Add, Subtract and Simplify Rational Expressions - Grade 11 Math Questions With Detailed Solutions

Detailed solutions to questions in How to Add, Subtract and Simplify Rational Expressions are presented.



Add, subtract and simplify the following with detailed solutions

a) \( \dfrac{7}{6} + \dfrac{1}{18} - \dfrac{5}{24} \)

Solution:

The three denominators in the fractions above are different and therefore we need to find a common denominator.

We first find the lowest common multiple (LCM) of the two denominators 6, 18 and 24.


6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 80,...

18: 18, 36, 54, 72, 90,...

24: 24, 48, 72, 96...

The lowest common denominator is 72 and we now convert all 3 denominators to the common denominator 72 and simplify as follows:

\( \dfrac{7}{6} + \dfrac{1}{18} - \dfrac{5}{24} = \dfrac{7 \cdot \color{\red}{12}}{6 \cdot \color{\red}{12}} + \dfrac{1 \cdot \color{\red}{4}}{18 \cdot \color{\red}{4}} - \dfrac{5 \cdot \color{\red}{3}}{24 \cdot \color{\red}{3}} = \dfrac{84}{72} + \dfrac{4}{72} - \dfrac{15}{72} = \dfrac{84+4-15}{72} = \dfrac{73}{72}\)


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

Solution:

The two rational expressions have different denominators. In order to add the rational expressions above, we need to convert them to a common denominator. The two denominators x + 5 and x + 2 have no common factors hence their LCM is given by:

LCM = (x + 5)(x + 2)

We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.

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

We now expand, simplify and factor the numerator if possible .

\( \dfrac{x^2+5x+6 + x^2+2x-15}{(x+5)(x+2)} = \dfrac{2 x^2+7x-9}{(x+5)(x+2)} = \dfrac{(x-1)(2x+9)}{(x+5)(x+2)} \)



c) \( \dfrac{-3}{x-4} + x + 4 \)

Solution:

In order to add a rational expression with an expression without denominator, we convert the one without denominator into a rational expression then add them.

\( \dfrac{-3}{x-4} + x + 4 = \dfrac{-3}{x-4} + (x + 4) \color{red} {\cdot \dfrac{x-4}{x-4}} = \)

The two rational expressions have the same denominator and they are added as follows:

\( \dfrac{-3+(x + 4)(x - 4)}{x-4} = \dfrac{x^2-19}{x+4} \)


d) \( \dfrac{x-4}{x^2-3x+2}-\dfrac{x+5}{x^2+2x-3} \)

Solution:

The two rational expressions have different denominators. In order to add the rational expressions above, we need to convert them to a common denominator. We first factor completely the two denominators x 2 - 3x + 2 and x 2 + 2 x - 3 and find the LCM of Expressions.

x 2 - 3x + 2 = (x - 1) (x - 2)

x 2 + 2 x - 3 = (x - 1)(x + 3)

LCM = (x - 1)(x - 2)(x + 3)

We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.

\( \dfrac{x-4}{x^2-3x+2}-\dfrac{x+5}{x^2+2x-3} = \dfrac{x-4}{(x-1)(x-2)} - \dfrac{x+5}{(x-1)(x+3)} = \)

\( \dfrac{x-4}{(x-1)(x-2)} \color{red}{\cdot \dfrac{(x+3)}{(x+3)}} - \dfrac{x+5}{(x-1)(x+3)}\color{red}{\cdot \dfrac{(x-2)}{(x-2)}} = \)

\( \dfrac{(x-4)(x+3)}{(x - 1)(x - 2)(x + 3)} - \dfrac{(x+5)(x-2)}{(x - 1)(x + 3)(x - 2)} = \)

We now add the numerators expand and simplify.

\( \dfrac{(x-4)(x+3)-(x+5)(x-2)}{(x+1)(x+3)(x-2)} = \dfrac{x^2-x-12-(x^2+3x-10)}{(x+1)(x+3)(x-2)} = \dfrac{-4x-2}{(x+1)(x+3)(x-2)} \)





e) \( \dfrac{x^2-x-6}{x^2-2x-3}-\dfrac{x^2+5x-6}{x^2+9x+18} \)

Solution:

The two rational expressions have different denominators. In order to add the rational expressions above, we need to convert them to a common denominator. We factor completely the two denominators x 2 - 2 x - 3 and x 2 + 9 x + 18 and find the LCM.

x 2 - 2 x - 3 = (x - 3)(x + 1)

x 2 + 9 x + 18 = (x + 3)(x + 6)

LCM = (x - 3)(x + 1)(x + 3)(x + 6)

Let us write the given expression with the denominators in factored form.

