Solve the following pair of equations by reducing them to a pair of linear equations: $\frac{4}{x} + 3 y = 14, \frac{3}{x} - 4 y = 23$

x = 7, y = 1

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x = - 3, y = - 1

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x = \frac{1}{5}, y = - 2

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x = \frac{7}{2}, y = 7

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Solution

The correct option is C $x = \frac{1}{5}, y = - 2$
Given
$\frac{4}{x} + 3 y = 14$

$\frac{3}{x} - 4 y = 23$

Let $\frac{1}{x} = a$

$4 a + 3 y = 14....... e q 1$
$3 a - 4 y = 23........ e q 2$
multiply eq1 by 4 and eq2 by 3 we get
$16 a + 12 y = 56...... e q 3$
$9 a - 12 y = 69......... e q 4$

Adding eq3 and eq4 we get
$25 a = 125$
$a = 5$

Substituting $a = 5$ in eq1 we get
$4 \times 5 + 3 y = 14$
$3 y = 14 - 20$
$3 y = - 6$
$y = - 2$

Hence $\frac{1}{x} = 5$ then $x = \frac{1}{5}, y = - 2$

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Question 1 (vi)
Solve the following pairs of equations by reducing them to a pair of linear equations:
6x + 3y = 6xy
2x + 4y = 5xy

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