The co-ordinates of the point of intersection of the lines $3 x + 4 y - 18 = 0$ and $4 x - 5 y + 7 = 0$ are

(- 2, - 3)

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(2, 3)

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(2, - 3)

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(- 2, 3)

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Solution

The correct option is D $(2, 3)$
$3 x + 4 y - 18 = 0 ⟹ 12 x + 16 y = 72$ ...(1)
$4 x - 5 y + 7 = 0 ⟹ 12 x - 15 y = - 21 = 0$ ....(2)
Subtracting equation 2 from equation 1, we get,
$31 y = 93$
$y = 3$
Putting this value in equation 1, we get,
$x = 2$
Hence, the required point is $(2, 3)$ .

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The co-ordinates of the point of intersection of the lines 3x+4y−18=0 and 4x−5y+7=0 are

The correct option is D (2,3)3x+4y−18=0⟹12x+16y=72 ...(1)4x−5y+7=0⟹12x−15y=−21=0 ....(2)Subtracting equation 2 from equation 1, we get,31y=93y=3Putting this value in equation 1, we get,x=2Hence, the required point is (2,3).

The co-ordinates of the point of intersection of the lines $3 x + 4 y - 18 = 0$ and $4 x - 5 y + 7 = 0$ are

The correct option is D $(2, 3)$
$3 x + 4 y - 18 = 0 ⟹ 12 x + 16 y = 72$ ...(1)
$4 x - 5 y + 7 = 0 ⟹ 12 x - 15 y = - 21 = 0$ ....(2)
Subtracting equation 2 from equation 1, we get,
$31 y = 93$
$y = 3$
Putting this value in equation 1, we get,
$x = 2$
Hence, the required point is $(2, 3)$ .