Question

1/3 x + y = 10/3 ; 2x + 1/4 y = 11/4​

Answer 1

Answer:

$\frac{x}{3} + y = \frac{10}{3} = > y = \frac{10}{3} - \frac{x}{3}$

$2x + \frac{y}{4} = \frac{11}{4} = > 2x + \frac{ \frac{10}{3} - \frac{x}{3} }{4} = \frac{11}{4}$

$8x + \frac{10}{3} - \frac{x}{3} = 11$

$24x + 10 - x = 33$

$23x = 23 = > x = 1 = > y = 3$

Answer 2

★ Solution :-

$\sf \leadsto \dfrac{1}{3}x + y = \dfrac{10}{3} \: \: --- (i)$

$\sf \leadsto 2x + \dfrac{1}{4}y = \dfrac{11}{4} \: \: --- (ii)$

By first equation,

$\sf \leadsto \dfrac{1}{3}x + y = \dfrac{10}{3}$

$\sf \leadsto \dfrac{1}{3}x = \dfrac{10}{3} - y$

$\sf \leadsto \dfrac{1}{3}x = \dfrac{10 - 3y}{3}$

$\sf \leadsto x = \dfrac{\dfrac{10 - 3y}{3}}{\dfrac{1}{3}}$

$\sf \leadsto x = \dfrac{10 - 3y}{3} \times 3$

$\sf \leadsto x = \dfrac{30 - 9y}{3}$

Now, we can find the original value of y.

$\sf \leadsto 2x + \dfrac{1}{4}y = \dfrac{11}{4}$

$\sf \leadsto 2 \bigg( \dfrac{30 - 9y}{3} \bigg) + \dfrac{1}{4}y = \dfrac{11}{4}$

$\sf \leadsto \dfrac{60 - 18y}{3} + \dfrac{1y}{4} = \dfrac{11}{4}$

$\sf \leadsto \dfrac{240 - 72y + 3y}{12} = \dfrac{11}{4}$

$\sf \leadsto \dfrac{240 - 69y}{12} = \dfrac{11}{4}$

$\sf \leadsto 240 - 69y = 12 \bigg( \dfrac{11}{4} \bigg)$

$\sf \leadsto 240 - 69y = 33$

$\sf \leadsto -69y = 33 - 240$

$\sf \leadsto -69y = -207$

$\sf \leadsto y = \dfrac{-207}{-69}$

$\sf \leadsto y = 3$

Now, we can find the original value of x.

$\sf \leadsto \dfrac{1}{3}x + y = \dfrac{10}{3}$

$\sf \leadsto \dfrac{x}{3} + 3 = \dfrac{10}{3}$

$\sf \leadsto \dfrac{x + 9}{3} = \dfrac{10}{3}$

$\sf \leadsto x + 9 = 10$

$\sf \leadsto x = 10 - 9$

$\sf \leadsto x = 1$

Therefore, the values of x and y are 1 and 3 respectively.