Solve 1/3(x-3)^2+1/6(x+2)^2=(1/2x-1)(1/2x+1)+frac{1}{4}(x-2)^2

$$\frac { 1 } { 3 } ( x - 3 ) ^ { 2 } + \frac { 1 } { 6 } ( x + 2 ) ^ { 2 } = ( \frac { 1 } { 2 } x - 1 ) ( \frac { 1 } { 2 } x + 1 ) + \frac { 1 } { 4 } ( x - 2 ) ^ { 2 }$$

Solve for x

$$x=11$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\frac{1}{3}\left(x^{2}-6x+9\right)+\frac{1}{6}\left(x+2\right)^{2}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Use the distributive property to multiply \frac{1}{3} by x^{2}-6x+9.

$$\frac{1}{3}x^{2}-2x+3+\frac{1}{6}\left(x+2\right)^{2}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

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

$$\frac{1}{3}x^{2}-2x+3+\frac{1}{6}\left(x^{2}+4x+4\right)=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Use the distributive property to multiply \frac{1}{6} by x^{2}+4x+4.

$$\frac{1}{3}x^{2}-2x+3+\frac{1}{6}x^{2}+\frac{2}{3}x+\frac{2}{3}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Combine \frac{1}{3}x^{2} and \frac{1}{6}x^{2} to get \frac{1}{2}x^{2}.

$$\frac{1}{2}x^{2}-2x+3+\frac{2}{3}x+\frac{2}{3}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Combine -2x and \frac{2}{3}x to get -\frac{4}{3}x.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+3+\frac{2}{3}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Add 3 and \frac{2}{3} to get \frac{11}{3}.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right)+\frac{1}{4}\left(x-2\right)^{2}$$

Consider \left(\frac{1}{2}x-1\right)\left(\frac{1}{2}x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\left(\frac{1}{2}x\right)^{2}-1+\frac{1}{4}\left(x-2\right)^{2}$$

Expand \left(\frac{1}{2}x\right)^{2}.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\left(\frac{1}{2}\right)^{2}x^{2}-1+\frac{1}{4}\left(x-2\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\frac{1}{4}x^{2}-1+\frac{1}{4}\left(x-2\right)^{2}$$

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

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\frac{1}{4}x^{2}-1+\frac{1}{4}\left(x^{2}-4x+4\right)$$

Use the distributive property to multiply \frac{1}{4} by x^{2}-4x+4.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\frac{1}{4}x^{2}-1+\frac{1}{4}x^{2}-x+1$$

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

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\frac{1}{2}x^{2}-1-x+1$$

Add -1 and 1 to get 0.

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}=\frac{1}{2}x^{2}-x$$

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

$$\frac{1}{2}x^{2}-\frac{4}{3}x+\frac{11}{3}-\frac{1}{2}x^{2}=-x$$

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

$$-\frac{4}{3}x+\frac{11}{3}=-x$$

Add x to both sides.

$$-\frac{4}{3}x+\frac{11}{3}+x=0$$

Combine -\frac{4}{3}x and x to get -\frac{1}{3}x.

$$-\frac{1}{3}x+\frac{11}{3}=0$$

Subtract \frac{11}{3} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{3}x=-\frac{11}{3}$$

Multiply both sides by -3, the reciprocal of -\frac{1}{3}.

$$x=-\frac{11}{3}\left(-3\right)$$

Multiply -\frac{11}{3} and -3 to get 11.

$$x=11$$