Chapter R - Section R.2 - Real Numbers and Their Properties - R.2 Exercises - Page 17: 32

$-\dfrac{5}{8}$

\begin{array}{l} \text{ To evaluate the given expression, perform the following steps:}\\ {\textstyle \begin{array}{|l l|l|} \hline =\left[ -\dfrac{5}{8} +\dfrac{2}{5}\right] -\left(\dfrac{3}{2} -\dfrac{11}{10}\right) & & \text{Apply the rule} \ a-( -b) \ =\ a+b\\ & & \\ \hline \begin{array}{l} =\left[ -\dfrac{5( 5)}{8( 5)} +\dfrac{2( 8)}{5( 8)}\right] -\left(\dfrac{3}{2} -\dfrac{11}{10}\right)\\ \\ =\left[ -\dfrac{25}{40} +\dfrac{16}{40}\right] -\left(\dfrac{3}{2} -\dfrac{11}{10}\right) \end{array} & & \begin{array}{l} \text{The factors of } 8\ \text{are} \ 2^{3}\text{ and the factors of five are}\\ 5\text{ and } 1\text{. Thus the LCM of } 8 \text{ and } 5\text{ is } 2^{3}(5)\text{ or } 40\\ \text{which is the LCD of the first fraction grouping.}\\ \text{We can multiply top and bottom of five-eights }\\ \text{by } 5\text{ and multiply top and bottom of two-fifths}\\ \text{by } 8\text{ to express both fractions in terms of their} \ \ \\ \text{common denominator. } \end{array}\\ & & \\ \hline \begin{array}{l} =\left[ -\dfrac{25}{40} +\dfrac{16}{40}\right] -\left(\dfrac{3( 5)}{2( 5)} -\dfrac{11}{10}\right)\\ \\ =\left[ -\dfrac{25}{40} +\dfrac{16}{40}\right] \ -\left(\dfrac{15}{10} -\dfrac{11}{10}\right) \end{array} & & \begin{array}{l} \text{The LCM of } 2\text{ and } 10\text{ is ten which is the LCD of}\\ \text{the 2nd fraction grouping. We can multiply top }\\ \text{and bottom of three-halves by 5 to express }\\ \text{three-halfs and eleven-tenths in terms of a }\\ \text{common denominator.} \end{array}\\ & & \\ \hline \begin{array}{l} =\left[\dfrac{16}{40} -\dfrac{25}{40}\right] -\left(\dfrac{15}{10} -\dfrac{11}{10}\right)\\ \\ =\left[\dfrac{16-25}{40}\right] -\left(\dfrac{15-11}{10}\right) \end{array} & & \begin{array}{l} \text{Re-arrange the first grouping by applying the }\\ \text{commutative property then apply the following }\\ \text{fraction rule to both groupings} :\\ \\ \dfrac{a}{c} -\dfrac{b}{c} =\dfrac{a-b}{c} \end{array}\\ & & \\ \hline \begin{array}{l} =\left[\dfrac{-9}{40}\right] -\left(\dfrac{4}{10}\right)\\ \\ =\dfrac{-9}{40} -\dfrac{4( 4)}{10( 4)}\\ \\ =\dfrac{-9}{40} -\dfrac{16}{40} \end{array} & & \begin{array}{l} \text{The LCM of } 10\text{ and } 40\text{ is forty which is the LCD }\\ \text{of the remaining fraction pair. Multiply top and }\\ \text{bottom of four-tenths by } 4\text{ to express both }\\ \text{fractions in terms of their common denominator}\\ \text{then apply the }\dfrac{a}{c} -\dfrac{b}{c} =\dfrac{a-b}{c}\text{ rule once more.} \ \end{array}\\ & & \\ \hline \begin{array}{l} =\dfrac{-9-16}{40}\\ \\ =\dfrac{-25}{40} =-\dfrac{25}{40} \end{array} & & \text{Simplify.}\\\\ \hline \begin{array}{l} =-\left(\dfrac{5( 5)}{8( 5)}\right)\\ \\ =-\dfrac{5}{8} \end{array} & & \begin{array}{l} \text{Observe that five is common to } 25\text{ and} \ 40\text{.}\\ \text{Since} \ \dfrac{5}{5} =1\text{ we can reduce the fraction by a }\\ \text{factor of one.} \end{array}\\\\ \hline \end{array}} \end{array}