solving system of linear equations

Mar 2016
8
0
New York
need help in solving this equation by means of elimination
x+4y-z=12
3x+8y-2z=4

I don't know what to do because the number of unknown is not equal to the number of linear equation...
please help me(Thinking)
 
Feb 2014
1,748
651
United States
If you have a system of two equations with three unknowns, you can reduce it to a system of one equation in two unknowns. In other words, you cannot get a numeric answer, but you can get an algebraic answer.

Do you see how to proceed now?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Assuming the two equations are independent (I didn't check) you can solve for two of the unknowns in terms of the other, say, solving for x and y in terms of z. Letting z= t, we have the parametric equations, x= x(t), y= y(t), z= t. Since that depends on one parameter, it describes a "one dimensional" geometric object. Since the equations are linear, It is a straight line.
 
Feb 2015
2,255
510
Ottawa Ontario
Well, turns out that x = -20 (as stated in my other post),
but that z = 4y - 32, so solutions galore...
 

Debsta

MHF Helper
Oct 2009
1,313
599
Brisbane
You are, after all, finding the equation of the intersection of two planes. You should expect a line, which you have, rather than a single point.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Well, turns out that x = -20 (as stated in my other post),
but that z = 4y - 32, so solutions galore...
Yes, that is what you have been told in these responses. Taking y= t as parameter, you have x= -20, y=t, z= 4t- 32, the parametric equations of a line.
 
Feb 2015
2,255
510
Ottawa Ontario
Purty evident anyway, ain't it?

x+4y-z=12
3x+8y-2z=4

same as:
x + k = 12
3x + 2k = 4