# Take away:(i) $\frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x$ from $\frac{x^{3}}{3}-\frac{5}{2} x^{2}+$ $\frac{3}{5} x+\frac{1}{4}$(ii) $\frac{5 a^{2}}{2}+\frac{3 a^{3}}{2}+\frac{a}{3}-\frac{6}{5}$ from $\frac{1}{3} a^{3}-\frac{3}{4} a^{2}-$ $\frac{5}{2}$(iii) $\frac{7}{4} x^{3}+\frac{3}{5} x^{2}+\frac{1}{2} x+\frac{9}{2}$ from $\frac{7}{2}-\frac{x}{3}-$ $\frac{x^{2}}{5}$(iv) $\frac{y^{3}}{3}+\frac{7}{3} y^{2}+\frac{1}{2} y+\frac{1}{2}$ from $\frac{1}{3}-\frac{5}{3} y^{2}$(v) $\frac{2}{3} a c-\frac{5}{7} a b+\frac{2}{3} b c$ from $\frac{3}{2} a b-\frac{7}{4} a c-$ $\frac{5}{6} b c$

To do:

We have to take away the given algebraic expressions.

Solution:

(i) $(\frac{x^{3}}{3}-\frac{5}{2} x^{2}+\frac{3}{5} x+\frac{1}{4})-(\frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x)=\frac{x^{3}}{3}-\frac{5}{2} x^{2}+\frac{3}{5} x+\frac{1}{4}-\frac{6}{5} x^{2}+\frac{4}{5} x^{3}-\frac{5}{6}-\frac{3}{2} x$

$=\frac{x^{3}}{3}+\frac{4}{5} x^{3}-\frac{5}{2} x^{2}-\frac{6}{5} x^{2}+\frac{3}{5} x-\frac{3}{2} x+\frac{1}{4}-\frac{5}{6}$

$=(\frac{1}{3}+\frac{4}{5}) x^{3}+(\frac{-5}{2}-\frac{6}{5}) x^{2}+(\frac{3}{5}-\frac{3}{2}) x+(\frac{1}{4}-\frac{5}{6})$

$=\frac{5+12}{15} x^{3}+\frac{-25-12}{10} x^{2}+\frac{6-15}{10} x+\frac{3-10}{12}$

$=\frac{17}{15} x^{3}-\frac{37}{10} x^{2}-\frac{9}{10} x-\frac{7}{12}$

(ii) $(\frac{1}{3} a^{3}-\frac{3}{4} a^{2}-\frac{5}{2})-(\frac{5 a^{2}}{2}+\frac{3 a^{3}}{2}+\frac{a}{3}-\frac{6}{5})=\frac{1}{3} a^{3}-\frac{3}{4} a^{2}-\frac{5}{2}-\frac{5}{2} a^{2}-\frac{3}{2} a^{3}-\frac{a}{3}+\frac{6}{5}$

$=\frac{1}{3} a^{3}-\frac{3}{2} a^{3}-\frac{3}{4} a^{2}-\frac{5}{2} a^{2}-\frac{a}{3}+\frac{6}{5}-\frac{5}{2}$

$=(\frac{1}{3}-\frac{3}{2}) a^{3}-(\frac{3}{4}+\frac{5}{2}) a^{2}-(\frac{1}{3}) a+(\frac{6}{5}-\frac{5}{2})$

$=\frac{2-9}{6} a^{3}-\frac{3+10}{4} a^{2}-\frac{1}{3} a+\frac{12-25}{10}$

$=\frac{-7}{6} a^{3}-\frac{13}{4} a^{2}-\frac{1}{3} a-\frac{13}{10}$

(iii) $(\frac{7}{2}-\frac{x}{3}-\frac{x^{2}}{5})-(\frac{7}{4} x^{3}+\frac{3}{5} x^{2}+\frac{1}{2} x+\frac{9}{2})=\frac{7}{2}-\frac{x}{3}-\frac{x^{2}}{5}-\frac{7}{4} x^{3}-\frac{3}{5} x^{2}-\frac{1}{2} x-\frac{9}{2}$

$=\frac{-7}{4} x^{3}-\frac{x^{2}}{5}-\frac{3}{5} x^{2}-\frac{x}{3}-\frac{1}{2} x+\frac{7}{2}-\frac{9}{2}$

$=\frac{-7}{4} x^{3}-(\frac{1}{5}+\frac{3}{5}) x^{2}-(\frac{1}{3}+\frac{1}{2}) x+\frac{7-9}{2}$

$=\frac{-7}{4} x^{3}-\frac{1+3}{5} x^{2}-\frac{2+3}{6} x-\frac{2}{2}$

$=\frac{-7}{4} x^{3}-\frac{4}{5} x^{2}-\frac{5}{6} x-1$

(iv) $(\frac{1}{3}-\frac{5}{3} y^{2})-(\frac{y^{3}}{3}+\frac{7}{3} y^{2}+\frac{1}{2} y+\frac{1}{2})=\frac{1}{3}-\frac{5}{3} y^{2}-\frac{y^{3}}{3}-\frac{7}{3} y^{2}-\frac{1}{2} y-\frac{1}{2}$

$=\frac{-y^{3}}{3}-(\frac{5}{3}+\frac{7}{3}) y^{2}-\frac{1}{2} y-\frac{1}{2}+\frac{1}{3}$

$=\frac{-y^{3}}{3}-\frac{5+7}{3} y^{2}-\frac{1}{2} y-\frac{3-2}{6}$

$=\frac{-y^{3}}{3}-\frac{12}{3} y^{2}-\frac{1}{2} y-\frac{1}{6}$

$=\frac{-1}{3} y^{3}-4 y^{2}-\frac{1}{2} y-\frac{1}{6}$

(v) $(\frac{3}{2} a b-\frac{7}{4} a c-\frac{5}{6} b c)-(\frac{2}{3} a c-\frac{5}{7} a b+\frac{2}{3} b c)=(\frac{3}{2} a b-\frac{7}{4} a c-\frac{5}{6} b c-\frac{2}{3} a c+\frac{5}{7} a b-\frac{2}{3} b c)$

$=(\frac{3}{2} a b+\frac{5}{7} a b-\frac{5}{6} b c-\frac{2}{3} b c-\frac{7}{4} a c-\frac{2}{3} a c)$

$=(\frac{3}{2}+\frac{5}{7}) a b-(\frac{5}{6}+\frac{2}{3}) b c-(\frac{7}{4}+\frac{2}{3}) a c$

$=\frac{21+10}{14} a b-\frac{5+4}{6} b c-\frac{21+8}{12} a c$

$=\frac{31}{14} a b-\frac{9}{6} b c-\frac{29}{12} a c$

$=\frac{31}{14} a b-\frac{3}{2} b c-\frac{29}{12} a c$

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Updated on: 10-Oct-2022

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