Solve frac{sqrt{3}}{frac{sqrt{3}}{2}}=frac{b/sqrt{6-sqrt{2}}}{4}

$$\frac { \sqrt { 3 } } { \frac { \sqrt { 3 } } { 2 } } = \frac { \frac { b } { \sqrt { 6 } - \sqrt { 2 } } } { 4 }$$

Solve for b

$$b=8\left(\sqrt{6}-\sqrt{2}\right)\approx 8.282209443$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b}{\sqrt{6}-\sqrt{2}}$$

Rationalize the denominator of \frac{b}{\sqrt{6}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}+\sqrt{2}.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}$$

Consider \left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\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}.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Square \sqrt{6}. Square \sqrt{2}.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b\left(\sqrt{6}+\sqrt{2}\right)}{6-2}$$

Subtract 2 from 6 to get 4.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b\left(\sqrt{6}+\sqrt{2}\right)}{4}$$

Use the distributive property to multiply b by \sqrt{6}+\sqrt{2}.

$$\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}=\frac{b\sqrt{6}+b\sqrt{2}}{4}$$

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

$$\frac{b\sqrt{6}+b\sqrt{2}}{4}=\frac{8}{3}\times 3^{\frac{1}{2}}\sqrt{3}$$

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3\left(b\sqrt{6}+b\sqrt{2}\right)=32\times 3^{\frac{1}{2}}\sqrt{3}$$

Reorder the terms.

$$3\left(\sqrt{2}b+\sqrt{6}b\right)=32\sqrt{3}\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\left(\sqrt{2}b+\sqrt{6}b\right)=32\times 3$$

Use the distributive property to multiply 3 by \sqrt{2}b+\sqrt{6}b.

$$3\sqrt{2}b+3\sqrt{6}b=32\times 3$$

Multiply 32 and 3 to get 96.

$$3\sqrt{2}b+3\sqrt{6}b=96$$

Combine all terms containing b.

$$\left(3\sqrt{2}+3\sqrt{6}\right)b=96$$

Divide both sides by 3\sqrt{2}+3\sqrt{6}.

$$\frac{\left(3\sqrt{2}+3\sqrt{6}\right)b}{3\sqrt{2}+3\sqrt{6}}=\frac{96}{3\sqrt{2}+3\sqrt{6}}$$

Dividing by 3\sqrt{2}+3\sqrt{6} undoes the multiplication by 3\sqrt{2}+3\sqrt{6}.

$$b=\frac{96}{3\sqrt{2}+3\sqrt{6}}$$

Divide 96 by 3\sqrt{2}+3\sqrt{6}.

$$b=8\sqrt{6}-8\sqrt{2}$$