Solve the following system of equations of elimination by equating the coefficients method: $3 x - 4 y - 11 = 0, 5 x - 7 y + 4 = 0$ .

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Solution

Given $3 x - 4 y - 11 = 0 . . . . (1)$
$5 x - 7 y + 4 = 0 . . . (2)$
Multiply equations (1) and (2) by $5$ and $3$ respectively
We get
$15 x - 20 y - 55 = 0$ ----(1) and $15 x - 21 y + 12 = 0$ .........(ii)
subtract the eq(i) from eq(ii) we get the below equation,
$- 20 y - 55 = - 21 y + 12$
On solving we get $y = 67$
$x = 93$

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