How to multiply, divide and simplify rational expressions?

We multiply two rational expressions by multiplying their numerators and denominators as follows:

1)

\( \dfrac{A}{B} \cdot \dfrac{C}{D} = \dfrac{A \cdot C}{B \cdot D} \)

We divide two rational expressions by multiplying the first rational expression by the reciprocal of the second rational expression as follows:

2) \( \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C} \;\; \text{or} \;\; \dfrac{ \dfrac{A}{B} }{ \dfrac{C}{D}} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C} \)

Example 1: Multiply and simplify: \( \dfrac{2}{3x+6} \cdot \dfrac{x+2}{3x-7} \)

Solution:

Apply the multiplication rule (see above)

\( \dfrac{2}{3x+6} \cdot \dfrac{x+2}{3x-7} = \dfrac{2(x+2)}{(3x+6)(3x-7)} \)

Factor if possible

\( = \dfrac{2(x+2)}{3(x+2)(3x-7)} \)

Simplify if possible

\( = \dfrac{2{\colorcancel{red}{(x+2)}}}{3{\colorcancel{red}{(x+2)}}(3x-7)} = \dfrac{2}{3(3x-7)} , x \ne -2 \)

Example 2: Multiply and simplify: \( \dfrac{2x+4}{x-5} \cdot \dfrac{3x-15}{x+2} \)

Solution:

Apply the multiplication rule.

\( \dfrac{2x+4}{x-5} \cdot \dfrac{3x-15}{x+2} = \dfrac{(2x+4)(3x-15)}{(x-5)(x+2)} \)

Factor if possible

\( = \dfrac{2(x+2)3(x-5)}{(x-5)(x+2)} \)

Simplify if possible

\( = \dfrac{2{\colorcancel{red}{(x+2)}}3{\colorcancel{red}{(x-5)}}}{{\colorcancel{red}{(x-5)}}{{\colorcancel{red}{(x+2)}}}} = 6 \; \; \text{for} \; \; x \ne -2 \; \; \text{and} \; \; x \ne 5 \)

Example 3: Multiply and simplify: \( \dfrac{x+1}{4x^2-49y^2} \cdot \dfrac{2x-7y}{x^2-1} \).

Solution:

Multiply numerators and denominators (mulitplication rule)

\( \dfrac{x+1}{4x^2-49y^2} \cdot \dfrac{2x-7y}{x^2-1} = \dfrac{(x+1)(2x-7y)}{(4x^2-49y^2)(x^2-1)} \)

Factor the two terms in the denominator: 4 x^{ 2} - 49 y^{ 2} = (2x -7y)(2x + 7y) and x^{ 2} - 1 = (x - 1)(x + 1).

\( = \dfrac{(x+1)(2x-7y)}{(2x - 7y)(2x + 7y)(x - 1)(x + 1)} \)

Simplify if possible

\( = \dfrac{{\colorcancel{red}{(x + 1)}}{\colorcancel{red}{(2x-7y)}}}{{\colorcancel{red}{(2x-7y)}}(2x + 7y)(x - 1){\colorcancel{red}{(x + 1)}}} = \dfrac{1}{(2x + 7y)(x - 1)} \; \; \text{for} \; \; x \ne - 1 \; \; \text{and} \; \; x \ne -7y/2 \)

Example 4: Divide and simplify: \( \dfrac{x+2}{x+1} \div \dfrac{x+3}{x+1} \).

Solution:

The division of two rational expressions is done by multiplying the first rational expression by the reciprocal of the second rational expression as follows (see divison rule above). Hence

\( \dfrac{x+2}{x+1} \div \dfrac{x+3}{x+1} = \dfrac{x+2}{x+1} \cdot \dfrac{x+1}{x+3} \)

Multiply numerators and denominators (multiplication rule) but do not expand as we might be able to simplifty.

\( = \dfrac{(x+2)(x+1)}{(x+1)(x+3)} \)

Simplify if possible

\( = \dfrac{(x+2){\colorcancel{red}{(x+1)}}}{{\colorcancel{red}{(x+1)}}(x+3)} \)
\( = \dfrac{x+2}{x+3} \; \; \text{for} \; \; x \ne - 1 \)

Example 5: Divide and simplify: \( \dfrac{2x^2-3x-2}{x^2+x-6} \div \dfrac{x^2-2x-15}{x^2-x-20} \).

