Solve the following pairs of equations.
$2 x = y + 1$ ; $x + 2 y - 8 = 0$

x = 2, y = 3

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x = 3, y = 3

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x = 2, y = 2

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None of these

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Solution

The correct option is A $x = 2, y = 3$
$2 x = y + 1$ .....(ii) ;
and
$x + 2 y = 8$
$x = 8 - 2 y$ .....(i)
Put (i) in (ii)
$16 - 4 y = y + 1$
$y = 3$
$y = 3$ & $x = 2$

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Similar questions

Solve the following pair of equations.
$\frac{1}{x} + \frac{3}{y} = 1$
$\frac{6}{x} - \frac{12}{y} = 2$
(where $x \neq 0, y \neq 0)$

x^{2} y^{2} - 2 x + y^{2} = 0, 2 x^{2} - 4 x + 3 + y^{3} = 0

x + 2 y \leq 8

2 x + y \leq 8

x \geq 0

y \geq 0

\frac{2 x + y}{x + y} = \frac{3}{2} a n d \frac{x - y}{2 x - y} = \frac{- 3}{10}, x + y \neq 0, 2 x - y \neq 0

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