\( \dfrac{x^2-x-6}{x^2-2x-3}-\dfrac{x^2+5x-6}{x^2+9x+18} = \dfrac{x^2-x-6}{(x - 3)(x + 1)} - \dfrac{x^2+5x-6}{(x + 3)(x + 6)} \)

Before we convert the expressions to a common denominator, we examine if the numerators do not have common factors with the denominators which might lead to simplications. We factor the numerators.

x 2 - x - 6 = (x - 3)(x + 2)

x 2 + 5 x - 6 = (x + 6)(x - 1)

We rewrite the given expression with numerators and denominators in factored form and simplify if possible.

\( \dfrac{x^2-x-6}{x^2-2x-3}-\dfrac{x^2+5x-6}{x^2+9x+18} = \dfrac{(x + 2)(x - 3)}{(x + 1)(x - 3)} - \dfrac{(x + 6)(x - 1)}{(x + 6)(x + 3)} = \)

We cancel common factors.

\( \dfrac{{\colorcancel{red}{(x-3)}}(x+2)}{{\colorcancel{red}{(x-3)}}(x+1)} - \dfrac{{\colorcancel{red}{(x+6)}}(x - 1)}{{\colorcancel{red}{(x+6)}}(x+3)} =

\dfrac{x+2}{x+1} - \dfrac{x-1}{x+3} \)

The two denominators x + 1 and x + 3 have no common factors and therefore their LCM is (x + 1)(x + 3). We rewrite the above with the common factor (x + 1)(x + 3) as follows:

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

Expand and simplify.

\( \dfrac{5x+7}{(x + 1)(x + 3)} \)



f) \( \dfrac{2}{2x-1}+\dfrac{x+8}{2x^2+9x-5} - \dfrac{5}{2x+10} \)

Solution:

The three rational expressions have different denominators. In order to subtract/add the rational expressions above, we need to convert them to a common denominator.List and factor completely the three denominators 2x - 1 , 2 x 2 + 9 x - 5 and 2 x + 10 and find the LCM.

2x - 1 = 2x - 1

2 x 2 + 9 x - 5 = (2x - 1)(x + 5)

2x+10 = 2(x + 5)

LCM = 2(2x - 1)(x + 5)

We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.

\( \dfrac{2}{2x-1}+\dfrac{x+8}{2x^2+9x-5} - \dfrac{5}{2x+10} = \dfrac{2}{2x-1} \color{red}{\cdot \dfrac{2(x + 5)}{2(x + 5)}} +\dfrac{x+8}{(2x - 1)(x + 5)} \color{red}{\cdot \dfrac{2}{2}} - \dfrac{5}{2(x + 5)} \color{red}{\cdot \dfrac{(2x-1)}{(2x-1)}} = \)

We now add the numerators and simplify.

\( \dfrac{2 \cdot 2(x + 5) +2(x +8) - 5(2x-1)}{2(2x - 1)(x + 5) } = \dfrac{-(4x - 41)}{2(2x - 1)(x + 5)} \)



g) \( \dfrac{y-2}{y(xy-y+3x-3)}+\dfrac{x}{2x-2} \)

Solution:

The two rational expressions have different denominators. In order to subtract/add the rational expressions above, we need to convert them to a common denominator. List and factor completely the two denominators y(x y - y + 3 x - 3) and 2 x - 2 and find the LCM.

y(x y - y + 3 x - 3) = y( y(x - 1) + 3 (x - 1)) = y(x - 1)(y + 3)

2 x - 2 = 2(x - 1)

LCM = 2 y (x - 1)(y + 3)

We now use the LCM as the common denominator and rewrite the rational expressions with the same denominator as follows.

\( \dfrac{y-2}{y(xy-y+3x-3)}+\dfrac{x}{2x-2} = \dfrac{y-2}{y(x - 1)(y + 3)}+\dfrac{x}{2(x - 1)} = \dfrac{y-2}{y(x - 1)(y + 3)}\color{red}{\cdot \dfrac{2}{2}} + \dfrac{x}{2(x - 1)}\color{red}{\cdot \dfrac{y (y + 3)}{y (y + 3)}} = \)

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

Expand and simplify.

\(\dfrac{xy^2+3xy+2y-4}{2 x y (x - 1)(y + 2)} \)


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Updated: 20 January 2017 (A Dendane)