Adding, Subtracting, Multiplying, and Dividing Rational Expressions

Adding, Subtracting, Multiplying, and Dividing Rational Expressions exercise

• Explain how to add or subtract rational expressions.

  Hints

  We have that $\frac{5}{6x}+\frac3{2x^2}=\frac{5x+9}{6x^2}$.

  Let's have a look at another example:

  $\frac32-\frac53$.

  You can only combine (add or subtract) fraction with the same denominator. So in order to make these two fractions have the same denominator, expand the first fraction by $3$ and the second by $2$:

  $\frac32-\frac53=\frac{(3)(3)}{(2)(3)}-\frac{(5)(2)}{(3)(2)}=\frac{9}{6}-\frac{10}{6}=-\frac16$.

  Here is an example for factoring a given term.

  Solution

  To add or subtract rational expressions we first have to find the LCD, or

  the Least Common Denominator.

  How can we find it?

  1. First we factor the denominators of the given fractions.
  2. Next we find the LCD by taking all given factors of the denominators. But only take factors which appear in both denominators only once.

  Let's have a look at an example:

  $\frac{5}{6x}+\frac3{2x^2}$

  1. $6x=(2)(3)(x)$ and $2x^2=(2)(x)(x)$, so the factors $2$ as well as $x$ appear in both denominators. So we take each of these only once.
  2. The corresponding LCD is given by $(2)(3)(x)(x)=6x^2$.

  Now we expand each fraction such that they have a common denominator, namely the LCD:

  • We expand the first fraction by $x$ to get $\frac{5}{6x}=\frac{(5)(x)}{(6x)(x)}=\frac{5x}{6x^2}$
  • We expand the second one by $3$, which results in $\frac{3}{2x^2}=\frac{(3)(3)}{(2x^2)(3)}=\frac{9}{6x^2}$

  We can add or subtract fractions with the same denominator by adding or subtracting the numerators:

  $\frac{5}{6x}+\frac3{2x^2}=\frac{5x}{6x^2}+\frac{9}{6x^2}=\frac{5x+9}{6x^2}$.

• Multiply and divide the following rational expressions.

  Hints

  Here is an example of multiplying two fractions:

  To multiply two binomials use the FOIL method:

  $(x+1)(x+2)=x\times x+x\times 2+1\times x+1\times 2=x^2+3x+2$.

  Here is an example of the dividing two fractions:

  $\begin{array}{rclll} \frac{2}{3}\div\frac{5}{4}&=&~\frac23\times \frac45&|&\text{multiply with the flipped second fraction}\\ &=&~\frac{(2)(4)}{(3)(5)}\\ &=&~\frac{8}{15} \end{array}$

  Solution

  Multiplying or dividing two fractions is a little bit easier than adding or subtracting. You don't have to determine the least common denominator.

  In order to multiply, you only have to multiply the two numerators as well as the denominators:

  $\frac{5}{8x^2}\times\frac{2x^3}{3}=\frac{(5)(2x^3)}{(8x^2)(3)}$.

  Next list the factors of the numerator as well as of the denominator:

  $\frac{(5)(2x^3)}{(8x^2)(3)}=\frac{(5)(2)(x)(x)(x)}{(2)(2)(2)(x)(x)(3)}$,

  and cancel out those which appear in both:

  $\frac{(5)(\not 2)(\not x)(\not x)(x)}{(2)(2)(\not 2)(\not x)(\not x)(3)}=\frac{(5)(x)}{(2)(2)(3)}=\frac{5x}{12}$.

  For dividing two fractions you first have to flip the second one and then you can multiply in the same way like before:

  $\frac{4x^3}{x+1}\div\frac{x+2}{x}=\frac{4x^3}{x+1}\times\frac{x}{x+2}$.

  Multiply the numerators as well as the denominators:

  $\frac{4x^3}{x+1}\times\frac{x}{x+2}=\frac{(4x^3)(x)}{(x+1)(x+2)}$.

  Because we can't simplify this fraction by cancelling out we still have to multiply the binomials in the denominator:

  $\frac{(4x^3)(x)}{(x+1)(x+2)}=\frac{4x^4}{x^2+3x+2}$.

