Solve for $x$ and $y$ :
$(a + 2 b) x + (2 a - b) y = 2, (a - 2 b) x + (2 a + b) y = 3$

x = \frac{1}{a} + \frac{1}{15 b}, y = \frac{1}{2 a} - \frac{1}{20 b}

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x = \frac{1}{2 a} + \frac{1}{5 b}, y = \frac{1}{a} - \frac{1}{10 b}

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x = \frac{1}{a} - \frac{1}{5 b}, y = \frac{1}{2 a} + \frac{1}{10 b}

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x = \frac{1}{2 a} - \frac{1}{5 b}, y = \frac{1}{a} + \frac{1}{10 b}

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Solution

The correct option is D $x = \frac{1}{2 a} - \frac{1}{5 b}, y = \frac{1}{a} + \frac{1}{10 b}$
$(a + 2 b) x + (2 a - b) y = 2$ ........... (1)
$(a - 2 b) x + (2 a + b) y = 3$ ............ (2)

Adding (1) and (2), we get

$2 a x + 4 a y = 5$

$2 x + 4 y = \frac{5}{a}$ .......... (3)

Subtracting (2) from (1), we get

$4 b x - 2 b y = - 1$

$4 x - 2 y = - \frac{1}{b}$ ........... (4)

Multiply equation (3) by 2 and subtract (4) from the result,

$2 (2 x + 4 y) - (4 x - 2 y) = \frac{10}{a} + \frac{1}{b}$

$10 y = \frac{10}{a} + \frac{1}{b}$

$y = \frac{1}{a} + \frac{1}{10 b}$ ............ (5)

Now, from (3),

$2 x = \frac{5}{a} - 4 y$

$2 x = \frac{5}{a} - 4 (\frac{1}{a} + \frac{1}{10 b})$

$2 x = \frac{1}{a} - \frac{2}{5 b}$

$x = \frac{1}{2 a} - \frac{1}{5 b}$ .......... (6)

Suggest Corrections

Similar questions

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

\frac{6}{x - 1} - \frac{3}{y - 2} = 1

x \neq 1, y \neq 2)

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

\frac{6}{x - 1} - \frac{3}{y - 2} = 1

x \neq 1, y \neq 2)

Solve the following pair of equations:
$\frac{5}{x - 1} + \frac{1}{y - 2} = 2$
$\frac{6}{x - 1} - \frac{3}{y - 2} = 1$
(where $x \neq 1, y \neq 2)$

If $\frac{2^{(2 a + 1)} \times 3^{(3 b + 2)}}{6^{2}} = 2^{x} \times 3^{y}$

Chose the correct option:

a x^{2} + b y^{2} - c z^{2}

x = 1

y = - 1

z = 2

a = - 2

b = 1

c = - 2

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