Solve the following system of equations.
$7 x - 2 y = 1$
$3 x + 4 y = 15$

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Solution

Given:
$7 x - 2 y = 1$ ...(1)
$3 x + 4 y = 15$ .... (2)
First multiply the first equation by $2$ :
We got, $14 x - 4 y = 2$

Now add the first and second equation:
$14 x - 4 y + 3 x + 4 y = 15 + 2$
$17 x = 17$
so, $x = 1$
By putting $x$ in any equation we got;
$y = 3$

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Solve the following systems of equations:

$\frac{7 x - 2 y}{x y} = 5$

$\frac{8 x + 7 y}{x y} = 15$

\frac{1}{3 x} - \frac{1}{4 y} + 1 = 0; \frac{1}{5 x} + \frac{1}{2 y} = \frac{4}{15}

3 x + 2 y + 11 = 0; 7 x - 4 y = 9

x - 2 y = 0

3 x + 4 y = 20

\frac{2}{3 x + 2 y} + \frac{3}{3 x - 2 y} = \frac{17}{5}

\frac{5}{3 x + 2 y} + \frac{1}{3 x - 2 y} = 2

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