SOLUTION: Can you please help me with this. The directions say solve each system algebraically. {{{2x-y+z=11}}} {{{x+2y+3z=8}}} {{{3x-4y-5z=-2}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can you please help me with this. The directions say solve each system algebraically. {{{2x-y+z=11}}} {{{x+2y+3z=8}}} {{{3x-4y-5z=-2}}}      Log On


   



Question 195166This question is from textbook
: Can you please help me with this.
The directions say solve each system algebraically.
2x-y%2Bz=11
x%2B2y%2B3z=8
3x-4y-5z=-2
This question is from textbook

Answer by anantha(86) About Me  (Show Source):
You can put this solution on YOUR website!
sol: 2x-y+z=11 this is 1st equation
x+2y+3z=8 this is 2nd equation
3x-4y-5z=-2 this is 3rd equation
let us solve 1st and 2nd equations
2x-y+z=11
x+2y+3z=8
multiplying 2nd equation by 2 on both sides we get
2x+4y+6z=16
subtract this value from 1st equation means change the symbols(i.e; write - instead of +)
2x-y+z=11
-2x-4y-6z= -16
---------------------
0x-5y-5z=-5 this is 4th euation
let us solve 2nd and 3rd equations
x+2y+3z=8
3x-4y-5z=-2
multiplying 2nd equation by 3
3x+6y+9z=24
3x-4y-5z=-2
subtracting these two equations means change the symbols of 3rd equation
3x+6y+9z=24
-3x+4y+5z=+2
----------------
0x+10y+14z=26 this is 5 th equation
now solving 4th and 5th equations
multiplying 4t equation by 2
0x-10y-10z= -10
0x+10y+14z= 26
------------------ adding these two we get
0x+0y+4z=16
4z=16
dividing both sides by 4
4z/4=16/4
z=4
substitute this value in 4th equation
0x-5y-5z=-5
0x-5y-5*(4)=-5
-5y-20=-5
add 20 on both sides
-5y-20+20=-5+20
-5y=-15
5y=15
dividing by 5 on both sides
5y/5=15/5
y=3
substitute y and z values in 1st equation
2x-y+z=11
2x-3+4=11
2x+1=11
subtract 1 on both sides
2x+1-1=11-1
2x=10
dividing by 2 on both sides
2x/2=10/2
x=5
x=5,y=3,z=4
check these values satisfy the equation or not
2x-y=z=11
2*5-3+4=11
10-3+4=11
7+4=11
11=11
these values satisfy the equation