Systems of equations with double denominators

[rimeyy]

Hello, while practicing today for the upcoming school year, I found this interesting system of equations:
[math]\frac{5}{3x}+\frac{2}{5y}=7[/math][math]\frac{7}{6x}-\frac{1}{10y}=3[/math]I notice that the denominators on the bottom equation are double that of the top equation, so I assumed that there needs to be a way to equalize them:
First I transformed the fractions into V (x) and U (y) variables:
[math]\frac{1}{3x}=V[/math][math]\frac{1}{5y}=U[/math]Then I applied it to the system:
5V + 2U =7
7V- 1U = 3
Then I multiplied the top one with 2:
10V + 4U =14
7V - 1U = 3
Then I multiplied the bottom one with 4 to do the elimination:
10V+ 4U = 14
28V - 4U = 12
After that I added them together to get:
38V = 26
[math]V=\frac{26}{38}[/math](I divide them both my 2)
[math]V=\frac{13}{19}[/math]which is a unusually large number for V, and it also isn't the correct solution for it.
I feel like I am missing something very obvious that we haven't mentioned in class.
Either way, thank you for your time.

 

[Deleted member 4993]

Hello, while practicing today for the upcoming school year, I found this interesting system of equations:
[math]\frac{5}{3x}+\frac{2}{5y}=7[/math][math]\frac{7}{6x}-\frac{1}{10y}=3[/math]I notice that the denominators on the bottom equation are double that of the top equation, so I assumed that there needs to be a way to equalize them:
First I transformed the fractions into V (x) and U (y) variables:
[math]\frac{1}{3x}=V[/math][math]\frac{1}{5y}=U[/math]Then I applied it to the system:
5V + 2U =7
7V- 1U = 3
Then I multiplied the top one with 2:
10V + 4U =14
7V - 1U = 3
Then I multiplied the bottom one with 4 to do the elimination:
10V+ 4U = 14
28V - 4U = 12
After that I added them together to get:
38V = 26
[math]V=\frac{26}{38}[/math](I divide them both my 2)
[math]V=\frac{13}{19}[/math]which is a unusually large number for V, and it also isn't the correct solution for it.
I feel like I am missing something very obvious that we haven't mentioned in class.
Either way, thank you for your time.

Your second equation should be:

7V - 1U = 6

Fix it and continue ......

 

[canvas]

Your second equation should be:

7V - 1U = 6

Fix it and continue ......

It's not true... [math]\text{Since that}\,\,\,u=\frac{1}{3x}\,\,\,\text{and}\,\,\,v=\frac{1}{5y}\,\,\,\text{we should get system:} \begin{cases}5v+2u=7\\\frac{7}{2}v-\frac{1}{2}u=3 \end{cases} \Rightarrow \begin{cases}u=1\\v=1\end{cases}[/math]

 

[canvas]

It's not true... [math]\text{Since that}\,\,\,u=\frac{1}{3x}\,\,\,\text{and}\,\,\,v=\frac{1}{5y}\,\,\,\text{we should get system:} \begin{cases}5v+2u=7\\\frac{7}{2}v-\frac{1}{2}u=3 \end{cases} \Rightarrow \begin{cases}u=1\\v=1\end{cases}[/math]

I replied to the wrong post, sorry sir. I wanted to reply to @rimeyy

 

[lev888]

It's not true... [math]\text{Since that}\,\,\,u=\frac{1}{3x}\,\,\,\text{and}\,\,\,v=\frac{1}{5y}\,\,\,\text{we should get system:} \begin{cases}5v+2u=7\\\frac{7}{2}v-\frac{1}{2}u=3 \end{cases} \Rightarrow \begin{cases}u=1\\v=1\end{cases}[/math]

It is true. If you multiply your second equation by 2 you'll get 7V - 1U = 6.

 

[JeffM]

Hello, while practicing today for the upcoming school year, I found this interesting system of equations:
[math]\frac{5}{3x}+\frac{2}{5y}=7[/math][math]\frac{7}{6x}-\frac{1}{10y}=3[/math]I notice that the denominators on the bottom equation are double that of the top equation, so I assumed that there needs to be a way to equalize them:
First I transformed the fractions into V (x) and U (y) variables:
[math]\frac{1}{3x}=V[/math][math]\frac{1}{5y}=U[/math]Then I applied it to the system:
5V + 2U =7
7V- 1U = 3
Then I multiplied the top one with 2:
10V + 4U =14
7V - 1U = 3
Then I multiplied the bottom one with 4 to do the elimination:
10V+ 4U = 14
28V - 4U = 12
After that I added them together to get:
38V = 26
[math]V=\frac{26}{38}[/math](I divide them both my 2)
[math]V=\frac{13}{19}[/math]which is a unusually large number for V, and it also isn't the correct solution for it.
I feel like I am missing something very obvious that we haven't mentioned in class.
Either way, thank you for your time.