Solution:

The division of two rational expressions is done by multiplying the first by the reciprocal of the second as follows (see divison rule above). Hence

\( \dfrac{2x^2-3x-2}{x^2+x-6} \div \dfrac{x^2-2x-15}{x^2-x-20} = \dfrac{2x^2-3x-2}{x^2+x-6} \cdot \dfrac{x^2-x-20}{x^2-2x-15} \)

Multiply numerators and denominators (multiplication rule) but do not expand.

\( = \dfrac{(2x^2-3x-2)(x^2-x-20)}{(x^2+x-6)(x^2-2x-15)} \)

Factor the terms included in the numerator and denominator (if possible):

2 x^{ 2} - 3 x - 2 = (2x + 1)(x - 2) ; x^{ 2} - x - 20 = (x + 4)(x - 5)

x^{ 2} + x - 6 = (x + 3)(x - 2) ; x^{ 2} - 2 x - 15 = (x + 3)(x - 5)

and use the factored form in the rational expression to simplify

\( = \dfrac{(2x + 1)(x - 2)(x + 4)(x - 5)}{(x + 3)(x - 2)(x + 3)(x - 5)} \)
\( = \dfrac{(2x + 1){\colorcancel{red}{(x-2)}}(x + 4){\colorcancel{red}{(x-5)}}}{(x + 3){\colorcancel{red}{(x-2)}}(x + 3){\colorcancel{red}{(x-5)}}} \)
\( = \dfrac{(2x + 1)(x + 4)}{(x + 3)^2} \; \; \text{for} \; \; x \ne 2 \; \; \text{and} \; \; x \ne 5\)

Example 6: Divide and simplify: \( \dfrac{-2x+4}{x-1} \div (x - 2) \).

Solution:

We first convert (x - 2) into a rational expression. Hence

\( \dfrac{-2x+4}{x-1} \div (x - 2) = \dfrac{-2x+4}{x-1} \div \dfrac{x - 2}{1} \)

The division of two rational expressions is done by multiplying the first by the reciprocal of the second as follows (see divison rule above). Hence

\( = \dfrac{-2x+4}{x-1} \cdot \dfrac{1}{x - 2} \)

Multiply numerators and denominators (multiplication rule) but do not expand.

\( = \dfrac{(-2x+4) \cdot 1}{(x-1)(x-2)} \)

Factor the terms - 2 x + 4 included in the numerator as follows:

- 2 x + 4 = -2(x - 2)

and use - 2 x + 4 in factored form in the rational expression to simplify

\( = \dfrac{-2(x - 2) \cdot 1}{(x-1)(x-2)} \)
\( = \dfrac{-2{\colorcancel{red}{(x-2)}} \cdot 1}{(x-1){\colorcancel{red}{(x-2)}}} \)
\( = \dfrac{-2}{x - 1} \; \; \text{for} \; \; x \ne 2 \)

More Questions: Multiply and/or divide and simplify the given rational expressions.

Detailed Solutions and explanations to these questions.

a) \( \dfrac{-3}{2} \div \dfrac{6x-9}{2x - 3} \)

b) \( \dfrac{2x-5}{2x+2} \cdot \dfrac{10x+10}{4x-10} \)

c) \( \dfrac { \dfrac{2x^2-7x-15}{x^2+3x-4} }{ \dfrac{x^2+x-30}{x^2-1} } \)

d) \( (\dfrac{x-1}{x+2} \cdot \dfrac{x^2-4}{x^2-1}) \div \dfrac{x-2}{x+5} \)

e) \( \dfrac{ \dfrac{x^3-27}{x+3} }{ \dfrac{x-3}{(x+3)^2}} \)

f) \( \dfrac{2y-x}{4x^2-9y^2} \cdot \dfrac{4x+6y}{y-\dfrac{1}{2}x} \)

Detailed Solutions and explanations to these questions.