• Find the expressions that can be simplified.

  Hints

  Let's have a look at the following example $\frac46$.

  The numerator contains the factor $2$ as well as the denominator. So you can cancel it out.

  Here you see how to cancel a common factor out.

  In general, factor both the numerator as well as the denominator. Check factors in common and cancel those common factors out.

  Solution

  How can we decide if a given fraction could be simplified?

  First factor the numerator as well as the denominator. If both have one or more factors in common, then those can be cancelled out. This is how you simplify an expression. For

  $\frac{2x^2}{3x}=\frac{(2)(x)(x)}{(3)(x)}$,

  we can find the factor $x$ in the numerator as well as in the denominator.

  On the other hand, $\frac{2x^2}{3}=\frac{(2)(x)(x)}{3}$ has no factors in common and thus can't be simplified.

  There could also be more than one variable:

  $\frac{6x}{4a}=\frac{(2)(3)(x)}{(2)(2)(a)}$.

  Here we have the common factor $2$, which could be cancelled out.

  Remember that we can only cancel out factors.

  For $\frac{6x+1}{4a}=\frac{6x+1}{(2)(2)(a)}$ there is no common factor.

  Lastly let's have a look at an example where more than one factor which could be cancelled out:

  $\begin{array}{rclll} \frac{12x}{6x^2}&=&~\frac{(2)(2)(3)(x)}{(2)(3)(x)(x)}&|&\text{factoring}\\ &=&~\frac{(\not 2)(2)(\not 3)(\not x)}{(\not 2)(\not 3)(\not x)(x)}&|&\text{cancelling out}\\ &=&~\frac2x \end{array}$

• Determine the following expression.

  Hints

  To multiply two fractions multiply their numerators as well as their denominators.

  Here you see how to divide fractions. Just multiply with the flipped fraction.

  Simplify the fractions if it's possible. That means cancel common factors in the numerator and the denominator out.

  You could add or subtract from left to the right according to PEMDAS:

  Start with $\frac{5}{12x}+\frac{x}{3}$. First determine the LCD:

  • $12x=(2)(2)(3)(x)$
  • $3=(3)$

  The LCD is $(2)(2)(3)(x)=12x$. So you have to expand the second fraction by $4x$ to get

  $\frac{5}{12x}+\frac{x}{3}=\frac{5}{12x}+\frac{x(4x)}{(3)(4x)}=\frac{5}{12x}+\frac{4x^2}{12x}=\frac{4x^2+5}{12x}$.

  Solution

  This term looks a little bit more complicated.

  First keep PEMDAS in mind: let's sketch the calculation:

  1. $\frac{4}{x^3} \times \frac{x}{18}$
  2. Divide the product by $\frac{2}{3x^3}$
  3. From left to the right do the addition (or subtraction).

  For the first step, we multiply the numerators of the fractions as well as the denominators of the fractions:

  $\begin{array}{rclll} \frac{4}{x^3} \times \frac{x}{18}&=&~\frac{(4)(x)}{(x^3)(18)}&|&\text{factoring}\\ &=&~\frac{(2)(2)(x)}{(2)(3)(3)(x)(x)(x)}&|&\text{cancelling out}\\ &=&~\frac{2}{9x^2} \end{array}$

  Next divide the resulting fraction by $\frac{2}{3x^3}$. For this we multiply the resulting fraction with the flip of the fraction we are dividing by:

  $\begin{array}{rclll} \frac{2}{9x^2}\div \frac{2}{3x^3}&=&~\frac{2}{9x^2}\times \frac{3x^3}{2}&|&\text{multiply with the flipped fraction}\\ &=&~\frac{(2)(3x^3)}{(9x^2)(2)}&|&\text{factoring}\\ &=&~\frac{(2)(3)(x)(x)(x)}{(2)(3)(3)(x)(x)}&|&\text{cancelling out}\\ &=&~\frac{x}{3} \end{array}$

  Now we can add or subtract them from left to the right:

  $\frac{5}{12x}+\frac{x}{3}-\frac{x}{3}=\frac{5}{12x}$.

• Explain the meaning of LCD.

  Hints

  To determine the LCD first factor, as you can see pictured.

