Solve a(b-4)-a(b-1)=a(ay-b)

$$a ( b - 4 ) - a ( b - 1 ) = a ( a y - b )$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=ay+3\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by b-4.

$$ab-4a-a\left(b-1\right)=a\left(ay-b\right)$$

Use the distributive property to multiply a by b-1.

$$ab-4a-\left(ab-a\right)=a\left(ay-b\right)$$

To find the opposite of ab-a, find the opposite of each term.

$$ab-4a-ab+a=a\left(ay-b\right)$$

Combine ab and -ab to get 0.

$$-4a+a=a\left(ay-b\right)$$

Combine -4a and a to get -3a.

$$-3a=a\left(ay-b\right)$$

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

$$-3a=ya^{2}-ab$$

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

$$ya^{2}-ab=-3a$$

Subtract ya^{2} from both sides.

$$-ab=-3a-ya^{2}$$

Reorder the terms.

$$-ab=-ya^{2}-3a$$

The equation is in standard form.

$$\left(-a\right)b=-ya^{2}-3a$$

Divide both sides by -a.

$$\frac{\left(-a\right)b}{-a}=-\frac{a\left(ay+3\right)}{-a}$$

Dividing by -a undoes the multiplication by -a.

$$b=-\frac{a\left(ay+3\right)}{-a}$$

Divide -a\left(ya+3\right) by -a.

$$b=ay+3$$

Solve for b

$$\left\{\begin{matrix}\\b=ay+3\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by b-4.

$$ab-4a-a\left(b-1\right)=a\left(ay-b\right)$$

Use the distributive property to multiply a by b-1.

$$ab-4a-\left(ab-a\right)=a\left(ay-b\right)$$

To find the opposite of ab-a, find the opposite of each term.

$$ab-4a-ab+a=a\left(ay-b\right)$$

Combine ab and -ab to get 0.

$$-4a+a=a\left(ay-b\right)$$

Combine -4a and a to get -3a.

$$-3a=a\left(ay-b\right)$$

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

$$-3a=ya^{2}-ab$$

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

$$ya^{2}-ab=-3a$$

Subtract ya^{2} from both sides.

$$-ab=-3a-ya^{2}$$

Reorder the terms.

$$-ab=-ya^{2}-3a$$

The equation is in standard form.

$$\left(-a\right)b=-ya^{2}-3a$$

Divide both sides by -a.

$$\frac{\left(-a\right)b}{-a}=-\frac{a\left(ay+3\right)}{-a}$$

Dividing by -a undoes the multiplication by -a.

$$b=-\frac{a\left(ay+3\right)}{-a}$$

Divide -a\left(ya+3\right) by -a.

$$b=ay+3$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a=\frac{b-3}{y}\text{, }&y\neq 0\\a\in \mathrm{C}\text{, }&b=3\text{ and }y=0\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a=\frac{b-3}{y}\text{, }&y\neq 0\\a\in \mathrm{R}\text{, }&b=3\text{ and }y=0\end{matrix}\right.$$