# Thread: Find x AND y

1. ## Find x AND y

It's been a long time since I had to do any algebra, and nothing I've found on Google seems to be pitched at the right level for me. I'm looking to find a number for 'x' and a number for 'y' (consistant across these two equations):

(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

2. Originally Posted by Dufus
It's been a long time since I had to do any algebra, and nothing I've found on Google seems to be pitched at the right level for me. I'm looking to find a number for 'x' and a number for 'y' (consistant across these two equations):

(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

Is this a set of equations you need to solve simultaneously?

3. Originally Posted by Prove It
Is this a set of equations you need to solve simultaneously?
Yes, x is the same in both equations and y is the same in both equations.

4. Originally Posted by Dufus
(1/3) * (5/113) * (x/y) = 0.05
HINT: above results in 100x = 339y

5. Originally Posted by Wilmer
HINT: above results in 100x = 339y
I appreciate the helpful nudge, but I'm getting more and more bogged down. Here's the rabbit hole I've followed from that:

100x = 339y
x = 3.39y

(1/3) * (5/113) * (3.39) = 0.05
2/3 + ((1/3) * (5/113) * ((y-(3.39*y))/y)) = 0.63

((1/3) * (5/113) * (3.39)) - (2/3) = -0.61666666666666666666666666666667
(1/3) * (5/113) * ((y-(3.39*y))/y) = -0.036666666666666666666666666666667

(((1/3) * (5/113) * (3.39)) - (2/3)) / (1/3) = -1.850000000000000000000000000001
(5/113) * ((y-(3.39*y))/y) = -0.11

(((1/3) * (5/113) * (3.39)) - (2/3)) / (1/3) / (5/113) = -41.810000000000000000000000000023
(y-(3.39*y))/y = -2.486

I've come unstuck at this point.

Edit:

I think I've narrowed it down this far:

x = 3.39y
(y-x)/y = (0.63-(2/3)) / (1/3) / (5/113)

I just don't know how to untangle that "(y-x)/y" beyond substituting x:

(y-(3.39*y))/y = (0.63-(2/3)) / (1/3) / (5/113)

And that doesn't seem to help much, assuming I'm even on the right track.

6. Hello, Dufus!

Who created this problem? . . . What a horrible way to write it!

And don't use decimals!

Besides, the problem has no solution . . .

Solve the system: . $\displaystyle \begin{array}{ccccc}\dfrac{1}{3}\!\cdot\!\dfrac{5} {113}\!\cdot\!\dfrac{x}{y} &=& \dfrac{1}{20} & {\color{blue}[1]} \\ \\[-3mm] \dfrac{2}{3} + \dfrac{1}{3}\!\cdot\!\dfrac{5}{113}\!\cdot\!\dfrac {y-x}{y} &=& \dfrac{63}{100} & {\color{blue}[2]} \end{array}$

As Wilmer pointed out, [1] becomes: .$\displaystyle 100x \:=\:339y \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{339}{100}$ .(a)

[2] simplifies to: .$\displaystyle \frac{2}{3} + \frac{5}{339}\cdot\frac{y-x}{y} \:=\:\frac{63}{100}$

. . . . . . . . . . . $\displaystyle \frac{2}{3} + \frac{5}{339}\left(1 - {\color{red}\frac{x}{y}}\right) \;=\;\frac{63}{100}$

Substitute (a): . $\displaystyle \frac{2}{3} + \frac{5}{339}\left(1 - {\color{red}\frac{339}{100}}\right) \;=\;\frac{63}{100}$

. . . . . . . . . . . . . $\displaystyle \frac{2}{3} + \frac{5}{339}\left(-\frac{239}{100}\right) \;=\;\frac{63}{100}$

. . . . . . . . . . . . . . . . .$\displaystyle \frac{2}{3} - \frac{1196}{33,900} \;=\;\frac{63}{100}$

. . . . . . . . . . . . . . . . . . . $\displaystyle \frac{21,405}{33,900} \;=\;\frac{63}{100}$

. . And we have: . . . . . . . . $\displaystyle \frac{1427}{2260} \;{\color{red}\neq}\;\frac{63}{100}$

The system is inconsistent . . . It has no solution.

