Solve 3/sqrt{3-sqrt{2}}=asqrt{3}-bsqrt{2}

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

Solve for a

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

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3}{\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{2}.

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Use the distributive property to multiply 3 by \sqrt{3}+\sqrt{2}.

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

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

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

Add b\sqrt{2} to both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

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

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

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

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

Solve for b

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

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3}{\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{2}.

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Use the distributive property to multiply 3 by \sqrt{3}+\sqrt{2}.

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

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

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

Subtract a\sqrt{3} from both sides.

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

The equation is in standard form.

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

Divide both sides by -\sqrt{2}.

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

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

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

Divide \left(3+\sqrt{6}-a\right)\sqrt{3} by -\sqrt{2}.

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