# Sequence question

1. Aug 6, 2014

1. The problem statement, all variables and given/known data
The last two terms, are added together to produce the next term

$-33,x,y,z,88$
$\text{Find }x,y\text{ and }z$
2. Relevant equations
$y=x-33$
$x+y=z$
$y+z=88$
3. The attempt at a solution

By substituting the expression for y in the third equation for the first equation, we get:
$88-z=x-33 \rightarrow x=121-z$
Again, by substituting the expression for y in the first equation for the second equation we get:
$x+(x-33)=z \rightarrow 2x-33=z$

So we get two equations,
$x=121-z$
$2x-33=z$

Solving both gives the right answers but I remember getting the wrong answer by not including one of the equations there.
Is there any easy way to solve all three equations at once?

Last edited: Aug 6, 2014
2. Aug 6, 2014

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by "solve all three at once". I get this image of just immediately writing down the values for x, y, and z! I could not do that!:tongue:

But you can rewrite the first equation as y- x= -33. Adding that to x+ y= z gives 2y= z- 33. The last equation, y+ z= 88, is equivalent to y= -z+ 88. Adding those equations, 3y= 55.

That's about as simple as I can make it.

3. Aug 6, 2014

Oh, I must have made some silly mistake in the previous calculations. Thanks :)

Haha, I mean like how you solve two simultaneous equation, is there a method for solving three simultaneous equations like this? Other than substituting the values?

4. Aug 6, 2014

### HakimPhilo

This system of linear equations, like any other, can be solved by first writing it in upper-triangular form then using the method of back-substitution. Let's first define the terms we used: A system in upper-triangular form is one like: $$\left\{\begin{array}{rl}4x-3y+2z&=-5\\14y+2z&=18\\-4z&=3\end{array}\right.,\qquad\left\{\begin{array}{rl} 15x-2y+z&=1\\ 3z&=-8\end{array}\right..$$ As you can see, what's common about them is that if $x$ appears in no equation other than the first one and $y$ appears in no equation after the second... (It is possible that $y$ may not even appear in the second equation as in e.g.2)
You can easily solve systems in this type using the method of back-substitution. Take the first example, using the third equation you solve for $z$ to get $z=-\tfrac34$, you substitute that value in the second equation to get $14y-\tfrac32=18$ which can be used to solve for $y$, and when you get your value for $y$, you substitute it in the first equation along with your value for $z$ to solve for $x$.
What's interesting now is that you can try to transform any system you have in upper-triangular form, then you use back-substitution and you're done. The basic way we proceed with this task consists of applying the elementary operations which are: 1. multiplying both sides of an equation by a nonzero constant, 2. interchanging the order of two equations, 3. adding to one equation a multiple of another.
I think this will give you enough information to solve any system in 3 variables or more. If you need to know more then just ask away.

5. Aug 7, 2014