Solve 4a=4a-3ab+4b

$$4 a = 4 a - 3 a b + 4 b$$

Solve for a

$$\left\{\begin{matrix}\\a=\frac{4}{3}\approx 1.333333333\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&b=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 4a from both sides.

$$4a-4a=-3ab+4b$$

Combine 4a and -4a to get 0.

$$0=-3ab+4b$$

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

$$-3ab+4b=0$$

Subtract 4b from both sides. Anything subtracted from zero gives its negation.

$$-3ab=-4b$$

The equation is in standard form.

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

Divide both sides by -3b.

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

Dividing by -3b undoes the multiplication by -3b.

$$a=-\frac{4b}{-3b}$$

Divide -4b by -3b.

$$a=\frac{4}{3}$$

Solve for b

$$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=\frac{4}{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$4a-3ab+4b=4a$$

Subtract 4a from both sides.

$$-3ab+4b=4a-4a$$

Combine 4a and -4a to get 0.

$$-3ab+4b=0$$

Combine all terms containing b.

$$\left(-3a+4\right)b=0$$

The equation is in standard form.

$$\left(4-3a\right)b=0$$

Divide 0 by -3a+4.

$$b=0$$