Your approach is clever, but it is not general for equations with variables in the denominator and is, I suspect, slightly prone to error.

Here is a more general technique.

[math]\dfrac{5}{3x} + \dfrac{2}{5y} = 7 \implies \dfrac{25y + 6x}{15xy} = 7 \implies 25y + 6x = 105xy.[/math]
[math]\dfrac{7}{6x} - \dfrac{1}{10y} = 3 \implies \dfrac{70y - 6x}{60xy} = 3 \implies 70y - 6x = 180xy.[/math]
Now we see an easy elimination.

[math]25y + 6x + 70y - 6x = 105xy + 180xy \implies 95y = 285xy \implies x = \dfrac{95y}{285y} = \dfrac{1}{3}.[/math]
[math]\therefore 25y + 6x = 105xy \implies 25y + 2 = 35y \implies y = \dfrac{1}{5}.[/math]
Let’s check

[math]\dfrac{5}{3 * \dfrac{1}{3}} + \dfrac{2}{5 * \dfrac{1}{5}} = 5 + 2 = 7. \checkmark[/math]
[math]\dfrac{7}{6 * \dfrac{1}{3}} - \dfrac{1}{10 * \dfrac{1}{5}} = \dfrac{7}{2} - \dfrac{1}{2} = \dfrac{6}{2} = 3. \checkmark.[/math]
Your approach was elegant. But it depends on seeing that the same substitution of variables will work in both equations.

 

[rimeyy]

Yes, I am glad I was atleast going in a similar, albeit incorrect direction. I have fully understood the answer, but are there any other methods of solving this, particularily using a substitution similar to the one I've tried at the beginning? Thank you for your great answer!

 

[JeffM]

There probably are all kinds of special cases. But in general a system of n non-linear equations in n unknowns does not have a closed form solution. If, however, a system of 2 equations in two unknowns can be reduced to polynomial equations of degree 2, it can solved by substitution and the quadratic formula. Let's use this problem as an example.

[math]\dfrac{5}{3x} + \dfrac{5}{2y} = 7 \implies 25y + 6x = 105xy \implies y = \dfrac{6x}{105x - 25} \text { if } x \ne \dfrac{5}{21}.[/math]
We now should show that x = 5/21 is not a valid solution, but we know that from previous work so I ignore that.

[math]\therefore \dfrac{7}{6x} - \dfrac{1}{10y} = 3 \implies 70y - 6x = 180xy \implies[/math]
[math]\dfrac{420x}{105x - 25} - 6x = \dfrac{1080x^2}{105x - 25} \implies \dfrac{-630x^2 +570x}{105x - 25} = \dfrac{1080x^2}{105x - 25} \implies[/math]
[math]1710x^2 - 570x + 0 = 0 \implies x = \dfrac{570 \pm \sqrt{(-570)^2 - 4 * 1710 * 0}}{2 * 1710} = 0 \text { or } \dfrac{1}{3}.[/math]
We would realize that x = 0 is a spurious solution because 5/(3 * 0) is not defined. And so we reach the answer x = 1/3 and can then find y from

[math]y = \dfrac{6x}{105x - 25} = \dfrac{2}{35 - 25} = \dfrac{1}{5}.[/math]
It may be ugly, but you can solve any problem of the type you showed by this method.

 

[Dr.Peterson]

Yes, I am glad I was atleast going in a similar, albeit incorrect direction. I have fully understood the answer, but are there any other methods of solving this, particularily using a substitution similar to the one I've tried at the beginning? Thank you for your great answer!

I would probably have used either u = 1/x, v = 1/y for simplicity, or, after more thought, u = 1/(6x), v = 1/(10y), so that all coefficients would be integers. These would lead to essentially the same work, so they're hardly different.

There are many, many other ways; the best way is the one you think of, especially if you have enough experience to look ahead and avoid pitfalls. And the way to get that experience is to try solving the same problem in many ways! (That's not the same thing as asking someone else to show you many ways.)