2元1次方程

x/2 - y/3 = 4 --------- (1)
x/4 + y/2 = 8 -------- (2)

(1) * 6, 3x - 2y = 24 ------- (3)
(2) * 4, x + 2y = 32 ------- (4)
(3) + (4), 4x = 56
x = 14 ----------- (5)

sub (5) into (4),
2y = 18
y = 9

so, x = 14, y = 9

x/2 - y/3 = 4.........(1)
x/4 + y/2 = 8 ..............(2)
(1), x/2 - y/3 = 4
6(x/2 - y/3)=6(4)
3x-2y=24
3x=24+2y
x=(24+2y)/3.........(3)
(3)to(2), ((24+2y)/3)/4+y/2=8
4(((24+2y)/3)/4+y/2)=4(8)
(24+2y)/3+2y=32
3((24+2y)/3+2y)=32
24+2y+6y=32
24+8y=32
8y=32-24
8y=8
y=1
y=1 to (1), x/2 - 1/3 = 4
x/2=4+1/3
x/2=13/3
x=26/3

(x/2) - (y/3) = 4 and (x/4) + (y/2) = 8
3x - 2y = 24 and x + 2y = 32
4x = 56 ==> x = 14
2y = 32 - 14 =====> y = 9

x/2 - y/3 = 4
x/4 + y/2 = 8
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