Solve N=1/7+4sqrt{3}

$$N = \frac { 1 } { 7 + 4 \sqrt { 3 } }$$

Solve for N

$$N=7-4\sqrt{3}\approx 0.07179677$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{1}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$N=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$N=\frac{7-4\sqrt{3}}{7^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$N=\frac{7-4\sqrt{3}}{49-\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$N=\frac{7-4\sqrt{3}}{49-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$N=\frac{7-4\sqrt{3}}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$N=\frac{7-4\sqrt{3}}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$N=\frac{7-4\sqrt{3}}{49-48}$$

Subtract 48 from 49 to get 1.

$$N=\frac{7-4\sqrt{3}}{1}$$

Anything divided by one gives itself.

$$N=7-4\sqrt{3}$$

Assign N

$$N≔7-4\sqrt{3}$$