a) Solve the system:

$$\begin{cases}x-y+z=10 \\ 2x+3y-2z=-21 \\ \frac{1}{2}x+ \frac{2}{5}y+ \frac{1}{4}z=-\frac{1}{2} \end{cases}$$

enter image description here

b) Give a geometric interpretation of the solution(s).

I tried doing part a) but I instantly got confused as I don't know what the question is asking me. For example: How would I solve this? What would I need to do first? Is there more than one solution to this?

As I tried to solve part b), I realized that I have solve to part a) before getting to part b). As well, I got more confused on the question of part b) as well. If anyone can help me out, that would be great. I am a new user so if this question is not that good of a question, then I apologize. Thanks.

Here is how I tried to get the answer. I still haven't gotten the answer because I feel that the way I'm doing it is incorrect. Can anyone check it over, and please help me out? enter image description here

Thanks again.

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    $\begingroup$ Try substitution (successively rewriting one variable in terms of the others, until you're left with a linear equation in one variable) or elimination (multiplying and subtracting equations wholesale to get to a linear equation in one variable). Once you know one variable, back-substitute to find the others. Geometrically, what do the equations represent? What shapes are they? What would it mean to be in/on all three? $\endgroup$ – Integrand Jun 5 '20 at 3:04
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    $\begingroup$ Btw, the answer is $(1,-5,4)$ $\endgroup$ – Alexey Burdin Jun 5 '20 at 3:21
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    $\begingroup$ You might find it helpful to get some experience with a simpler problem first, then you can see the approach to try for the system you have. For example, consider a system of two lines: $y=2x+3$ and $y=4x+1$. Rewriting them like above gives $-2x+y=3, -4x+y=1$. Solving it means finding the $x$ and $y$ that satisfies both equations. In this example, the solution is $x=1,y=5$. Now, geometrically, if you sketched the lines, what is the significance of the solution point (1,5)? Now you can think about the 3D problem you have in the question in a similar, but 3D way. I hope this helps. $\endgroup$ – ad2004 Jun 5 '20 at 3:41
  • $\begingroup$ @AlexeyBurdin I was wondering if you can explain how you got the answer? I've tried lots of ways and I couldn't get the answer at all. $\endgroup$ – Panda14 Jun 5 '20 at 5:38
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    $\begingroup$ Maybe your confusion about (b) is, that there can be infinitly many solutions. Which form a Plane or a Line in the 3-dimensional space. In (b) they basically asking: Is the solution a plane, a line, a point or not existing. Eventhough the existence of the (b) questions kind of gives a little hint, that there exists atleast one solution. But thats just me guessing ;) $\endgroup$ – CoffeeArabica Jun 5 '20 at 7:08


You should study a more systematic way using Gaussian elimination.

I see that you are trying to use substitution, that's fine too, but you should not substitute everything simultaneously as that would not get rid of variables. Let's do it one by one.


Now substitute this into the second and third equation.



Now you have two equations and two unknown, try to solve them, either by another substitution or elimination.


It is a point


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