
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:
\( \dfrac{7}{6} + \dfrac{1}{18}  \dfrac{5}{24} = \dfrac{7 \cdot \color{\red}{12}}{6 \cdot \color{\red}{12}} + \dfrac{1 \cdot \color{\red}{4}}{18 \cdot
\color{\red}{4}}  \dfrac{5 \cdot \color{\red}{3}}{24 \cdot \color{\red}{3}} =
\dfrac{84}{72} + \dfrac{4}{72}  \dfrac{15}{72} = \dfrac{84+415}{72} = \dfrac{73}{72}\)
b) \( \dfrac{x+3}{x+5} + \dfrac{x3}{x+2} \)
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.
\( \dfrac{x+3}{x+5} + \dfrac{x3}{x+2} = \dfrac{x+3}{x+5} \color{red}{\cdot \dfrac{x+2}{x+2}} + \dfrac{x3}{x+2}\color{red}{\cdot \dfrac{x+5}{x+5}} = \dfrac{ (x+3)(x+2)+(x3)(x+5) }{(x+5)(x+2)} = \)
We now expand, simplify and factor the numerator if possible .
\( \dfrac{x^2+5x+6 + x^2+2x15}{(x+5)(x+2)} = \dfrac{2 x^2+7x9}{(x+5)(x+2)} = \dfrac{(x1)(2x+9)}{(x+5)(x+2)} \)
c) \( \dfrac{3}{x4} + x + 4 \)
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.
\( \dfrac{3}{x4} + x + 4 = \dfrac{3}{x4} + (x + 4) \color{red} {\cdot \dfrac{x4}{x4}} = \)
The two rational expressions have the same denominator and they are added as follows:
\( \dfrac{3+(x + 4)(x  4)}{x4} = \dfrac{x^219}{x+4} \)
d) \( \dfrac{x4}{x^23x+2}\dfrac{x+5}{x^2+2x3} \)
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.
\(
\dfrac{x4}{x^23x+2}\dfrac{x+5}{x^2+2x3} = \dfrac{x4}{(x1)(x2)}  \dfrac{x+5}{(x1)(x+3)} = \)
\( \dfrac{x4}{(x1)(x2)} \color{red}{\cdot \dfrac{(x+3)}{(x+3)}}  \dfrac{x+5}{(x1)(x+3)}\color{red}{\cdot \dfrac{(x2)}{(x2)}} = \)
\( \dfrac{(x4)(x+3)}{(x  1)(x  2)(x + 3)}  \dfrac{(x+5)(x2)}{(x  1)(x + 3)(x  2)} = \)
We now add the numerators expand and simplify.
\( \dfrac{(x4)(x+3)(x+5)(x2)}{(x+1)(x+3)(x2)} = \dfrac{x^2x12(x^2+3x10)}{(x+1)(x+3)(x2)} = \dfrac{4x2}{(x+1)(x+3)(x2)} \)
e) \( \dfrac{x^2x6}{x^22x3}\dfrac{x^2+5x6}{x^2+9x+18} \)
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.
\( \dfrac{x^2x6}{x^22x3}\dfrac{x^2+5x6}{x^2+9x+18} = \dfrac{x^2x6}{(x  3)(x + 1)}  \dfrac{x^2+5x6}{(x + 3)(x + 6)} \)
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.
\( \dfrac{x^2x6}{x^22x3}\dfrac{x^2+5x6}{x^2+9x+18} = \dfrac{(x + 2)(x  3)}{(x + 1)(x  3)}  \dfrac{(x + 6)(x  1)}{(x + 6)(x + 3)} = \)
We cancel common factors.
\( \dfrac{{\colorcancel{red}{(x3)}}(x+2)}{{\colorcancel{red}{(x3)}}(x+1)}  \dfrac{{\colorcancel{red}{(x+6)}}(x  1)}{{\colorcancel{red}{(x+6)}}(x+3)} =
\dfrac{x+2}{x+1}  \dfrac{x1}{x+3}
\)
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:
\( = \dfrac{(x+2)(x+3)}{(x + 1)(x + 3)}  \dfrac{(x1)(x+1)}{(x + 3)(x + 1)} = \dfrac{(x+2)(x+3)  (x1)(x+1)}{(x + 3)(x + 1)} = \)
Expand and simplify.
\( \dfrac{5x+7}{(x + 1)(x + 3)} \)
f) \( \dfrac{2}{2x1}+\dfrac{x+8}{2x^2+9x5}  \dfrac{5}{2x+10} \)
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.
\(
\dfrac{2}{2x1}+\dfrac{x+8}{2x^2+9x5}  \dfrac{5}{2x+10} = \dfrac{2}{2x1} \color{red}{\cdot \dfrac{2(x + 5)}{2(x + 5)}} +\dfrac{x+8}{(2x  1)(x + 5)} \color{red}{\cdot \dfrac{2}{2}}  \dfrac{5}{2(x + 5)} \color{red}{\cdot \dfrac{(2x1)}{(2x1)}} = \)
We now add the numerators and simplify.
\( \dfrac{2 \cdot 2(x + 5) +2(x +8)  5(2x1)}{2(2x  1)(x + 5) } = \dfrac{(4x  41)}{2(2x  1)(x + 5)} \)
g) \( \dfrac{y2}{y(xyy+3x3)}+\dfrac{x}{2x2} \)
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.
\( \dfrac{y2}{y(xyy+3x3)}+\dfrac{x}{2x2} = \dfrac{y2}{y(x  1)(y + 3)}+\dfrac{x}{2(x  1)} = \dfrac{y2}{y(x  1)(y + 3)}\color{red}{\cdot \dfrac{2}{2}} + \dfrac{x}{2(x  1)}\color{red}{\cdot \dfrac{y (y + 3)}{y (y + 3)}} = \)
\( \dfrac{2(y2) + x y (y + 3)}{2 y (x  1)(y + 3)} = \)
Expand and simplify.
\(\dfrac{xy^2+3xy+2y4}{2 x y (x  1)(y + 2)} \)

