## Annoying problem (simultaneous equations)

Discussions concerned with knowledge of measurement, properties, and relations quantities, theoretical or applied.  ### Annoying problem (simultaneous equations)

This is really simple, but for some reason I am just stuck on this (Feel pretty stupid at the moment!)

I want a step by step solution to this problem by SUBSTITUTION. I can easily do this by elimination, but for some bizarre reason my attempts at SUBSTITUTION keep failing.

1) (3/x) + (2/y) = 14

2) (5/x) - (3/y) = -2

This looks pretty simple to me yet somewhere along the way I am making some drastic error when number crunching. It's gotten to the point where I've started getting several different answers and I cannot see where I am going wrong.

I have attempted ... oh shit! Just spotted a stupid error!! haha!! Now I get no answer!

I am right in thinking the solution cannot be found unless I use elimination? If not how else could I find the solution? Resident Member

Posts: 5570
Joined: 14 Mar 2012    ### Re: Annoying problem (simultaneous equations)

BadgerJelly » November 1st, 2017, 9:59 pm wrote:
1) (3/x) + (2/y) = 14

2) (5/x) - (3/y) = -2

From (1) we have 3y + 2x = 14xy and from (2) we have 5y - 3x = -2xy. Let's see, we can multiply (2) by 7 to get 35y - 21x = -14xy. Adding that to (1) makes the xy term go away.

Working out the rest of this, after adding the equations we get 38y - 19x = 0. It's always good when things divide evenly, that's a clue that we're doing this right. We have 38y = 19x and 2y = x.

Now substituting back in to (1) we have 3/(2y) + 2/y = 14, then multiplying by 2y gives 3 + 4 = 28y or 7 = 28y and y = 1/4.

Now plugging y = 1/4 into (1), we have 3/x + 2/(1/4) = 14, or 3/x + 8 = 14, or 3/x = 6 and x = 1/2.

Does that look right? Yes, if we plug our values for x and y back into (1) and (2) they both check out.

I'm a little too long out of school so I'm not sure what's substitution versus elimination. I just think of this as hacking away at the algebra till you get the answer. The main tricks were first, getting the variables out of the denominators; and then getting rid of that pesky xy term.
someguy1
Member

Posts: 753
Joined: 08 Nov 2013    ### Re: Annoying problem (simultaneous equations)

Adding or subtraction is elimination. You used elimination. The problem I find with pure subtraction, as you show yourself, is that the xy values only seem to be able to resolved by elimination (which is what I am trying to avoid.)

I have managed to do it another way (using ratios), but that is really tiresome and nothing other than elimination really. At least it was a different approach. I think there is a way (possibly?) by using quadratic equations, but obviously that would involve a +/- answer.

I guess there is value in making errors. Helps you look at problems in completely different ways. I never looked at the importance of an unknown denominator before. Back in school/college I just number crunched without much thought about why it worked.

note: to anyone who cares, and doesn't know, the quickest way is to just subst in a=(1/x) and b=(1/y). I was just playing around with ways where I avoided this initial step because I forgot about it! Resident Member

Posts: 5570
Joined: 14 Mar 2012    ### Re: Annoying problem (simultaneous equations)

BadgerJelly » November 1st, 2017, 10:59 pm wrote:I am right in thinking the solution cannot be found unless I use elimination? If not how else could I find the solution?

No. Ultimately it is equivalent. I think you can change an elimination solution into a substitution solution with a little rearranging.

That's not how I did it though.

I took the second equation and solved for y = 3( 1/(5/x+2))
Then I substituted this in for 2/y = (2/3)(5+2x)/x = (2/3)(5/x +2)
Now add 3/x to get the first equation 3/x +2/y = 3/x +10/(3x) +4/3 =14
Combine the x terms to get 19/3x +4/3 =14
19/3x = 38/3
x = 1/2
plug this into the second equation to get 10 - 3/y = -2 or -3/y = -12 or y = 1/4

You can see the same numbers popping up as when you do the problem by elimination because there is an equivalence here between the two solutions.

There is also another way to do the problem.... change of variables.
let a = 1/x and b = 1/y then the two equations become
1) 3a + 2b =14
2) 5a + 3b = -2
Then is it easy to solve this for a = 2 and b = 4. Then plug these in to get x and y.

Hint: The thing is, you need to avoid overly complicated fractions if at all possible because there are two many degrees of freedom in the structure of those things. There are endless ways of writing the same thing in complicated fractions, but difficult to see the relationship. So it is easy to get stuck going around in circles with those things. (e.g. Take a look at continued fractions. Ramanujan loved those things and came up with all kinds of things using them) mitchellmckain
Active Member

Posts: 1300
Joined: 27 Oct 2016
 TheVat liked this post  