.

$$x=\frac{-9}{-9}$$

Divide -9 by -9 to get 1.

$$x=1$$