Solve 2/3a-b/b=a+b=

$$\frac { 2 } { 3 } a - \frac { b } { b } = a + b =$$

Solve for a

$$a=-3b-3$$

$$b\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3b, the least common multiple of 3,b.

$$\frac{2}{3}a\times 3b-3b=3ba+3bb$$

Multiply \frac{2}{3} and 3 to get 2.

$$2ab-3b=3ba+3bb$$

Multiply b and b to get b^{2}.

$$2ab-3b=3ba+3b^{2}$$

Subtract 3ba from both sides.

$$2ab-3b-3ba=3b^{2}$$

Combine 2ab and -3ba to get -ab.

$$-ab-3b=3b^{2}$$

Add 3b to both sides.

$$-ab=3b^{2}+3b$$

The equation is in standard form.

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

Divide both sides by -b.

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

Dividing by -b undoes the multiplication by -b.

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

Divide 3b\left(1+b\right) by -b.

$$a=-3b-3$$

Solve for b

$$b=-\frac{a}{3}-1$$

$$a\neq -3$$