Solve sqrt{3}y+16y/8-4sqrt{3}=120

$$\sqrt { 3 } y + \frac { 16 y } { 8 - 4 \sqrt { 3 } } = 120$$

Solve for y

$$y = \frac{600 \sqrt{3} - 960}{11} \approx 7.202771322$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{16y}{8-4\sqrt{3}} by multiplying numerator and denominator by 8+4\sqrt{3}.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{\left(8-4\sqrt{3}\right)\left(8+4\sqrt{3}\right)}=120$$

Consider \left(8-4\sqrt{3}\right)\left(8+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}.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{8^{2}-\left(-4\sqrt{3}\right)^{2}}=120$$

Calculate 8 to the power of 2 and get 64.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{64-\left(-4\sqrt{3}\right)^{2}}=120$$

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

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{64-\left(-4\right)^{2}\left(\sqrt{3}\right)^{2}}=120$$

Calculate -4 to the power of 2 and get 16.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{64-16\left(\sqrt{3}\right)^{2}}=120$$

The square of \sqrt{3} is 3.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{64-16\times 3}=120$$

Multiply 16 and 3 to get 48.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{64-48}=120$$

Subtract 48 from 64 to get 16.

$$\sqrt{3}y+\frac{16y\left(8+4\sqrt{3}\right)}{16}=120$$

Cancel out 16 and 16.

$$\sqrt{3}y+y\left(8+4\sqrt{3}\right)=120$$

Use the distributive property to multiply y by 8+4\sqrt{3}.

$$\sqrt{3}y+8y+4y\sqrt{3}=120$$

Combine \sqrt{3}y and 4y\sqrt{3} to get 5\sqrt{3}y.

$$5\sqrt{3}y+8y=120$$

Combine all terms containing y.

$$\left(5\sqrt{3}+8\right)y=120$$

Divide both sides by 5\sqrt{3}+8.

$$\frac{\left(5\sqrt{3}+8\right)y}{5\sqrt{3}+8}=\frac{120}{5\sqrt{3}+8}$$

Dividing by 5\sqrt{3}+8 undoes the multiplication by 5\sqrt{3}+8.

$$y=\frac{120}{5\sqrt{3}+8}$$

Divide 120 by 5\sqrt{3}+8.

$$y=\frac{600\sqrt{3}-960}{11}$$