  The LCD is $(2)(2)(2)(3)=24$.

  To check it, you can divide $24$ by $8$ as well as by $12$.

  You can only add or subtract fractions with common denominators.

  To multiply two fractions just multiply the numerators as well as the denominators.

  The result of the addition above is given by

  $\begin{array}{rcl} \frac{5}{12}+\frac{3}{8}&=&~\frac{(5)(2)}{(12)(2)}+\frac{(3)(3)}{(8)(3)}\\ &=&~\frac{10}{24}+\frac{9}{24}\\ &=&~\frac{19}{24} \end{array}$

  Solution

  To add or subtract two fractions we first have to determine the LCD, or least common denominator. How can we determine it?

  Let's have a look at the following example: $\frac{5}{6x}+\frac3{2x^2}$.

  1. Factor both denominators $6x=(2)(3)(x)$ and $2x^2=(2)(x)(x)$ The factors $2$ as well as $x$ appear in both denominators. So we take them only once.
  2. The LCD is given by $(2)(3)(x)(x)=6x^2$.

  If we must multiply fractions, we don't need to determine the LCD. We still have to multiply the numerators as well as the denominators.

• Calculate the given expressions.

  Hints

  For adding or subtracting fractions you have to find the LCD of their denominators.

  Let's find the LCD of the denominators of the fractions in the expression $\frac{2}{3x}+\frac{4}{6}$:

  • $3x=(3)(x)$
  • $6=(2)(3)$

  The LCD is $(2)(3)(x)=6x$.

  Expand each fraction so that we have the LCD as the common denominator.

  To multiply two fractions, multiply the numerators as well as the denominators.

  To divide two fractions, $\frac ab\div \frac cd$, multiply the first one with the flip of the second one $\frac ab\times\frac dc$.

  Solution

  Here we practice each operation once.

  For the addition of fractions:

  $\frac{4x^2}{9}+\frac{2}{3x}$.

  For this we first determine the LCD as follows:

  • $9=(3)(3)$
  • $3x=(3)(x)$

  The LCD is $(3)(3)(x)$. Next we expand both fractions to get the LCD:

  $\begin{array}{rclll} \frac{4x^2}{9}+\frac{2}{3x}&=&~\frac{(4x^2)(x)}{(9)(x)}+\frac{(2)(3)}{(3x)(3)}\\ &=&~\frac{4x^3}{9x}+\frac{6}{9x}&|&\text{add the numerators}\\ &=&~\frac{4x^3+6}{9x} \end{array}$

  For subtraction we proceed in a similar manner:

  $\begin{array}{rclll} \frac{9}{4x^2}-\frac{4}{3x}&=&~\frac{(9)(3)}{(4x^2)(3)}-\frac{(4)(4x)}{(3x)(4x)}&|&\text{LCD: } 12x^2\\ &=&~\frac{27}{4x^2}-\frac{16x}{12x^2}&|&\text{subtract the numerators}\\ &=&~\frac{-16x+27}{12x^2} \end{array}$

  Keep in mind that terms get written in descending order of the exponents.

  For the multiplication of two fractions, multiply the numerators as well as the denominators.

  $\begin{array}{rclll} \frac{9}{4x^2}\times\frac{4}{3x}&=&~\frac{(9)(4)}{(4x^2)(3x)}&|&\text{factoring }\\ &=&~\frac{(2)(2)(3)(3)}{(2)(2)(3)(x)(x)(x)}&|&\text{cancelling out}\\ &=&~\frac{3}{x^3} \end{array}$

  Last but not least we look at the division of two fractions. For this multiply by the flip of the second one.

  $\begin{array}{rclll} \frac{15}{4x+4}\div\frac{21}{5x+5}&=&~\frac{15}{4x+4}\times\frac{5x+5}{21}\\ &=&~\frac{(15)(5x+5)}{(4x+4)(21)}&|&\text{factoring}\\ &=&~\frac{(3)(5)(5)(x+1)}{(2)(2)(3)(7)(x+1)}&|&\text{cancelling out}\\ &=&~\frac{(5)(5)}{(2)(2)(7)}=\frac{25}{28} \end{array}$