Solve the following simultaneous equations by substitution method.
$2 x + 3 y = - 4; x - 5 y = 11$

x = - 1, y = 2

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x = 1, y = - 2

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x = - 1, y = - 2

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x = 1, y = 2

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Solution

The correct option is B $x = 1, y = - 2$
$2 x + 3 y = - 4$ ..........(1)
$x - 5 y = 11$
$∴ x = 11 + 5 y$ .............(2)
Substituting eq. (2) in eq. (1), we get,
$2 (11 + 5 y) + 3 y = - 4$
$22 + 10 y + 3 y = - 4$
$22 + 4 = - 13 y$
$∴ y = - \frac{26}{13}$

$y = - 2$
Substituting $y = - 2$ in (1) we get,
$x = 1$

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x - y = 1

2 x - 3 y = 2

\frac{1}{2 x} + \frac{1}{3 y} = 2 (x \neq 0, y \neq 0) \frac{1}{3 x} + \frac{1}{y} = \frac{11}{3}

\frac{1}{x} + \frac{2}{y} = 9; \frac{2}{x} + \frac{1}{y} = 12 (x \neq 0, y \neq 0)

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

$3 x + 2 y = 11$

$2 x - 3 y + 10 = 0$

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