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.
[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.