Solve a+2sqrt{2}=3+bsqrt{2}

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

Solve for b

$$b=\frac{\sqrt{2}a+4-3\sqrt{2}}{2}$$

Steps for Solving Linear Equation

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

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

Subtract 3 from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{2}.

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

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

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

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

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

Solve for a

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