0
$\begingroup$

$$\begin{cases}\begin{align}x-4y+3z=5\\y+2z=6\end{align}\end{cases}$$ Given that system of equation, the question calls on to solve the system

I was very confused because with two variables on the bottom equation and three variables in this I had difficulties in solving for any of the variables and would like some help.

Currently I have subtracted the first equation with $4(y+2z)=24$ which cancels out the $y$ variable but that didn't help at all.

$\endgroup$
  • $\begingroup$ If one equation has three variables and the other has two, simply put a zero in front of the missing variable in the second equation. That is, consider the system: $$\begin{cases}x-4y+3z=5\\0x+y+2z=6\end{cases}$$ Are you familiar with row operations and matrices? $\endgroup$ – Dave Jun 8 '17 at 0:22
1
$\begingroup$

First, this system has many solutions. You can check that both $(x,y,z)=(29,6,0)$ and $(-4,0,3)$ are solutions. So the solution is not of the form that $x$, $y$ and $z$ equal to some particular numbers.

If you let $z=t$, then $y=-2t+6$ and therefore,

$$x-4(-2t+6)+3t=0$$

and hence $x=11t+24$.

$(x,y,z)=(11t+24,-2t+6,t)$ is a solution for any $t\in \mathbb{R}$. This is the general solution.

$\endgroup$
  • $\begingroup$ so does that mean that there are infinitely many solution sets to this system of equations? $\endgroup$ – John Rawls Jun 8 '17 at 0:36
  • $\begingroup$ Yes. Each $t$ is corresponding to a solution. The solutions form a line in $\mathbb{R}^3$ $\endgroup$ – CY Aries Jun 8 '17 at 1:43
0
$\begingroup$

I noticed in a comment that the OP wasn't sure if there were an infinite number of solutions. How about looking at something really simple to get some insight.

Here is a linear system with 2 equation and 3 unknowns.

$x + y = 1$
$z = 1$

The first equation is a line on the x,y-axis. The second equation is telling you that in the $x,y,z$ space, the line is up one unit off of the x,y plane. But it is a line.

How do you describe a line in 3-space? Well, if you want to see a line, you would like to see a 1:1 correspondence with the real number line. There is no unique way to do this, but this process is called 'parameterization'.

You can google this parameterization of a line in 3 space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.