# Rationalize the denominator and simplify. All variables represent positive real

Rationalize the denominator and simplify. All variables represent positive real numbers.
$$\displaystyle{\frac{{{3}\sqrt{{{y}}}}}{{{2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Isma Jimenez

Rationalization is used to eliminate root radical in the denominator, for this both the numerator and denominator is multiplied by algebraic conjugate of denominator .For example if the denominator of an expression is $$\displaystyle{a}+\sqrt{{{b}}}$$ then the algebraic conjugate is $$\displaystyle{a}-\sqrt{{{b}}}$$. Now conjugate is multiplied to both numerator and denominator to get rationalized form of the expression.
In the given expression $$\displaystyle{\frac{{{3}\sqrt{{{y}}}}}{{{2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}}}}$$ the conjugate of denominator $$\displaystyle{\left({2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}\right)}$$ is $$(2\sqrt{x}+3\sqrt{y})$$
To get the rationalize form multiply the numerator and denominator by $$\displaystyle{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}$$
Use the algebraic identity $$\displaystyle{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{2}}-{b}^{{2}}$$ to simplify the denominator
Apply product rule of exponent $$\displaystyle{a}^{{m}}\cdot{a}^{{n}}={a}^{{{m}+{n}}}$$
$$\displaystyle{\frac{{{3}\sqrt{{{y}}}}}{{{2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}}}}={\frac{{{3}\sqrt{{{y}}}{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}}}{{{\left({2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}\right)}{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}}}}$$
$$\displaystyle={\frac{{{6}\sqrt{{{x}{y}}}+{9}{y}}}{{{4}{x}-{9}{y}}}}$$