# solving system of linear equations

#### noface0711

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...

#### JeffM

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?

#### DenisB

x+4y-z=12
3x+8y-2z=4
Yer lucky!
Multiply 1st equation by -2 : -2x - 8y + 2z = -24

Now add the 2 equations to get x = -20

Continue...

#### HallsofIvy

MHF Helper
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.

#### DenisB

Well, turns out that x = -20 (as stated in my other post),
but that z = 4y - 32, so solutions galore...

#### Debsta

MHF Helper
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
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.

Thanks Halls.

#### DenisB

Purty evident anyway, ain't it?

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

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