I am stuck trying to get the values for x, y, and z. I keep moving variables around but I end up getting answers like x = x or z = z and I do not think that is what I want. It's really just algebra but I seem to have forgotten that.

(x/2) + (2z/3) = x

(x/2) + (y/3) = y

(2y/3) + (z/3) = z

I initially tried finding the values for z and y from the first two equations and plugged those into the third equation but ended up getting that x is equal to 0 and when I plug zero into the other equations to solve for the other three variables I get that x, y, and z are all equal to 0. I do not think I am tackling this correctly.

  • $\begingroup$ How is x=y=z=0 not a valid solution? $\endgroup$ – JB King Apr 24 '14 at 0:26
  • $\begingroup$ Well I am more concerned with how I am tackling the problem. If The way I have found the values is valid? @JBKing $\endgroup$ – userunknown Apr 24 '14 at 0:30

From the first equation, we have:

$$z = \dfrac{3x}{4}$$

From the second equation, we have:

$$y = \dfrac{3x}{4}$$

Substituting these into the third equation gives an identity. So, we have:

$$z = y = \dfrac{3x}{4},~ x ~~\mbox{is a free variable}$$

Note, you could have rewritten this system as:

$$3 x=4 z,3 x=4 y,y=z$$


Step 1: Move all unknowns to the left, in the same order, i.e. a*x + b*y + c*z = d

-1/2 * x + 0 * y + 2/3 * z = 0 (1)

1/2 * x - 2/3 * y + 0 * z = 0 (2)

0 * x + 2/3 * y - 2/3 * z = 0 (3)

This is a homogeneous linear system, and is normally expressed as the product of a matrix and vector. If you want a rigorous run through of how to solve a problem of this sort, look up Gaussian Elimination. I'll provide a quick run through in the context of separate equations, rather than matrix operations, though they are just different ways of expressing the same mathematical operations.

The overall aim is to modify equation (3) to look like 0 * x + 0 * y + a * z = b. That's one equation in one unknown, and can be solved. Next, if we can make modify equation (2) such that it looks like 0 * x + c*y + d*z = e, since we know z, we'll be able to find y. We can then use known values of y and z in equation (1). This is known as backward substitution.

Before we can do backward substitution, however, we need to make equations (2) and (3) look 'appropriate'.

Let's start off by trying to make the coefficients of the x terms in equations (2) and (3) zero. (3) already has a zero coefficient of x (though this isn't always the case), so that leaves just (2).

In systems like this, we're able to add any number of one equation to another, without changing the solution. How many lots of equation (1) should we add to equation (2) such that we get an equation with a zero x coefficient? 1!

Replace (2) with (2) + (1)

-1/2 * x + 0 * y + 2/3 * z = 0 (1)

0 * x - 2/3 * y + 2/3 * z = 0 (2)

0 * x + 2/3 * y - 2/3 * z = 0 (3)

That's looking better - equation (2) is now in the correct form. Next, we want to make the coefficient of y in (3) zero. We could try and use equation (1) for this, but if we did this, we'd be adding a non-zero x-term - something we don't want! If, on the other hand, we use equation (2), we know we won't be adding a non-zero x-term, since we know the x-term in equation (2) is zero (a result of the previous step).

How many lots of equation (2) do we add to equation (3) to get a zero y-coefficient? 1 again!

Replace (3) with (3) + (2)

-1/2 * x + 0 * y + 2/3 * z = 0 (1)

0 * x - 2/3 * y + 2/3 * z = 0 (2)

0 * x + 0 * y + 0 * z = 0 (3)

Now we're ready to backward substitute.

Looking at equation (3), we see it's now meaningless. Any value of x, y and z will satisfy equation 3, so we effectively only have two meaningful equations with 3 unknowns. This means we'll have infinitely many solutions - if you give me a specific value of z, I can give you values of x and y that will make each of (1) and (2) hold.

In this situation, we say we let z be free - often by writing something along the lines, "let z = t", just so it is clearly identified as something we're not trying to solve for, rather than something we simply haven't solved for yet, and move on.

The next step of backward substitution, as the name implies, is to go up an equaiton ('backwards') and substitute values we know. Equation (2) thus becomes

-2/3 y + 2/3 t = 0

=> y = t

This might not look like much of an answer, but it's still meaningful. This says, whatever value z is, y is the same. They're both free, but they're not free independent of each other. If you give me a specific value of z, there's only one possible value of y that will satisfy these equations.

Now that we've processed 2, we continue backward substitution by substituting in known values to (1)

-1/2*x + 0 * t + 2/3*t = 0

=> x = 4/3 * t

We now have values of x, y and z, dependent on some free parameter, t.

x = 4/3 t

y = t

z = t

You give me a value of t - and it can be -any- value - and I'll give you values of x, y and z that satisfy those equations.

E.g if t = 3,

x = 4

y = 3

z = 3

satisfies the original equations. Since there are an infinite number of possible values of t, there are an infinite number of solutions. When t = 0, we get what's known as the trivial solution - x = y = z = 0 - but this obviously isn't the only solution.


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