Question

Solve x/3+y/2=4 2/3x-y/6=1 by elemenation method

Answer 1

★ Solution :-

$\sf \leadsto \dfrac{x}{3} + \dfrac{y}{2} = 4 \: \: - - - (i)$

$\sf \leadsto \dfrac{2}{3}x + \dfrac{y}{6} = 1 \: \: - - - (ii)$

Now, by multiplying the first equation by 1/3.

$\sf \leadsto \dfrac{1}{3} \bigg( \dfrac{x}{3} + \dfrac{y}{2} = 4 \bigg)$

$\sf \leadsto \dfrac{x}{9} + \dfrac{y}{6} = \dfrac{4}{3}$

Now, by subtracting both the equations,

$\sf \: \: \: \: \: \dfrac{x}{9} + \dfrac{y}{6} = \dfrac{4}{3} \\ \sf - \: \dfrac{2x}{3} + \frac{y}{6} = 1$

$\sf \leadsto \dfrac{ - 5x}{9} + 0 = \dfrac{1}{3}$

$\sf \leadsto \dfrac{ - 5x}{9} = \dfrac{1}{3}$

$\sf \leadsto 3( - 5x) = 9(1)$

$\sf \leadsto - 15x = 9$

$\sf \leadsto x = \dfrac{9}{ - 15}$

$\sf \leadsto x = \dfrac{ - 3}{5}$

Now, by first equation,

$\sf \leadsto \dfrac{x}{3} + \dfrac{y}{2} = 4$

$\sf \leadsto \dfrac{\dfrac{ - 3}{5}}{3} + \dfrac{y}{2} = 4$

$\sf \leadsto \dfrac{ - 1}{5} + \dfrac{y}{2} = 4$

$\sf \leadsto \dfrac{ - 2 + 5y}{5} = 4$

$\sf \leadsto - 2 + 5y = 4(5)$

$\sf \leadsto - 2 + 5y = 20$

$\sf \leadsto 5y = 20 + 2$

$\sf \leadsto 5y = 22$

$\sf \leadsto y = \dfrac{22}{5}$

Therefore, the values of x and y are -3/5 and 22/5 respectively.