Question

solve it
3x/2 - 5y/3 = -2
x/3 + y/2 = 13/6
Find the value of x and y​

Answer 1

$\Large\bf\blue{\underline{\underline{\bf{\red{Given}}}}}$

$\sf \Large\frac{3x}{2}-\frac{5y}{3}=-2$

$\sf \Large\frac{x}{3}+\frac{y}{2}=\Large\frac{13}{6}$

$\Large\bf\blue{\underline{\underline{\bf{\red{To\:find}}}}}$

Find the value of x and y

$\Large\bf\blue{\underline{\underline{\bf{\red{Solution}}}}}$

$\large{\boxed{\bf{\green{Method\::\:1}}}}$

Solve by substitution method

• $\sf \large\frac{3x}{2}-\frac{5y}{3}=-2$

$\implies\sf \frac{9x-10y}{6}=-2$

$\implies\sf 9x-10y=-12$ ----(i)

• $\sf \large\frac{x}{3}+\frac{y}{2}=\large\frac{13}{6}$

$\implies\sf \large\frac{2x+3y}{6}=\large\frac{13}{6}$

$\implies\sf 2x+3y=13$ ----(ii)

From (ii) equation

$\implies\sf 2x+3y-13=0$

$\implies\sf 3y=13-2x$

$\therefore$$\sf y=\large\frac{13-2x}{3}$

Substituting the value of $\sf y=\large\frac{13-2x}{3}$ in equation (i)

$\implies\sf 9x-10y+12=0$

$\implies\sf 9x-\large\frac{10(13-2x)}{3}+12=0$

$\implies\sf 9x-\large\frac{130+20x}{3}+12=0$

Write in this way also

$\implies\sf 9x-\large\frac{130}{3}+\large\frac{20x}{3}+12=0$

$\implies\sf 9x-\large\frac{20x}{3}-\large\frac{130}{3}+12=0$

$\implies\sf \large\frac{27x+20x}{3}+\large\frac{36-130}{3}=0$

$\implies\sf 47x=94$

$\therefore$$\sf x=\Large\cancel\frac{94}{47}=2$

Again substituting the value of x =2 in equation (i)

$\implies\sf 9x-10y=-12$

$\implies\sf 9\times{2}-10y=-12$

$\implies\sf 18-10y=-12$

$\implies\sf -10y=-12-18$

$\implies\sf -10y=-30$

$\therefore\sf y=\Large\cancel\frac{30}{10}=3$

______________________________________

$\large{\boxed{\bf{\green{Method\::\:2}}}}$

Solve by elimination method

$\sf 9x-10y=-12$ ----(i)

$\sf 2x+3y=13$ ----(ii)

Multiply (i) by 3 & multiply (ii) by 10

$\implies\sf 27x-30y=-36$ ---(iii)

$\implies\sf 20x+30y=130$ ----(iv)

Adding (iii) and (iv)

$\implies\sf (27x-30y)+(20x+30y)=-36+130$

$\implies\sf 27x-30y+20x+30y=94$

$\implies\sf 47x=94$

$\therefore$$\sf x=\Large\cancel\frac{94}{47}=2$

By substitution method

Substitute the value of x = 2 in equation (iii)

$\implies\sf 27x-30y=-36$

$\implies\sf 27\times{2}-30y=-36$

$\implies\sf 54-30y=-36$

$\implies\sf -30y=-36-54$

$\implies\sf -30y=-90$

$\therefore$$\sf y=\Large\cancel\frac{90}{30}=3$

Required value

$\large{\boxed{\bf{x=2}}}$

$\large{\boxed{\bf{y=3}}}$

$\Large\bf\blue{\underline{\underline{\bf{\red{Verification}}}}}$

Substituting x=2 and y = 3 in the both equations

$\sf \Large\frac{3x}{2}-\frac{5y}{3}+2=0$

LHS

$\sf \Large\frac{3x}{2}-\frac{5y}{3}+2$

$\sf =\Large\frac{3×2}{2}-\Large\frac{5×3}{3}+2$

$\sf =3-5+2=0$=RHS verified

$\sf \Large\frac{x}{3}+\frac{y}{2}-\Large\frac{13}{6}=0$

LHS

$\sf \Large\frac{x}{3}+\frac{y}{2}-\Large\frac{13}{6}$

$\sf =\Large\frac{2}{3}+\Large\frac{3}{2}-\Large\frac{13}{6}$

$\sf =\Large\frac{4+9-13}{6}$

$\sf =\Large\frac{13-13}{6}$ = RHS verified

Both equations are satisfied

$\Large\bf\blue{\underline{\underline{\bf{\red{Note}}}}}$

★ Substitution method ★

In this method substitute the value of one variable by expressing it in the terms of the other one variable

★ Elimination method ★

In this method, removing one variable to get a linear equation in one variable