Question 2240: Ok this is my last problem. That you so much!
Solve for X,Y,and Z in the following systems of three equations:
A) X+2Y+Z =6
X+Y=4
3X+Y+Z=8
B) 10X+Y+Z=12
8X+2Y+Z=11
20X-10Y-2Z=8
C) 22X+5Y+7Z=12
10X+3Y+2Z=5
9X+2Y+12Z=14
Answer by DeepaEvangeline(15)   (Show Source):
You can put this solution on YOUR website! a)
X+2Y+Z=6 - (1)
X+Y=4 - (2)
3X+Y+Z=8 - (3)
Eqn(3) - Eqn(1) gives
2X-Y=2 - (4)
Adding Eqn(2) & (4) gives
3X=6 -> X = 2
Substituting the value of X in eqn(2) gives
X+Y=4
2+Y=4 -> Y = 2
Sub the value of X=2 and Y=2 in eqn (1) gives
2+2.2+Z=6
2+4+Z=6 -> Z = 0
X = 2, Y = 2, Z = 0
b)
10X+Y+Z=12 - (1)
8X+2Y+Z=11 - (2)
20X-10Y-2Z=8 - (3)
Subtracting eqn(2) from eqn(1) gives
2X-Y=1 - (4)
Eqn(2) x 2 gives
16X+4Y+2Z=22 - (5)
Eqn(3) + Eqn(5) gives
36X-6Y=30 - (6)
Eqn(4) x 6 gives
12X-6Y=6 - (7)
Eqn(6) - Eqn(7) gives
24X=24 => X = 1
Substituting X = 1 in eqn(4) gives
2.1-Y=1 => Y = 1
Sub the value of X = 1 and Y = 1 in eqn(1) gives
10.1 + 1 + Z = 12
11 + Z = 12 => Z = 1
X = 1, Y = 1, Z = 1
c)
22X+5Y+7Z=12 - (1)
10X+3Y+2Z=5 - (2)
9X+2Y+12Z=14 - (3)
Eqn(1) x 3 gives
66X+15Y+21Z=36 - (4)
Eqn(2) x 5 gives
50X+15Y+10Z=25 - (5)
Eqn(4) - eqn(5) gives
16X+11Z=11 - (6)
Eqn(2) x 2 gives
20X+6Y+4Z=10 - (7)
Eqn(3) x 3 gives
27X+6Y+36Z=42 - (8)
Eqn(8) - eqn(7) gives
7X+32Z=32 - (9)
Solving eqn(6) & (9),
Eqn(6) x 7 gives
112X+77Z=77 - (10)
Eqn(9) x 16 gives
112X+512Z=512 - (11)
Eqn(11) - eqn(10) gives
435Z=435 => Z = 1
Substituting Z = 1 in eqn(9) gives
7X+32=32 => X = 0
Sub X = 0 and Z = 1 in eqn(2) gives
10.0 + 3Y + 2.1 = 5
3Y = 3 => Y = 1
X = 0, Y = 1, Z = 1
|
|
|
