Here are the steps to solve the system of linear equations:
The given system of equations is:
- 4x1+x2−5x3=8
- −2x1+3x2+x3=12
- 3x1+x2+4x3=5
Step 1: Eliminate x1 from equations (1) and (2).
Multiply equation (2) by 2:
2(−2x1+3x2+x3)=2(12)
−4x1+6x2+2x3=24(Equation 4)
Add Equation (1) to Equation (4):
(4x1+x2−5x3)+(−4x1+6x2+2x3)=8+24
7x2−3x3=32(Equation 5)
Step 2: Eliminate x1 from equations (1) and (3).
Multiply equation (1) by 3:
3(4x1+x2−5x3)=3(8)
12x1+3x2−15x3=24(Equation 6)
Multiply equation (3) by 4:
4(3x1+x2+4x3)=4(5)
12x1+4x2+16x3=20(Equation 7)
Subtract Equation (6) from Equation (7):
(12x1+4x2+16x3)−(12x1+3x2−15x3)=20−24
12x1+4x2+16x3−12x1−3x2+15x3=−4
x2+31x3=−4(Equation 8)
Step 3: Solve the system of equations (5) and (8) for x2 and x3.
The new system is:
5) 7x2−3x3=32
8) x2+31x3=−4
From Equation (8), express x2 in terms of x3:
x2=−4−31x3
Substitute this expression for x2 into Equation (5):
7(−4−31x3)−3x3=32
−28−217x3−3x3=32
−220x3=32+28
−220x3=60
x3=−22060
x3=−113
Step 4: Substitute the value of x3 into Equation (8) to find x2.
x2=−4−31(−113)
x2=−4+1193
x2=11−4×11+93
x2=11−44+93
x2=1149
Step 5: Substitute the values of x2 and x3 into Equation (1) to find x1.
4x1+x2−5x3=8
4x1+1149−5(−113)=8
4x1+1149+1115=8
4x1+1149+15=8
4x1+1164=8
4x1=8−1164
4x1=118×11−64
4x1=1188−64
4x1=1124
x1=4×1124
x1=116
The solution to the system of equations is:
x1=116,x2=1149,x3=−113
x1=116,x2=1149,x3=−113
3 done, 2 left today. You're making progress.