Solve y=(x-1)^2-3x-(x-1)*(x+1)

$$y = ( x - 1 ) ^ { 2 } - 3 x - ( x - 1 ) \cdot ( x + 1 )$$

Solve for x

$$x=\frac{2-y}{5}$$

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}.

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

Combine -2x and -3x to get -5x.

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

Consider \left(x-1\right)\left(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.

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

To find the opposite of x^{2}-1, find the opposite of each term.

$$y=x^{2}-5x+1-x^{2}+1$$

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

$$y=-5x+1+1$$

Add 1 and 1 to get 2.

$$y=-5x+2$$

Swap sides so that all variable terms are on the left hand side.

$$-5x+2=y$$

Subtract 2 from both sides.

$$-5x=y-2$$

Divide both sides by -5.

$$\frac{-5x}{-5}=\frac{y-2}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x=\frac{y-2}{-5}$$

Divide y-2 by -5.

$$x=\frac{2-y}{5}$$

Solve for y

$$y=2-5x$$

Solution Steps

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

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

Combine -2x and -3x to get -5x.

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

Consider \left(x-1\right)\left(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.

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

To find the opposite of x^{2}-1, find the opposite of each term.

$$y=x^{2}-5x+1-x^{2}+1$$

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

$$y=-5x+1+1$$

Add 1 and 1 to get 2.

$$y=-5x+2$$