Question

solve by gauss elimination method x+3y+2z=19, 2x+y+z=13,4x+2y+3z=31​

Answer 1

Answer:
x=3,y=2,z=5

Explanation:
x+3y+2z=19 —————— 1
2x+y+z=13 ——————- 2
4x+2y+3z=31 —————- 3
Multiplying eq 1 by 2 and eq 2 by 1
2x+6y+4z=38 ——————- 4
2x+y+z=13 —————- 5
Subtracting eq 4 and 5
5y+3z=25—————— 6
Now,
Multiplying eq 1 by 4 and eq 3 by 1
4x+12y+8z=76 ————- 7
4x+2y+3z=31 ————- 8
Subtracting eq 7 and 8
10y+5z=45 ————— 9
Multiplying eq 6 by 2 and eq 9 by 1
10y+6z=50—————- 10
10y+5z=45—————- 11
Subtracting eq 10 and 11
6z-5z=50-45
z=5
Substituting z=5 on eq 9
10y+5*5=45
10y+25=45
10y=45-25
10y=20
y=2
Substituting y=2 and z=5 on the first equation(eq 1)
x+3*2+2*5=19
x+6+10=19
x+16=19
x=3