Solve the system by addition or substitution.
3x – 4y = 8
6x – 2y = 10
3x-4y = 8.........(1)
6x- 2y = 10.......(2)
From eq'n (2) y = (10-6x)/-2 = -5+3x
substituing the value of y in eq'n (1) we get 3x-4(-5+3x) = 8
3x+20-12x = 8
-9x = 8-20 = -12
x = -12/-9 = 4/3
substitute the value of x in eq'n (1) we get 3(4/3)-4y = 8
4-4y = 8
-4y = 8-4 = 4
y = 4/-4 = -1
The values of x = 4/3 and y = -1 are the solutions for the eq'ns
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