7. Originally Posted by Soroban
Who created this problem? . . . What a horrible way to write it!

The system is inconsistent . . . It has no solution.
That'll be my fault. It actually began life as a computer programming problem, but I thought I'd found the best way to represent it in mathematical terms. Evidently not, sorry!

Rather than go into the programming code, I'll explain the problem like this:

There are 113 marbles in a jar. Five of them are red, the rest are blue.

There's a one in three chance you'll be selected to play the game.

If you are, then blindfolded, you shake the jar and draw one marble completely at random.

If it's blue, you keep it and the game is over.

If it's red, there's an [x/y] chance I'll let you keep it, or a [1-(x/y)] chance I'll make you put it back. Whether you get to keep it or not, the game is over.

I want there to be a 5% probability of you drawing a red marble and keeping it. So what should x and y be?

8. Originally Posted by Dufus
(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63
Probem is the 2nd equation should be:
2/3 + (1/3) * (5/113) * ((y-x)/y) = 1427/2260

1427/2260 = .63141..... ; that was "rounded" to .63 : a NO-NO!

And since we have (from 1st equation) x = 3.39y,
then ANY value can be assigned to y!

9. Originally Posted by Dufus
There are 113 marbles in a jar. Five of them are red, the rest are blue.
There's a one in three chance you'll be selected to play the game.
If you are, then blindfolded, you shake the jar and draw one marble completely at random.
If it's blue, you keep it and the game is over.
If it's red, there's an [x/y] chance I'll let you keep it, or a [1-(x/y)] chance I'll make you put it back. Whether you get to keep it or not, the game is over.
I want there to be a 5% probability of you drawing a red marble and keeping it. So what should x and y be?
I must be missing something: but that makes no sense to me.
The probability of drawing a red is 5/113 ; that's a ~4.42% probability.
Whatever you do AFTER the draw will certainly not INCREASE that.

If you said "I want there to be a 3% probability.....", then we have:
(let k = x/y)
5k/113 = 3/100
k = 339/500

10. Hello, Dufus!

Wilmer is absolutely correct.
There is still something terribly wrong with the problem . . .

There are 113 marbles in a jar; 5 are red, the rest blue.

There's a one-in-three chance you'll be selected to play the game.

If you are, you draw one marble completely at random.

If it's blue, you keep it and the game is over.

If it's red, there's an $\displaystyle \tfrac{x}{y}$ chance that I'll let you keep it,
. . and a $\displaystyle 1 -\tfrac{x}{y}$ chance I'll make you put it back.
Whether you get to keep it or not, the game is over.

I want there to be a 5% probability of you drawing a red marble and keeping it.
So what should $\displaystyle x$ and $\displaystyle y$ be?
To get a red marble and keep it, three events must occur:

. . $\displaystyle \text{[1] You get to play: }\;P(\text{play}) \:=\:\frac{1}{3}$

. . $\displaystyle \text{[2] You draw a red marble: }\;P(\text{red}) \:=\:\frac{5}{113}$

. . $\displaystyle \text{[3] You get to keep it: }\;P(\text{keep}) \:=\:\frac{x}{y}$

$\displaystyle \text{Hence: }\;P(\text{play} \wedge \text{red} \wedge \text{keep}) \;=\;\frac{1}{3}\cdot\frac{5}{113}\cdot\frac{x}{y}$
. . This is to be .$\displaystyle 5\% \,=\,\frac{1}{20}$

Hence, we have: .$\displaystyle \frac{5x}{339y} \:=\:\frac{1}{20} \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{339}{100} \;=\;339\%$ .??

. . A probability cannot be greater than 100%.

$\displaystyle P(\text{play} \wedge \text{red} \wedge \text{keep}) \;=\;\frac{1}{3}\cdot\frac{5}{113}\cdot\frac{x}{y}$

No matter what x/y is (provided it's a probability, ie: y is greater than x), the result can never be greater than the multiplication of the first two fractions. The penny's dropped.

So it's never going to reach 5%, even if, in the marbles analogy, you were allowed to keep the red marble every time you pulled one out of the jar. The maximum chance is still less than 1.5%. It's stupidly obvious really...

Well thank you everyone for your help and your patience with me. I'm sorry it's turned out to be such a wild goose chase! It has refreshed my memory a little on how this all works though, so it hasn't been a complete waste of time. Thanks again.