X+2y+z=7;x+3z=11;2x-3y=1
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Hi friend,
x+2y+z = 7
x+3z = 11
2x-3y = 1
→ x+3z = 11
x = 11-3z
→ 2x-3y = 1
2(11-3z) = 1+3y
22-6z = 1+3y
3y = 21-6z
y = 3(7-2z)/3
y = 7-2z
Substitute x and y values,
x+2y+z = 7
11-3z+2{7-2z}+z = 7
11-3z+14-4z+z = 7
25-6z = 7
6z = 25-7
6z = 18
z = 18/6 = 3
x = 11-3z = 11-3(3) = 11-9 = 2
y = 7-2z = 7-2(3) = 7-6 = 1
Therefore, x = 2 , y = 1 and z = 3
Hope it helps......
x+2y+z = 7
x+3z = 11
2x-3y = 1
→ x+3z = 11
x = 11-3z
→ 2x-3y = 1
2(11-3z) = 1+3y
22-6z = 1+3y
3y = 21-6z
y = 3(7-2z)/3
y = 7-2z
Substitute x and y values,
x+2y+z = 7
11-3z+2{7-2z}+z = 7
11-3z+14-4z+z = 7
25-6z = 7
6z = 25-7
6z = 18
z = 18/6 = 3
x = 11-3z = 11-3(3) = 11-9 = 2
y = 7-2z = 7-2(3) = 7-6 = 1
Therefore, x = 2 , y = 1 and z = 3
Hope it helps......
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