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Mathematics

Subtract :

(i) 35\dfrac{3}{5} from 12\dfrac{1}{2}

(ii) −47\dfrac{-4}{7} from 23\dfrac{2}{3}

(iii) −56\dfrac{-5}{6} from −34\dfrac{-3}{4}

(iv) −79\dfrac{-7}{9} from 0

(v) 4 from −611\dfrac{-6}{11}

(vi) 38\dfrac{3}{8} from −56\dfrac{-5}{6}

Rational Numbers

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Answer

(i) 35\dfrac{3}{5} from 12\dfrac{1}{2}

We have:

=(12−35)=12+(additive inverse of 35)=12+−35\phantom{=} \Big(\dfrac{1}{2} - \dfrac{3}{5}\Big) \\[1em] = \dfrac{1}{2} + \Big(\text{additive inverse of } \dfrac{3}{5}\Big) \\[1em] = \dfrac{1}{2} + \dfrac{-3}{5}

The L.C.M. of 2 and 5 is 10.

Now, expressing each fraction with denominator 10:

=1×52×5+−3×25×2=510+−610=5+(−6)10=−110= \dfrac{1 \times 5}{2 \times 5} + \dfrac{-3 \times 2}{5 \times 2} \\[1em] = \dfrac{5}{10} + \dfrac{-6}{10} \\[1em] = \dfrac{5 + (-6)}{10} \\[1em] = \dfrac{-1}{10}

Hence, the answer is −110\dfrac{-1}{10}

(ii) −47\dfrac{-4}{7} from 23\dfrac{2}{3}

We have:

=(23−−47)=23+(additive inverse of −47)=23+47\phantom{=} \Big(\dfrac{2}{3} - \dfrac{-4}{7}\Big) \\[1em] = \dfrac{2}{3} + \Big(\text{additive inverse of } \dfrac{-4}{7}\Big) \\[1em] = \dfrac{2}{3} + \dfrac{4}{7}

The L.C.M. of 3 and 7 is 21.

Now, expressing each fraction with denominator 21:

=2×73×7+4×37×3=1421+1221=14+1221=2621= \dfrac{2 \times 7}{3 \times 7} + \dfrac{4 \times 3}{7 \times 3} \\[1em] = \dfrac{14}{21} + \dfrac{12}{21} \\[1em] = \dfrac{14 + 12}{21} \\[1em] = \dfrac{26}{21}

Hence, the answer is 2621\dfrac{26}{21}

(iii) −56\dfrac{-5}{6} from −34\dfrac{-3}{4}

We have:

=(−34−−56)=−34+(additive inverse of −56)=−34+56\phantom{=} \Big(\dfrac{-3}{4} - \dfrac{-5}{6}\Big) \\[1em] = \dfrac{-3}{4} + \Big(\text{additive inverse of } \dfrac{-5}{6}\Big) \\[1em] = \dfrac{-3}{4} + \dfrac{5}{6}

The L.C.M. of 4 and 6 is 12.

Now, expressing each fraction with denominator 12:

=−3×34×3+5×26×2=−912+1012=−9+1012=112= \dfrac{-3 \times 3}{4 \times 3} + \dfrac{5 \times 2}{6 \times 2} \\[1em] = \dfrac{-9}{12} + \dfrac{10}{12} \\[1em] = \dfrac{-9 + 10}{12} \\[1em] = \dfrac{1}{12}

Hence, the answer is 112\dfrac{1}{12}

(iv) −79\dfrac{-7}{9} from 0

We have:

=(0−−79)=0+(additive inverse of −79)=0+79=79\phantom{=} \Big(0 - \dfrac{-7}{9}\Big) \\[1em] = 0 + \Big(\text{additive inverse of } \dfrac{-7}{9}\Big) \\[1em] = 0 + \dfrac{7}{9} \\[1em] = \dfrac{7}{9}

Hence, the answer is 79\dfrac{7}{9}

(v) 4 from −611\dfrac{-6}{11}

we have:

=(−611−4)=−611+(additive inverse of 4)=−611+−41\phantom{=} \Big(\dfrac{-6}{11} - 4\Big) \\[1em] = \dfrac{-6}{11} + \Big(\text{additive inverse of } 4\Big) \\[1em] = \dfrac{-6}{11} + \dfrac{-4}{1}

The L.C.M. of 11 and 1 is 11.

Now, expressing each fraction with denominator 11:

=−611+−4×111×11=−611+−4411=−6+(−44)11=−5011= \dfrac{-6}{11} + \dfrac{-4 \times 11}{1 \times 11} \\[1em] = \dfrac{-6}{11} + \dfrac{-44}{11} \\[1em] = \dfrac{-6 + (-44)}{11} \\[1em] = \dfrac{-50}{11}

Hence, the answer is −5011\dfrac{-50}{11}

(vi) 38\dfrac{3}{8} from −56\dfrac{-5}{6}

we have:

=(−56−38)=−56+(additive inverse of 38)=−56+−38\phantom{=} \Big(\dfrac{-5}{6} - \dfrac{3}{8}\Big) \\[1em] = \dfrac{-5}{6} + \Big(\text{additive inverse of } \dfrac{3}{8}\Big) \\[1em] = \dfrac{-5}{6} + \dfrac{-3}{8}

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

Now, expressing each fraction with denominator 24:

=−5×46×4+−3×38×3=−2024+−924=−20+(−9)24=−2924= \dfrac{-5 \times 4}{6 \times 4} + \dfrac{-3 \times 3}{8 \times 3} \\[1em] = \dfrac{-20}{24} + \dfrac{-9}{24} \\[1em] = \dfrac{-20 + (-9)}{24} \\[1em] = \dfrac{-29}{24}

Hence, the answer is −2924\dfrac{-29}{24}

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