Simultaneous Eqns solving by sub.: 6/(x-2y)-15/(x+y)=0.5, 12/(x-2y)-9/(x+y)=-0.4

[MathsWannabe]

Hi

I have the following problem and have spent 3 days on it.

Determine x and y

(6/(x-2y)) - (15/(x+y)) = 0.5

(12/(x-2y)) - (9/(x+y)) = - 0.4

The answer is x=-14 and y=-1

I have tried to simplify and got as far as

-9x + 36y = 0.5/(x+y)(x-2y)

3x + 30y = -0.4(x+y)(x-2y)

however, No matter how I substitute, I cannot seem to get rid of the (x+y)(x-2y).

Please help as I cannot solve.

Many Thanks

Mark

 

[stapel]

Determine x and y

(6/(x-2y)) - (15/(x+y)) = 0.5

(12/(x-2y)) - (9/(x+y)) = - 0.4

The answer is x=-14 and y=-1

I have tried to simplify and got as far as

-9x + 36y = 0.5/(x+y)(x-2y)

3x + 30y = -0.4(x+y)(x-2y)

however, No matter how I substitute, I cannot seem to get rid of the (x+y)(x-2y).

Yeah, those are kinda nasty. But if you'll note, they're the same in each of the two equations. So let's replace them, shall we? And make things a little easier for us, at least to start. So:

. . . . .\(\displaystyle \dfrac{1}{x\, -\, 2y}\, =\, A\)

. . . . .\(\displaystyle \dfrac{1}{x\, +\, y}\, =\, B\)

Then our equations become:

. . . . .\(\displaystyle 6A\, -\, 15B\, =\, \dfrac{1}{2}\)

. . . . .\(\displaystyle 12A\, -\, 9B\, =\, -\, \dfrac{2}{5}\)

Now apply the usual techniques for solving linear systems. In this case, minus twice the first equation gives us something that cancels nicely:

. . . . .\(\displaystyle -12A\, +\, 30B\, =\, -1\)

. . . . .\(\displaystyle +12A\, -\, 9B\, =\, -\, \dfrac{2}{5}\)

. . . . .-------------------------------

. . . . .\(\displaystyle +0A\, +\, 21B\, =\, -\, \dfrac{7}{5}\)

This is our new Row 2. So the new system is:

. . . . .\(\displaystyle 6A\, -\, 15B\, =\, \dfrac{1}{2}\)

. . . . .\(\displaystyle +0A\, +\, 21B\, =\, -\, \dfrac{7}{5}\)

And so forth. Solve for the values of A and B; you'll be getting fractions.

Then back-substitute, replacing A and B with their x- and y-based expressions. After rearrangement, you'll get another system to solve, this time in x and y, but also linear, so it'll be simple to deal with. After completing the computations, I do get the answer they gave you.

If you get stuck, please write back, showing all of your steps, starting from the last system shown above. Thank you!

 

[Ishuda]

Hi

I have the following problem and have spent 3 days on it.

Determine x and y

(6/(x-2y)) - (15/(x+y)) = 0.5

(12/(x-2y)) - (9/(x+y)) = - 0.4

The answer is x=-14 and y=-1

I have tried to simplify and got as far as

-9x + 36y = 0.5/(x+y)(x-2y)

3x + 30y = -0.4(x+y)(x-2y)

however, No matter how I substitute, I cannot seem to get rid of the (x+y)(x-2y).

Please help as I cannot solve.

Many Thanks

Mark

Same sort of thing:
u = x + y
v = x - 2y

6/v - 15/u = 1/2
Multiply through by 2 uv
[1] 12 u - 30 v = uv

12/v - 9/u = - 0.4
Multiply through by 2.5uv
[2]30 u - 22.5 v = -uv

Add [1] and 2] and simplify
u = (3/4) v

You now have a very simple quadratic in v (or u if you so choose) using either [1] or [2]. One solution is inadmissible.

EDIT: Dumb mistake, see(ing) red.

 

[MathsWannabe]

Wow! Many thanks Stapel that worked for me !!!
Also thanks to Ishuda for the alternative.

Funny how a small nudge in the right direction always helps

 

[Deleted member 4993]

Ishuda, there will be no quadratic:
u and v can be directly solved (2 equations, 2 unknowns).
Then x and y from the result.

Agree?

In Ishuda's approach, the equations are non-linear ('uv' term present).

Thus - "2 equations - 2 unknowns" → 2 answers - principle "may not" hold good.

Stapel's approach avoids the non-linearity.

 

[Ishuda]

Ishuda, there will be no quadratic:
u and v can be directly solved (2 equations, 2 unknowns).
Then x and y from the result.

Agree?

Certainly you can make it a straight linear system, stapel showed that with his solution. My approach was just different, that's all.

 

[Ishuda]

Ishuda, I wasn't trying to spoil your day.
I was confuded by your results:

To start I disagree with u = (3/4)v:
shouldn't that be u = (5/4)v?

Then I couldn't quite see what you meant by a quadratic;
I see it leads to v^2 + 12v = 0.
I simply divided by v: v + 12 = 0.

Anyhoo...you need to join me and Jomo in the corner
(for that grave (3/4) error); may I ask you to keep
a close eye on Jomo; we're playing X and O's: but the
sob CHEATS: he erased one of my X's and changed
it to an O

Yeah, I think I subtracted when I should have added or maybe half and half or ... I'm going to the corner right now. Oh, and that dividing through by v to make it linear. Only if v isn't zero so two answers 0 and whatever but 0 not allowed because of original equations

 

[Steven G]

Ishuda, I wasn't trying to spoil your day.
I was confuded by your results:

To start I disagree with u = (3/4)v:
shouldn't that be u = (5/4)v?

Then I couldn't quite see what you meant by a quadratic;
I see it leads to v^2 + 12v = 0.
I simply divided by v: v + 12 = 0.

Anyhoo...you need to join me and Jomo in the corner
(for that grave (3/4) error); may I ask you to keep
a close eye on Jomo; we're playing X and O's: but the
sob CHEATS: he erased one of my X's and changed
it to an O

I cheated and still lost! (I think that maybe Denis cheats better than me)

 

[Otis]

... maybe Denis cheats better than me

Of course he does! He used to work in the financial industry, after all. :twisted: