Question

3/2 x + 2/3 y = 13 1/3 ; 2/3 x + 3/2 y = 8 1/3​

Answer 1

$\huge \fbox \red{Answer : x = 8 and y = 2}$

Solution :

Given that :

$\frac{3x}{2} + \frac{2y}{3} = 13\frac{1}{3} = > \frac{9x + 4y}{6} = \frac{40}{3} \\ \\ = > 9x + 4y = 80...(1)$

$\frac{2x}{3} + \frac{3y}{2} = 8\frac{1}{3} = > \frac{4x + 9y}{6} = \frac{25}{3} \\ \\ = > 4x + 9y = 50...(2)$

Now solving equation (1) and (2) by elimination method :

$9x + 4y = 80...(1) \times 4$

$4x + 9y = 50...(2) \times 9$

$36x + 16y = 320...(3)$

$36x + 81y = 450...(4)$

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On subtracting :

$- 65y = - 130 \\ \\ = > y = \frac{- 130}{ - 65} = 2$

On putting the value of y in equation (1)

$9x + 4y = 80 \\ \\ = > 9x + 4 \times 2 = 80 \\ \\ = > 9x = 80 - 8 = 72\\ \\ = > x = \frac{72}{9} = 8$

So, required values : x = 8 and y = 2

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Hope it helps ☺

Fóllòw Më ❤