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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