Solve 3a+ab=4ac+1

$$3 a + a b = 4 a c + 1$$

Solve for a

$$a=\frac{1}{b-4c+3}$$

$$b\neq 4c-3$$

Steps for Solving Linear Equation

Subtract 4ac from both sides.

$$3a+ab-4ac=1$$

Combine all terms containing a.

$$\left(3+b-4c\right)a=1$$

The equation is in standard form.

$$\left(b-4c+3\right)a=1$$

Divide both sides by b-4c+3.

$$\frac{\left(b-4c+3\right)a}{b-4c+3}=\frac{1}{b-4c+3}$$

Dividing by b-4c+3 undoes the multiplication by b-4c+3.

$$a=\frac{1}{b-4c+3}$$

Solve for b

$$b=4c-3+\frac{1}{a}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Subtract 3a from both sides.

$$ab=4ac+1-3a$$

The equation is in standard form.

$$ab=4ac-3a+1$$

Divide both sides by a.

$$\frac{ab}{a}=\frac{4ac-3a+1}{a}$$

Dividing by a undoes the multiplication by a.

$$b=\frac{4ac-3a+1}{a}$$

Divide 4ac+1-3a by a.

$$b=4c-3+\frac{1}{a}$$