Question

Re-arrange suitably and find the sum of the following
$\frac{-4}{7}+\frac{7}{6}+\frac{2}{7}+3+\frac{-11}{6}$.

• A. $\frac{43}{21}$
• B. $\frac{43}{42}$
• C. $\frac{23}{42}$
• D. $\frac{23}{21}$

Answer 1

Answer: (A)

$\frac{43}{21}$

Re-arranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominators, we have

$\frac{-4}{7}+\frac{7}{6}+\frac{2}{7}+3+\frac{-11}{6}$

$=\left(\begin{array}{c}\frac{-4}{7}+\frac{2}{7}\end{array}\right)+\left(\begin{array}{c}\frac{7}{6}+\frac{-11}{6}\end{array}\right)+3$

$=\frac{(-4)+2}{7}+\frac{7+(-11)}{6}+3=\frac{-2}{7}+\frac{(-4)}{6}+3$

$=\frac{(-2)}{7}+\frac{(-2)}{3}+3\left[\begin{array}{c}\therefore \frac{-4}{6}=\frac{-2}{3}\end{array}\right]$

$=\left(\begin{array}{c}\frac{(-2)}{7}+\frac{(-2)}{3}\end{array}\right)+3$

$=\frac{(-2)\times 3+(-2)\times 7}{21}+3=\frac{(-6)+(-14)}{21}+3$

$=\frac{(-20)}{21}+\frac{3}{1}=\frac{(-20)+3\times 21}{21}$

$=\frac{(-20)+63}{21}=\frac{43}{21}$