Question

Solve for $x:$
$\frac{3x-4}{\sqrt{3x}+2}=2+\frac{\sqrt{3x}-2}{2}$

Answer 1

We rationalise the denominator of $\left(\frac{3x-4}{\sqrt{3x}+2}\right)$ as shown below:

$\frac{3x-4}{\sqrt{3x}+2}\times \frac{\sqrt{3x}-2}{\sqrt{3x}-2}=\frac{(3x-4)(\sqrt{3x}-2)}{(\sqrt{3x}+2)(\sqrt{3x}-2)}=\frac{(3x-4)(\sqrt{3x}-2)}{(\sqrt{3x}{)}^2-(2{)}^2}=\frac{(3x-4)(\sqrt{3x}-2)}{3x-4}=\sqrt{3x}-2$

Now, consider the given problem $\frac{3x-4}{\sqrt{3x}+2}=2+\frac{\sqrt{3x}-2}{2}$ that can be simplified as follows:

$\frac{3x-4}{\sqrt{3x}+2}=2+\frac{\sqrt{3x}-2}{2}\Rightarrow \sqrt{3x}-2=2+\frac{\sqrt{3x}-2}{2}\Rightarrow \sqrt{3x}-2-\frac{\sqrt{3x}-2}{2}=2\Rightarrow \frac{2(\sqrt{3x}-2)-\sqrt{3x}-2}{2}=2\Rightarrow \frac{\sqrt{3x}-2}{2}=2\Rightarrow \sqrt{3x}-2=2\times 2\Rightarrow \sqrt{3x}-2=4\Rightarrow \sqrt{3x}=4+2\Rightarrow \sqrt{3x}=6\Rightarrow (\sqrt{3x}{)}^2=6^2\Rightarrow 3x=36\Rightarrow x=12$

Hence, $x=12$