Solve a/2sqrt{3+4}=c/sqrt{31+16sqrt{3}}

$$\frac { a } { 2 \sqrt { 3 } + 4 } = \frac { c } { \sqrt { 31 + 16 \sqrt { 3 } } }$$

Solve for a

$$a=\frac{2\sqrt{2316\sqrt{3}+4825}c}{193}$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{a}{2\sqrt{3}+4} by multiplying numerator and denominator by 2\sqrt{3}-4.

$$\frac{a\left(2\sqrt{3}-4\right)}{\left(2\sqrt{3}+4\right)\left(2\sqrt{3}-4\right)}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Consider \left(2\sqrt{3}+4\right)\left(2\sqrt{3}-4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{a\left(2\sqrt{3}-4\right)}{\left(2\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{a\left(2\sqrt{3}-4\right)}{2^{2}\left(\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a\left(2\sqrt{3}-4\right)}{4\left(\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

The square of \sqrt{3} is 3.

$$\frac{a\left(2\sqrt{3}-4\right)}{4\times 3-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Multiply 4 and 3 to get 12.

$$\frac{a\left(2\sqrt{3}-4\right)}{12-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{a\left(2\sqrt{3}-4\right)}{12-16}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Subtract 16 from 12 to get -4.

$$\frac{a\left(2\sqrt{3}-4\right)}{-4}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Use the distributive property to multiply a by 2\sqrt{3}-4.

$$\frac{2a\sqrt{3}-4a}{-4}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Multiply both sides of the equation by -4.

$$2a\sqrt{3}-4a=-\frac{4}{193}\left(5983-3088\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}c$$

Reorder the terms.

$$2\sqrt{3}a-4a=-\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}c$$

Combine all terms containing a.

$$\left(2\sqrt{3}-4\right)a=-\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}c$$

The equation is in standard form.

$$\left(2\sqrt{3}-4\right)a=-\frac{4\sqrt{5983-3088\sqrt{3}}c}{193}$$

Divide both sides by 2\sqrt{3}-4.

$$\frac{\left(2\sqrt{3}-4\right)a}{2\sqrt{3}-4}=-\frac{\frac{4\sqrt{5983-3088\sqrt{3}}c}{193}}{2\sqrt{3}-4}$$

Dividing by 2\sqrt{3}-4 undoes the multiplication by 2\sqrt{3}-4.

$$a=-\frac{\frac{4\sqrt{5983-3088\sqrt{3}}c}{193}}{2\sqrt{3}-4}$$

Divide -\frac{4c\sqrt{-3088\sqrt{3}+5983}}{193} by 2\sqrt{3}-4.

$$a=\frac{2\sqrt{2316\sqrt{3}+4825}c}{193}$$

Solve for c

$$c=\frac{\sqrt{25-12\sqrt{3}}a}{2}$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{a}{2\sqrt{3}+4} by multiplying numerator and denominator by 2\sqrt{3}-4.

$$\frac{a\left(2\sqrt{3}-4\right)}{\left(2\sqrt{3}+4\right)\left(2\sqrt{3}-4\right)}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Consider \left(2\sqrt{3}+4\right)\left(2\sqrt{3}-4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{a\left(2\sqrt{3}-4\right)}{\left(2\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{a\left(2\sqrt{3}-4\right)}{2^{2}\left(\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a\left(2\sqrt{3}-4\right)}{4\left(\sqrt{3}\right)^{2}-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

The square of \sqrt{3} is 3.

$$\frac{a\left(2\sqrt{3}-4\right)}{4\times 3-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Multiply 4 and 3 to get 12.

$$\frac{a\left(2\sqrt{3}-4\right)}{12-4^{2}}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{a\left(2\sqrt{3}-4\right)}{12-16}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Subtract 16 from 12 to get -4.

$$\frac{a\left(2\sqrt{3}-4\right)}{-4}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

Use the distributive property to multiply a by 2\sqrt{3}-4.

$$\frac{2a\sqrt{3}-4a}{-4}=\frac{c}{\sqrt{31+16\sqrt{3}}}$$

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

$$\frac{c}{\sqrt{31+16\sqrt{3}}}=\frac{2a\sqrt{3}-4a}{-4}$$

Multiply both sides of the equation by -4.

$$-\frac{4}{193}\left(5983-3088\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}c=2a\sqrt{3}-4a$$

Reorder the terms.

$$-\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}c=2\sqrt{3}a-4a$$

The equation is in standard form.

$$\left(-\frac{4\sqrt{5983-3088\sqrt{3}}}{193}\right)c=2\sqrt{3}a-4a$$

Divide both sides by -\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}.

$$\frac{\left(-\frac{4\sqrt{5983-3088\sqrt{3}}}{193}\right)c}{-\frac{4\sqrt{5983-3088\sqrt{3}}}{193}}=\frac{2\left(\sqrt{3}-2\right)a}{-\frac{4\sqrt{5983-3088\sqrt{3}}}{193}}$$

Dividing by -\frac{4}{193}\sqrt{-3088\sqrt{3}+5983} undoes the multiplication by -\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}.

$$c=\frac{2\left(\sqrt{3}-2\right)a}{-\frac{4\sqrt{5983-3088\sqrt{3}}}{193}}$$

Divide 2a\left(\sqrt{3}-2\right) by -\frac{4}{193}\sqrt{-3088\sqrt{3}+5983}.

$$c=\frac{\sqrt{25-12\sqrt{3}}a}{2}$$