Question

Simultaneous Linear Equations​

Answer 1

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: here \: given \: linear \: equations \: are \: \\ \frac{x}{3} \: - \: \frac{2}{x + y} = 1 \: \\ \\ ie \: \: \frac{x}{3} \: - \: 2 \times \frac{1}{x + y} = 1 \\ \\ \frac{x}{4} \: + \: \frac{3}{x + y} = 3 \\ \\ ie \: \: \frac{x}{4} \: + \: 3 \times \frac{1}{x + y} = 3 \\ \\ so \: \: let \: \: \frac{1}{x + y} = k$

$hence \: accordingly \\ \frac{x}{3} - 2k = 1 \\ \\ \frac{ x - 2k \times 3}{3} = 1 \\ \\ x - 6k = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ \\ \frac{x}{4} + 3k = 3 \\ \\ \frac{x + 3k \times 4}{4} = 3 \\ \\ ie \: \: x + 12k = 12 \: \: \: \: \: (2)$

$subtracting \: (1) \: from \: (2) \\ we \: get \\ 18k = 9 \\ ie \: \: k = \frac{9}{18} \\ \\ k = \frac{1}{2} \\ \\ substitute \: value \: of \: k \: in \: (1) \\ we \: get \\ x = 6 \: \: \: \: \: \: \: \: \: \: (3) \\ \\ so \: resubstituting \: value \: of \: k \: as \: \frac{1}{x + y} \\ we \: get \\ \frac{1}{x + y} = \frac{1}{2 } \\ \\ x + y = 2 \\ 6 + y = 2 \: \: \: \: \: \: \: \: \: from \: (3) \\ ie \: \: y = - 4$

$hence \: (x,y) = (6, - 4) \: is \: solution \: of \: above \: linear \: equations \:$