# Fairly simple math problem **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [June 12, 2012, 8:51pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955 "2012-06-12T20:51:37Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 12, 2012, 8:51pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/1 "2012-06-12T20:51:37Z")
Please don’t worry and simply solve the following system of equations. (Would be best to solve it before looking at others’ answers…) 2x÷3y = 1/3 4x÷3y = 2/3 Thanks in advance for your patience. ---
**Author:** ![Sicks\_Ate](https://avatars.discourse-cdn.com/v4/letter/s/258eb7/32.png) [@Sicks\_Ate](https://boards.straightdope.com/u/Sicks_Ate)\ **Post date:** [June 12, 2012, 8:57pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/2 "2012-06-12T20:57:24Z")
I sucks at math. But I think that… x = .5 and y = 1 ??? ---
**Author:** ![ashtayk](https://avatars.discourse-cdn.com/v4/letter/a/e68b1a/32.png) [@ashtayk](https://boards.straightdope.com/u/ashtayk)\ **Post date:** [June 12, 2012, 8:59pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/3 "2012-06-12T20:59:32Z")
I am not sure you can solve it - the second equation is the first one multiplied by 2. No new info there to solve for x and y. I meant - you can’t solve for unique x and y values. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 12, 2012, 9:01pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/4 "2012-06-12T21:01:27Z")
> [@ashtayk](#): > > I am not sure you can solve it - the second equation is the first one multiplied by 2. No new info there to solve for x and y. > > I meant - you can’t solve for unique x and y values. The second one isn’t the first multiplied by two. ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [June 12, 2012, 9:16pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/5 "2012-06-12T21:16:19Z")
> [@Frylock](#): > > The second one isn’t the first multiplied by two. Look again. Any non-zero y that is twice x works. I suspect a typo in the OP. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [June 12, 2012, 9:23pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/6 "2012-06-12T21:23:37Z")
> [@ftg](#): > > Look again. > > Any non-zero y that is twice x works. > > I suspect a typo in the OP. Agreed. The equations are the same. ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [June 12, 2012, 9:25pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/7 "2012-06-12T21:25:03Z")
> [@Frylock](#): > > Please don’t worry and simply solve the following system of equations. (Would be best to solve it before looking at others’ answers…) > > 2x÷3y = 1/3 > > 4x÷3y = 2/3 > > Thanks in advance for your patience. Yup, those are divide signs there all right - would read more clearly as: 2x / 3y = 1/3 4x / 3y = 2/3 for which 2x = y with infinite solutions. Mind you, I’m kinda mentally putting brackets around the 2x and 3y etc, and by strict left-right evaluation with equal operator precedence, this would really be 2 times x, then divide by 3, then all times y, = 1/3 i.e. (2xy)/3 = 1/3. So xy = 1/2 Similarly 4xy/3 = 2/3 and again, xy = 1/2 Infinite solutions again. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 12, 2012, 9:26pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/8 "2012-06-12T21:26:51Z")
> [@ftg](#): > > Look again. > > Any non-zero y that is twice x works. > > I suspect a typo in the OP. No, not a typo, just me trying (for good but tedious reasons) to do this without actual fraction bars and getting screwed up in the head as a result. I’ll post a different system in another thread. Sorry! ---
**Author:** ![Asimovian](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/asimovian/32/172_2.png) [@Asimovian](https://boards.straightdope.com/u/Asimovian)\ **Post date:** [June 12, 2012, 9:50pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/9 "2012-06-12T21:50:20Z")
Post a link in this thread when you do. y=2x in both equations. ---
