Consider the lines $2 x + 3 y = 0$ , $5 x + 4 y = 7$ . Find the intersection point.

(3, - 2)

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(3, 2)

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(- 3, 2)

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(2, 3)

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Solution

The correct option is A $(3, - 2)$
The lines are $2 x + 3 y = 0$ ......(1)

$x = \frac{- 3 y}{2}$
$5 x + 4 y = 7$ ........(2)
$5 (\frac{- 3 y}{2}) + 4 y = 7$
$- 15 y + 8 y = 14$
$- 7 y = 14$
$y = - 2$
$x = \frac{- 3 (- 2)}{2} = 3$

$(x, y) = (3, - 2)$

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Similar questions

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Consider the lines 2x+3y=0, 5x+4y=7. Find the intersection point.

The correct option is A (3,−2)The lines are 2x+3y=0......(1)x=−3y25x+4y=7........(2)5(−3y2)+4y=7−15y+8y=14−7y=14y=−2x=−3(−2)2=3(x,y)=(3,−2)

Consider the lines $2 x + 3 y = 0$ , $5 x + 4 y = 7$ . Find the intersection point.

The correct option is A $(3, - 2)$
The lines are $2 x + 3 y = 0$ ......(1)

$x = \frac{- 3 y}{2}$
$5 x + 4 y = 7$ ........(2)
$5 (\frac{- 3 y}{2}) + 4 y = 7$
$- 15 y + 8 y = 14$
$- 7 y = 14$
$y = - 2$
$x = \frac{- 3 (- 2)}{2} = 3$

$(x, y) = (3, - 2)$