Solve (x-1)^2+(3+3x-5)^2=(x-1)^2+(2+3x-6)^2

$$( x - 1 ) ^ { 2 } + ( 3 + 3 x - 5 ) ^ { 2 } = ( x - 1 ) ^ { 2 } + ( 2 + 3 x - 6 ) ^ { 2 }$$

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(x-1\right)^{2}.

$$x^{2}-2x+1+\left(3+3x-5\right)^{2}=\left(x-1\right)^{2}+\left(2+3x-6\right)^{2}$$

Subtract 5 from 3 to get -2.

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

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

$$x^{2}-2x+1+4-12x+9x^{2}=\left(x-1\right)^{2}+\left(2+3x-6\right)^{2}$$

Add 1 and 4 to get 5.

$$x^{2}-2x+5-12x+9x^{2}=\left(x-1\right)^{2}+\left(2+3x-6\right)^{2}$$

Combine -2x and -12x to get -14x.

$$x^{2}-14x+5+9x^{2}=\left(x-1\right)^{2}+\left(2+3x-6\right)^{2}$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-14x+5=\left(x-1\right)^{2}+\left(2+3x-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$10x^{2}-14x+5=x^{2}-2x+1+\left(2+3x-6\right)^{2}$$

Subtract 6 from 2 to get -4.

$$10x^{2}-14x+5=x^{2}-2x+1+\left(-4+3x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4+3x\right)^{2}.

$$10x^{2}-14x+5=x^{2}-2x+1+16-24x+9x^{2}$$

Add 1 and 16 to get 17.

$$10x^{2}-14x+5=x^{2}-2x+17-24x+9x^{2}$$

Combine -2x and -24x to get -26x.

$$10x^{2}-14x+5=x^{2}-26x+17+9x^{2}$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-14x+5=10x^{2}-26x+17$$

Subtract 10x^{2} from both sides.

$$10x^{2}-14x+5-10x^{2}=-26x+17$$

Combine 10x^{2} and -10x^{2} to get 0.

$$-14x+5=-26x+17$$

Add 26x to both sides.

$$-14x+5+26x=17$$

Combine -14x and 26x to get 12x.

$$12x+5=17$$

Subtract 5 from both sides.

$$12x=17-5$$

Subtract 5 from 17 to get 12.

$$12x=12$$

Divide both sides by 12.

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

Divide 12 by 12 to get 1.

$$x=1$$