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Mathematics

Evaluate :

(i) 2−3+−49\dfrac{2}{-3} + \dfrac{-4}{9}

(ii) −12+−34\dfrac{-1}{2} + \dfrac{-3}{4}

(iii) 7−9+−56\dfrac{7}{-9} + \dfrac{-5}{6}

(iv) 2+−342 + \dfrac{-3}{4}

(v) 3+−563 + \dfrac{-5}{6}

(vi) −4+23-4 + \dfrac{2}{3}

Rational Numbers

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Answer

(i) 2−3+−49\dfrac{2}{-3} + \dfrac{-4}{9}

First, express 2−3\dfrac{2}{-3} with a positive denominator: 2×(−1)−3×(−1)=−23\dfrac{2 \times (-1)}{-3 \times (-1)} = \dfrac{-2}{3}.

Let us find LCM of denominators 3 and 9

33,931,31,1\begin{array}{r|l} 3 & 3, 9 \ \hline 3 & 1, 3 \ \hline & 1, 1 \end{array}

L.C.M. = 3 x 3 = 9

Now,

−23=−2×33×3=−69∴−69+−49=(−6)+(−4)9=−109\dfrac{-2}{3} = \dfrac{-2 \times 3}{3 \times 3} = \dfrac{-6}{9} \\[1em] \therefore \dfrac{-6}{9} + \dfrac{-4}{9} = \dfrac{(-6) + (-4)}{9} \\[1em] = \dfrac{-10}{9}

Hence, the answer is −109\dfrac{-10}{9}

(ii) −12+−34\dfrac{-1}{2} + \dfrac{-3}{4}

Let us find LCM of denominators 2 and 4.

22,421,21,1\begin{array}{r|l} 2 & 2, 4 \ \hline 2 & 1, 2 \ \hline & 1, 1 \end{array}

L.C.M. = 2 x 2 = 4

Now,

−12=−1×22×2=−24∴−24+−34=(−2)+(−3)4=−54\dfrac{-1}{2} = \dfrac{-1 \times 2}{2 \times 2} = \dfrac{-2}{4} \\[1em] \therefore \dfrac{-2}{4} + \dfrac{-3}{4} = \dfrac{(-2) + (-3)}{4} \\[1em] = \dfrac{-5}{4}

Hence, the answer is −54\dfrac{-5}{4}

(iii) 7−9+−56\dfrac{7}{-9} + \dfrac{-5}{6}

First, express 7−9\dfrac{7}{-9} with a positive denominator: 7×−1−9×−1=−79\dfrac{7 \times -1}{-9 \times -1} = \dfrac{-7}{9}.

Let us find LCM of denominators 9 and 6.

39,633,221,21,1\begin{array}{r|l} 3 & 9, 6 \ \hline 3 & 3, 2 \ \hline 2 & 1, 2 \ \hline & 1, 1 \end{array}

L.C.M. = 3 x 3 x 2 = 18

Now, expressing each fraction with denominator 18:

−79=−7×29×2=−1418−56=−5×36×3=−1518∴−1418+−1518=(−14)+(−15)18=−2918\dfrac{-7}{9} = \dfrac{-7 \times 2}{9 \times 2} = \dfrac{-14}{18} \\[1em] \dfrac{-5}{6} = \dfrac{-5 \times 3}{6 \times 3} = \dfrac{-15}{18} \\[1em] \therefore \dfrac{-14}{18} + \dfrac{-15}{18} \\[1em] = \dfrac{(-14) + (-15)}{18} \\[1em] = \dfrac{-29}{18}

Hence, the answer is −2918\dfrac{-29}{18}

(iv) 2+−342 + \dfrac{-3}{4}

Express 2 as 21\dfrac{2}{1}.

LCM of denominators 1 and 4 is 4.

Now,

21=2×41×4=84∴84+−34=8+(−3)4=54\dfrac{2}{1} = \dfrac{2 \times 4}{1 \times 4} = \dfrac{8}{4} \\[1em] \therefore \dfrac{8}{4} + \dfrac{-3}{4} \\[1em] = \dfrac{8 + (-3)}{4} \\[1em] = \dfrac{5}{4}

Hence, the answer is 54\dfrac{5}{4}

(v) 3+−563 + \dfrac{-5}{6}

Express 3 as 31\dfrac{3}{1}.

LCM of denominators 1 and 6 is 6.

Now,

31=3×61×6=186∴186+−56=18+(−5)6=136\dfrac{3}{1} = \dfrac{3 \times 6}{1 \times 6} = \dfrac{18}{6} \\[1em] \therefore \dfrac{18}{6} + \dfrac{-5}{6} \\[1em] = \dfrac{18 + (-5)}{6} \\[1em] = \dfrac{13}{6}

Hence, the answer is 136\dfrac{13}{6}

(vi) −4+23-4 + \dfrac{2}{3}

Express -4 as −41\dfrac{-4}{1}.

LCM of denominators 1 and 3 is 3.

Now,

−41=−4×31×3=−123∴−123+23=(−12)+23=−103\dfrac{-4}{1} = \dfrac{-4 \times 3}{1 \times 3} = \dfrac{-12}{3} \\[1em] \therefore \dfrac{-12}{3} + \dfrac{2}{3} \\[1em] = \dfrac{(-12) + 2}{3} \\[1em] = \dfrac{-10}{3}

Hence, the answer is −103\dfrac{-10}{3}

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