Question

Solve each of the following equations and also verify your solution :

$\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{(x+3)}{8}+\frac{(3x-1)}{14}$

Answer 1

$\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{(x+3)}{8}+\frac{(3x-1)}{14}=\frac{7(3x+1)+16(2x-3)=14(x+3)+8(3x-1)}{112}$
(L.C.M. of 16, 7, 8, 14 = 112)
$=21x+7+32x-48=14x+42+24x-8=21x+32x-14x-24x=42-8-7+48$
(By transposition)
$\Rightarrow 53x-38x=90-15\Rightarrow 15x=75\Rightarrow x=\frac{75}{15}\therefore x=5$
Verification :
$L.H.S.=\frac{(3x+1)}{16}+\frac{(2x-3)}{7}=\frac{3\times 5+1}{16}+\frac{2\times 5-3}{7}=\frac{16}{16}+\frac{7}{7}=1+1=2R.H.S.=\frac{x+3}{8}+\frac{3x-1}{11}=\frac{5+2}{8}+\frac{3\times 5-1}{14}=\frac{8}{8}+\frac{14}{14}=1+1=2\therefore L.H.S=R.H.S.$