Solve system of eqns: 4x+2y+3z=52, x+3y+2z=48, x+y+z= ?

[unknown621]

Hello. Can anyone help me with this one?

4x+2y+3z=52
x+3y+2z=48
x+y+z= ?

Sorry if I posted this in wrong category. Im in rush.

 

[Dr.Peterson]

Hello. Can anyone help me with this one?

4x+2y+3z=52
x+3y+2z=48
x+y+z= ?

Sorry if I posted this in wrong category. Im in rush.

The problem as written has no solution; were you in too much of a rush to check it, or is it just a bad problem?

The best you could do is to eliminate one variable and express x+y+z in terms of the other two variables (e.g. x-y+28).



Correction: You can express two variables in terms of one variable, e.g. y = x+8, z = 12-2x; then x+y+z = x + (x+8) + (12-2x) = 20.

The line of intersection of the two planes happens to lie in the plane x+y+z=20; I didn't go far enough. (My expression turns out to equal 20 for any x.)

Of course, if you happen to notice the trick, you can do it more quickly ...

 

[pka]

Hello. Can anyone help me with this one?
(1) 4x+2y+3z=52
(2) x+3y+2z=48
(3) x+y+z= ?

Add (1) & (2) to get \(\displaystyle 5x+5y+5z=100 \).
Now \(\displaystyle x+y+z=~? \)
HINT divide.