Solve (a+b)(a-b)=b(a-b)

$$( a + b ) ( a - b ) = b ( a - b )$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=a\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}-b^{2}=b\left(a-b\right)$$

Use the distributive property to multiply b by a-b.

$$a^{2}-b^{2}=ba-b^{2}$$

Subtract ba from both sides.

$$a^{2}-b^{2}-ba=-b^{2}$$

Add b^{2} to both sides.

$$a^{2}-b^{2}-ba+b^{2}=0$$

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

$$a^{2}-ba=0$$

Subtract a^{2} from both sides. Anything subtracted from zero gives its negation.

$$-ba=-a^{2}$$

Cancel out -1 on both sides.

$$ba=a^{2}$$

The equation is in standard form.

$$ab=a^{2}$$

Divide both sides by a.

$$\frac{ab}{a}=\frac{a^{2}}{a}$$

Dividing by a undoes the multiplication by a.

$$b=\frac{a^{2}}{a}$$

Divide a^{2} by a.

$$b=a$$

Solve for b

$$\left\{\begin{matrix}\\b=a\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}-b^{2}=b\left(a-b\right)$$

Use the distributive property to multiply b by a-b.

$$a^{2}-b^{2}=ba-b^{2}$$

Subtract ba from both sides.

$$a^{2}-b^{2}-ba=-b^{2}$$

Add b^{2} to both sides.

$$a^{2}-b^{2}-ba+b^{2}=0$$

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

$$a^{2}-ba=0$$

Subtract a^{2} from both sides. Anything subtracted from zero gives its negation.

$$-ba=-a^{2}$$

Cancel out -1 on both sides.

$$ba=a^{2}$$

The equation is in standard form.

$$ab=a^{2}$$

Divide both sides by a.

$$\frac{ab}{a}=\frac{a^{2}}{a}$$

Dividing by a undoes the multiplication by a.

$$b=\frac{a^{2}}{a}$$

Divide a^{2} by a.

$$b=a$$

Solve for a

$$a=b$$

$$a=0$$