Question

Question 1 (vi)
Solve the following pair of linear equations by the substitution method.
$\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Answer 1

$\frac{3x}{2}-\frac{5y}{3}=-2...\left(i\right)\frac{x}{3}+\frac{y}{2}=\frac{13}{6}...\left(ii\right)\text{From equation (i), we get,}9x-10y=-12x=-\frac{12}{9}+\frac{10y}{9}...\left(iii\right)\text{Putting this value in equation (ii), we get,}$
$\frac{\frac{-12+10y}{9}}{3}+\frac{y}{2}=\frac{13}{6}\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}\frac{-24+20y+27y}{54}=\frac{13}{6}47y=117+2447y=141y=3...\left(iv\right)\text{Substituting this value in equation (iii), we obtain,}x=\frac{-12+10\times 3}{9}=\frac{18}{9}=2Hence,x=2,y=3$