Solve 3a+ab+bsqrt{2}-3b=6

$$3 a + a b + b \sqrt { 2 } - 3 b = 6$$

Solve for a

$$a=-\frac{\sqrt{2}b-3b-6}{b+3}$$

$$b\neq -3$$

Steps for Solving Linear Equation

Subtract b\sqrt{2} from both sides.

$$3a+ab-3b=6-b\sqrt{2}$$

Add 3b to both sides.

$$3a+ab=6-b\sqrt{2}+3b$$

Reorder the terms.

$$ab+3a=-\sqrt{2}b+3b+6$$

Combine all terms containing a.

$$\left(b+3\right)a=-\sqrt{2}b+3b+6$$

Divide both sides by 3+b.

$$\frac{\left(b+3\right)a}{b+3}=\frac{-\sqrt{2}b+3b+6}{b+3}$$

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

$$a=\frac{-\sqrt{2}b+3b+6}{b+3}$$

Solve for b

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

$$a\neq 3-\sqrt{2}$$

Steps for Solving Linear Equation

Subtract 3a from both sides.

$$ab+b\sqrt{2}-3b=6-3a$$

Combine all terms containing b.

$$\left(a+\sqrt{2}-3\right)b=6-3a$$

Divide both sides by a+\sqrt{2}-3.

$$\frac{\left(a+\sqrt{2}-3\right)b}{a+\sqrt{2}-3}=\frac{6-3a}{a+\sqrt{2}-3}$$

Dividing by a+\sqrt{2}-3 undoes the multiplication by a+\sqrt{2}-3.

$$b=\frac{6-3a}{a+\sqrt{2}-3}$$

Divide 6-3a by a+\sqrt{2}-3.

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