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The point of ...

Question

The point of intersection of $3 x + 2 y = 11$ , $2 x + 3 y = 4$ lies in

A

Q_{1}

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B

Q_{2}

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C

Q_{3}

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D

Q_{4}

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Solution

The correct option is A $Q_{1}$
$3 x + 2 y = 11...... (1) \times 2$
$2 x + 3 y = 4 . . . . . . (2) \times 3$

Subtract equation (2) from equation (1)
$\Rightarrow y = 2$

Putting $y = 2$ in (1)

$3 x + 2 y = 11$
$\Rightarrow 3 x + 2 \times 2 = 11$
$\Rightarrow 3 x = 11 - 4$
$\Rightarrow x = \frac{7}{3}$

So point of intersection is $(\frac{7}{3}, 2)$
Therefore intersection point lies in first quadrant
I.e $Q_{1}$

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