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

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

Solve for b

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

Steps for Solving Linear Equation

Rationalize the denominator of \frac{\sqrt{3}}{3+\sqrt{3}} by multiplying numerator and denominator by 3-\sqrt{3}.

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

Consider \left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 3. Square \sqrt{3}.

$$a+b\sqrt{3}=\frac{\sqrt{3}\left(3-\sqrt{3}\right)}{9-3}$$

Subtract 3 from 9 to get 6.

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

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

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

The square of \sqrt{3} is 3.

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

Divide each term of 3\sqrt{3}-3 by 6 to get \frac{1}{2}\sqrt{3}-\frac{1}{2}.

$$a+b\sqrt{3}=\frac{1}{2}\sqrt{3}-\frac{1}{2}$$

Subtract a from both sides.

$$b\sqrt{3}=\frac{1}{2}\sqrt{3}-\frac{1}{2}-a$$

The equation is in standard form.

$$\sqrt{3}b=-a+\frac{\sqrt{3}}{2}-\frac{1}{2}$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+\frac{\sqrt{3}}{2}-\frac{1}{2}}{\sqrt{3}}$$

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

$$b=\frac{-a+\frac{\sqrt{3}}{2}-\frac{1}{2}}{\sqrt{3}}$$

Divide \frac{\sqrt{3}}{2}-\frac{1}{2}-a by \sqrt{3}.

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

Solve for a

$$a=\frac{-2\sqrt{3}b+\sqrt{3}-1}{2}$$