Solve (2+2sqrt{2})a+(3+sqrt{3})b+3=0

$$( 2 + 2 \sqrt { 2 } ) a + ( 3 + \sqrt { 3 } ) b + 3 = 0$$

Solve for a

$$a=-\frac{\sqrt{3}\left(\sqrt{2}-1\right)\left(\sqrt{3}b+b+\sqrt{3}\right)}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2+2\sqrt{2} by a.

$$2a+2\sqrt{2}a+\left(3+\sqrt{3}\right)b+3=0$$

Use the distributive property to multiply 3+\sqrt{3} by b.

$$2a+2\sqrt{2}a+3b+\sqrt{3}b+3=0$$

Subtract 3b from both sides. Anything subtracted from zero gives its negation.

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

Subtract \sqrt{3}b from both sides.

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

Subtract 3 from both sides.

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

Reorder the terms.

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

Combine all terms containing a.

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

Divide both sides by 2+2\sqrt{2}.

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

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

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

Divide -\sqrt{3}b-3b-3 by 2+2\sqrt{2}.

$$a=-\frac{\left(\sqrt{2}-1\right)\left(\sqrt{3}b+3b+3\right)}{2}$$

Solve for b

$$b=-\frac{\sqrt{3}\left(\sqrt{3}-1\right)\left(2\sqrt{2}a+2a+3\right)}{6}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2+2\sqrt{2} by a.

$$2a+2\sqrt{2}a+\left(3+\sqrt{3}\right)b+3=0$$

Use the distributive property to multiply 3+\sqrt{3} by b.

$$2a+2\sqrt{2}a+3b+\sqrt{3}b+3=0$$

Subtract 2a from both sides. Anything subtracted from zero gives its negation.

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

Subtract 2\sqrt{2}a from both sides.

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

Subtract 3 from both sides.

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

Combine all terms containing b.

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

The equation is in standard form.

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

Divide both sides by 3+\sqrt{3}.

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

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

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

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

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