Solve x/frac{sqrt{39-sqrt{3}}{8}}=6/frac{sqrt{39+sqrt{3}}{8}}

$$\frac{ x }{ \frac{ \sqrt{ 39 } - \sqrt{ 3 } }{ 8 } } = \frac{ 6 }{ \frac{ \sqrt{ 39 } + \sqrt{ 3 } }{ 8 } }$$

Solve for x

$$x=7-\sqrt{13}\approx 3.394448725$$

Steps for Solving Linear Equation

Divide x by \frac{\sqrt{39}-\sqrt{3}}{8} by multiplying x by the reciprocal of \frac{\sqrt{39}-\sqrt{3}}{8}.

$$\frac{x\times 8}{\sqrt{39}-\sqrt{3}}=\frac{6}{\frac{\sqrt{39}+\sqrt{3}}{8}}$$

Rationalize the denominator of \frac{x\times 8}{\sqrt{39}-\sqrt{3}} by multiplying numerator and denominator by \sqrt{39}+\sqrt{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{\left(\sqrt{39}-\sqrt{3}\right)\left(\sqrt{39}+\sqrt{3}\right)}=\frac{6}{\frac{\sqrt{39}+\sqrt{3}}{8}}$$

Consider \left(\sqrt{39}-\sqrt{3}\right)\left(\sqrt{39}+\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}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{\left(\sqrt{39}\right)^{2}-\left(\sqrt{3}\right)^{2}}=\frac{6}{\frac{\sqrt{39}+\sqrt{3}}{8}}$$

Square \sqrt{39}. Square \sqrt{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{39-3}=\frac{6}{\frac{\sqrt{39}+\sqrt{3}}{8}}$$

Subtract 3 from 39 to get 36.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{6}{\frac{\sqrt{39}+\sqrt{3}}{8}}$$

Divide 6 by \frac{\sqrt{39}+\sqrt{3}}{8} by multiplying 6 by the reciprocal of \frac{\sqrt{39}+\sqrt{3}}{8}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{6\times 8}{\sqrt{39}+\sqrt{3}}$$

Multiply 6 and 8 to get 48.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{48}{\sqrt{39}+\sqrt{3}}$$

Rationalize the denominator of \frac{48}{\sqrt{39}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{39}-\sqrt{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{48\left(\sqrt{39}-\sqrt{3}\right)}{\left(\sqrt{39}+\sqrt{3}\right)\left(\sqrt{39}-\sqrt{3}\right)}$$

Consider \left(\sqrt{39}+\sqrt{3}\right)\left(\sqrt{39}-\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}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{48\left(\sqrt{39}-\sqrt{3}\right)}{\left(\sqrt{39}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square \sqrt{39}. Square \sqrt{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{48\left(\sqrt{39}-\sqrt{3}\right)}{39-3}$$

Subtract 3 from 39 to get 36.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{48\left(\sqrt{39}-\sqrt{3}\right)}{36}$$

Divide 48\left(\sqrt{39}-\sqrt{3}\right) by 36 to get \frac{4}{3}\left(\sqrt{39}-\sqrt{3}\right).

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{4}{3}\left(\sqrt{39}-\sqrt{3}\right)$$

Use the distributive property to multiply \frac{4}{3} by \sqrt{39}-\sqrt{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{4}{3}\sqrt{39}+\frac{4}{3}\left(-1\right)\sqrt{3}$$

Multiply \frac{4}{3} and -1 to get -\frac{4}{3}.

$$\frac{x\times 8\left(\sqrt{39}+\sqrt{3}\right)}{36}=\frac{4}{3}\sqrt{39}-\frac{4}{3}\sqrt{3}$$

Use the distributive property to multiply x\times 8 by \sqrt{39}+\sqrt{3}.

$$\frac{8x\sqrt{39}+8x\sqrt{3}}{36}=\frac{4}{3}\sqrt{39}-\frac{4}{3}\sqrt{3}$$

Multiply both sides of the equation by 36, the least common multiple of 36,3.

$$8x\sqrt{39}+8x\sqrt{3}=48\sqrt{39}-48\sqrt{3}$$

Combine all terms containing x.

$$\left(8\sqrt{39}+8\sqrt{3}\right)x=48\sqrt{39}-48\sqrt{3}$$

The equation is in standard form.

$$\left(8\sqrt{3}+8\sqrt{39}\right)x=48\sqrt{39}-48\sqrt{3}$$

Divide both sides by 8\sqrt{39}+8\sqrt{3}.

$$\frac{\left(8\sqrt{3}+8\sqrt{39}\right)x}{8\sqrt{3}+8\sqrt{39}}=\frac{48\sqrt{39}-48\sqrt{3}}{8\sqrt{3}+8\sqrt{39}}$$

Dividing by 8\sqrt{39}+8\sqrt{3} undoes the multiplication by 8\sqrt{39}+8\sqrt{3}.

$$x=\frac{48\sqrt{39}-48\sqrt{3}}{8\sqrt{3}+8\sqrt{39}}$$

Divide 48\sqrt{39}-48\sqrt{3} by 8\sqrt{39}+8\sqrt{3}.

$$x=7-\sqrt{13}$$