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Mathematics

Evaluate :

(i) 56−78\dfrac{5}{6} - \dfrac{7}{8}

(ii) 512−1718\dfrac{5}{12} - \dfrac{17}{18}

(iii) 1115−1320\dfrac{11}{15} - \dfrac{13}{20}

(iv) −59−−23\dfrac{-5}{9} - \dfrac{-2}{3}

(v) 611−−34\dfrac{6}{11} - \dfrac{-3}{4}

(vi) −23−34\dfrac{-2}{3} - \dfrac{3}{4}

Rational Numbers

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Answer

(i) 56−78\dfrac{5}{6} - \dfrac{7}{8}

We have:

=(56−78)=56+(additive inverse of 78)=56+−78\phantom{=} \Big(\dfrac{5}{6} - \dfrac{7}{8}\Big) \\[1em] = \dfrac{5}{6} + \Big(\text{additive inverse of } \dfrac{7}{8}\Big) \\[1em] = \dfrac{5}{6} + \dfrac{-7}{8}

L.C.M. of 6 and 8 is 24.

Now, expressing each fraction with denominator 24:

=5×46×4+−7×38×3=2024+−2124=20+(−21)24=−124= \dfrac{5 \times 4}{6 \times 4} + \dfrac{-7 \times 3}{8 \times 3} \\[1em] = \dfrac{20}{24} + \dfrac{-21}{24} \\[1em] = \dfrac{20 + (-21)}{24} \\[1em] = \dfrac{-1}{24}

Hence, the answer is −124\dfrac{-1}{24}

(ii) 512−1718\dfrac{5}{12} - \dfrac{17}{18}

We have:

=(512−1718)=512+(additive inverse of 1718)=512+−1718\phantom{=} \Big(\dfrac{5}{12} - \dfrac{17}{18}\Big) \\[1em] = \dfrac{5}{12} + \Big(\text{additive inverse of } \dfrac{17}{18}\Big) \\[1em] = \dfrac{5}{12} + \dfrac{-17}{18}

L.C.M. of 12 and 18 is 36.

Now, expressing each fraction with denominator 36:

=5×312×3+−17×218×2=1536+−3436=15+(−34)36=−1936= \dfrac{5 \times 3}{12 \times 3} + \dfrac{-17 \times 2}{18 \times 2} \\[1em] = \dfrac{15}{36} + \dfrac{-34}{36} \\[1em] = \dfrac{15 + (-34)}{36} \\[1em] = \dfrac{-19}{36}

Hence, the answer is −1936\dfrac{-19}{36}

(iii) 1115−1320\dfrac{11}{15} - \dfrac{13}{20}

we have:

=(1115−1320)=1115+(additive inverse of 1320)=1115+−1320\phantom{=} \Big(\dfrac{11}{15} - \dfrac{13}{20}\Big) \\[1em] = \dfrac{11}{15} + \Big(\text{additive inverse of } \dfrac{13}{20}\Big) \\[1em] = \dfrac{11}{15} + \dfrac{-13}{20}

L.C.M. of 15 and 20 is 60.

Now, expressing each fraction with denominator 60:

=11×415×4+−13×320×3=4460+−3960=44+(−39)60=560= \dfrac{11 \times 4}{15 \times 4} + \dfrac{-13 \times 3}{20 \times 3} \\[1em] = \dfrac{44}{60} + \dfrac{-39}{60}\\[1em] = \dfrac{44 + (-39)}{60} \\[1em] = \dfrac{5}{60}

Hence, the answer is 560\dfrac{5}{60}

(iv) −59−−23\dfrac{-5}{9} - \dfrac{-2}{3}

We have:

=(−59−−23)=−59+(additive inverse of −23)=−59+23\phantom{=} \Big(\dfrac{-5}{9} - \dfrac{-2}{3}\Big) \\[1em] = \dfrac{-5}{9} + \Big(\text{additive inverse of } \dfrac{-2}{3}\Big) \\[1em] = \dfrac{-5}{9} + \dfrac{2}{3}

L.C.M. of 9 and 3 is 9.

Now, expressing each fraction with denominator 9:

=−59+2×33×3=−59+69=−5+69=19= \dfrac{-5}{9} + \dfrac{2 \times 3}{3 \times 3} \\[1em] = \dfrac{-5}{9} + \dfrac{6}{9} \\[1em] = \dfrac{-5 + 6}{9} \\[1em] = \dfrac{1}{9}

Hence, the answer is 19\dfrac{1}{9}

(v) 611−−34\dfrac{6}{11} - \dfrac{-3}{4}

We have:

=(611−−34)=611+(additive inverse of −34)=611+34\phantom{=} \Big(\dfrac{6}{11} - \dfrac{-3}{4}\Big) \\[1em] = \dfrac{6}{11} + \Big(\text{additive inverse of } \dfrac{-3}{4}\Big) \\[1em] = \dfrac{6}{11} + \dfrac{3}{4}

L.C.M. of 11 and 4 is 44.

Now, expressing each fraction with denominator 44:

=6×411×4+3×114×11=2444+3344=24+3344=5744= \dfrac{6 \times 4}{11 \times 4} + \dfrac{3 \times 11}{4 \times 11} \\[1em] = \dfrac{24}{44} + \dfrac{33}{44} \\[1em] = \dfrac{24 + 33}{44} \\[1em] = \dfrac{57}{44}

Hence, the answer is 5744\dfrac{57}{44}

(vi) −23−34\dfrac{-2}{3} - \dfrac{3}{4}

We have:

=(−23−34)=−23+(additive inverse of 34)=−23+−34\phantom{=} \Big(\dfrac{-2}{3} - \dfrac{3}{4}\Big) \\[1em] = \dfrac{-2}{3} + \Big(\text{additive inverse of } \dfrac{3}{4}\Big) \\[1em] = \dfrac{-2}{3} + \dfrac{-3}{4}

L.C.M. of 3 and 4 is 12.

Now, expressing each fraction with denominator 12:

=−2×43×4+−3×34×3=−812+−912=−8+(−9)12=−1712= \dfrac{-2 \times 4}{3 \times 4} + \dfrac{-3 \times 3}{4 \times 3} \\[1em] = \dfrac{-8}{12} + \dfrac{-9}{12} \\[1em] = \dfrac{-8 + (-9)}{12} \\[1em] = \dfrac{-17}{12}

Hence, the answer is −1712\dfrac{-17}{12}

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