Find the point of intersection of the following lines 3x-2y=0 and 3x+4y-12=0

Solution

3x-2y=0 ………………………(1)

3x+4y-12=0………………….(2)

Subtract equation (2) from equation (1)

3x-2y=0

-( 3x+4y=12)

6y = 12

Y = 2

Substitute for y in equation (1)

3x – 2(2) = 0

3x = 4

X = 4/3

Point of intersection (x,y) = [ 4/3, 2]

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