**Author:** ![Indian](https://avatars.discourse-cdn.com/v4/letter/i/c0e974/32.png) [@Indian](https://boards.straightdope.com/u/Indian)\ **Post date:** [June 13, 2012, 3:39am UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/10 "2012-06-13T03:39:55Z")
I am getting 2x=y. What is the catch? ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [June 13, 2012, 12:52pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/11 "2012-06-13T12:52:08Z")
> [@Frylock](#): > > The second one isn’t the first multiplied by two. No? Sure looks as if it is. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [June 13, 2012, 12:53pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/12 "2012-06-13T12:53:55Z")
No unique solution. ---
**Author:** ![Sicks\_Ate](https://avatars.discourse-cdn.com/v4/letter/s/258eb7/32.png) [@Sicks\_Ate](https://boards.straightdope.com/u/Sicks_Ate)\ **Post date:** [June 13, 2012, 12:55pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/13 "2012-06-13T12:55:08Z")
> [@ultrafilter](#): > > No unique solution. Told you I sucks at math. ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [June 13, 2012, 1:03pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/14 "2012-06-13T13:03:15Z")
I get y=2x for both. ---
**Author:** ![Heart\_of\_Dorkness](https://avatars.discourse-cdn.com/v4/letter/h/4bbf92/32.png) [@Heart\_of\_Dorkness](https://boards.straightdope.com/u/Heart_of_Dorkness)\ **Post date:** [June 13, 2012, 1:39pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/15 "2012-06-13T13:39:46Z")
> [@Sicks\_Ate](#): > > Told you I sucks at math. No, you’re right - x = .5 and y = 1 is one \*possible \*solution. In other words: 2(.5)/3(1) = 1/3 4(.5)/3(1) = 2/3 But remember, if you multiply both the numerator and the denominator of a fraction by the same number, you get an equivalent fraction: 1/3 = 2/6 = 3/9 = 4/12… etc. 2/3 = 4/6 = 6/9 = 8/12… etc. So as long as your x and y values always have the same relationship (that is, 2x = y), there can be infinite solutions: x = 1 and y = 2: 2(1)/3(2) = 2/6 (which = 1/3) 4(1)/3(2) = 4/6 (which = 2/3) x = 2 and y = 4: 2(2)/3(4) = 4/12 (which = 1/3) 4(2)/3(4) = 8/12 (which = 2/3) …and so on. And yeah, if that’s not the answer **Frylock** is looking for, I’m also curious to know why. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [June 13, 2012, 3:04pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/16 "2012-06-13T15:04:11Z")
2x / 3y = 1/3 4x / 3y = 2/3 2x = 3y \* 1/3 2x = y x = y/2 2(y/2) / 3y = 1/3 y/3y = 1/3 y = 1 2x / 3 = 1/3 x = 1/3 \* 3/2 x = ½ Check: 2(1/2) / 3(1) = 1/3 1 / 3 = 1/3 4(1/2) / 3(1) = 2/3 2 / 3 = 2/3 As others noted, the answer is not unique. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [June 13, 2012, 3:14pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/17 "2012-06-13T15:14:42Z")
> [@Johnny\_L.A](#): > > 2(y/2) / 3y = 1/3 > y/3y = 1/3 > y = 1 This doesn’t work. ---
**Author:** ![Hogfather65](https://avatars.discourse-cdn.com/v4/letter/h/7bcc69/32.png) [@Hogfather65](https://boards.straightdope.com/u/Hogfather65)\ **Post date:** [June 13, 2012, 3:19pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/18 "2012-06-13T15:19:49Z")
> [@Asimovian](#): > > Post a link in this thread when you do. > > y=2x in both equations. ditto ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [June 13, 2012, 3:20pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/19 "2012-06-13T15:20:33Z")
Well, I suck at math. ---
**Author:** ![Chefguy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chefguy/32/138_2.png) [@Chefguy](https://boards.straightdope.com/u/Chefguy)\ **Post date:** [June 13, 2012, 5:32pm UTC](https://boards.straightdope.com/t/fairly-simple-math-problem/624955/20 "2012-06-13T17:32:48Z")
> [@Frylock](#): > > Please don’t worry and simply solve the following system of equations. (Would be best to solve it before looking at others’ answers…) > > 2x÷3y = 1/3 > > 4x÷3y = 2/3 > > Thanks in advance for your patience. Just doing it in my head, it would seem that x=.5 and y=1 Now looking at other answers, I see that any number works. [Next page](https://boards.straightdope.com/t/fairly-simple-math-problem/624955.md?page=2)