Question

algebraic expression plz plz solve question no 12 a and b with solution and send solution image

Answer 1

1)
$\frac{2}{3} m - \frac{4}{5} n + \frac{3}{5} p + ( - \frac{3}{4} m - \frac{5}{2} n + \frac{2}{3} p) + \frac{5}{2} m + \frac{3}{4} p - \frac{5}{6} n \\ \\ \frac{2}{3} m - \frac{4}{5} n + \frac{3}{5} p - \frac{3}{4} m + \frac{5}{2} n - \frac{2}{3} p+ \frac{5}{2} m + \frac{3}{4} p - \frac{5}{6} n \\ \\ ( \frac{2}{3} - \frac{3}{4} + \frac{5}{2}) m + ( \frac{5}{2} - \frac{4}{5} - \frac{5}{6} )n + ( \frac{3}{5} - \frac{2}{3} + \frac{3}{4} )p \\ \\ = (\frac{8 - 9 + 30}{12} )m + ( \frac{75 - 24 - 25}{30} )n + ( \frac{36 - 40 + 45}{60} )p \\ \\ = \frac{29}{12} m + \frac{26}{30} n + \frac{41}{60} p$
2)

$\frac{3}{2} {x}^{3} - \frac{1}{4} {x}^{2} + \frac{3}{8} + \frac{5}{4} {x}^{3} - \frac{3}{5} {x}^{2} + x - \frac{1}{5} + \frac{1}{6} {x}^{2} - \frac{5}{3} x + \frac{4}{15} \\ \\ (\frac{3}{2} + \frac{5}{4} ) {x}^{3} - ( \frac{1}{4} + \frac{3}{5} - \frac{1}{6} ) {x}^{2} + (1 - \frac{5}{3}) x + \frac{3}{8} - \frac{1}{5} + \frac{4}{15} \\ \\ \frac{6 + 5}{4} {x}^{3} - (\frac{30 + 72 - 20}{120} ) {x}^{2} - \frac{2}{3} x + ( \frac{45 - 24 + 32}{120} ) \\ \\ = \frac{11}{4} {x}^{3} - \frac{82}{120} {x}^{2} - \frac{2}{3} x + \frac{53}{120}$

Answer 2

hey mate

here's your answer,

$\frac{2m}{3}$ - $\frac{4n}{5}$ +$\frac{3p}{5}$ +( $\frac{-3m}{4}$ - $\frac{-5n}{2}$ + $\frac{2p}{3}$ ) + ($\frac{5m}{2}$ + $\frac{3p}{4}$ - $\frac{-5n}{6}$ )

$\frac{2m}{3}$ - $\frac{4n}{5}$ + $\frac{3p}{5}$ - $\frac{3m}{4}$ - $\frac{5n}{2}$ + $\frac{2p}{3}$ + $\frac{5m}{2}$ + $\frac{3p}{4}$ - $\frac{5n}{6}$

$\frac{2m}{3}$-$\frac{3m}{4}$+$\frac{5m}{2}$-$\frac{4n}{5}$-$\frac{5n}{2}$ -$\frac{5n}{6}$+$\frac{3p}{5}$+$\frac{2p}{3}$+$\frac{3p}{4}$

= $\frac{29m}{12}$ -$\frac{62n}{15}$+$\frac{121p}{60}$

hope it helps you mate!!!