Two algebraic problems I can't solve

  • Thread starter Thread starter stonecoldgen
  • Start date Start date
AI Thread Summary
The discussion revolves around two algebraic problems that a college freshman is struggling to solve. The first problem involves simplifying a complex fraction, where initial attempts to separate the numerator and denominator were unsuccessful. The second problem requires finding the sum of solutions from a system of equations, with attempts made to express one variable in terms of another but leading to confusion. Suggestions include using linear combinations and matrices to solve the equations more effectively. The participant expresses a desire to regain confidence in algebra as they begin their college studies.
stonecoldgen
Messages
108
Reaction score
0

Homework Statement


1) [(1/y2)-x2]/[(1/x2)-y2]=?

2) what is the sum of the solutions, x+y, in the following system of equations?
x/y=1/2
(x+2)/(y-9)=-4/5



The Attempt at a Solution



1) I started by separating the numerator of the initial big fractionk trying to simplify but failing. However, i did get to the point where I had to separated fractions with the same denominator, but it was a fail because I was just doing a round trip, lol.

2) I defined x in terms of y and plugged it in into the (x+2)/(y-9)=-4/5 equation, being as careful as I could with having all addingg/subtracting terms with the same denominator, careful about signs etc. But I'm just doing something wrong.



BTW, I'm already a college freshman, I already passed algebra back in 9th and 10th, but I am just loosing the rhythm for this stuff and I don't want that, so any help is appreciated, thanks (just tell me which is the best path to start with if you can, thanks).
 
Physics news on Phys.org
It would be nice of you to actually type out what you did!
 
stonecoldgen said:

Homework Statement


1) [(1/y2)-x2]/[(1/x2)-y2]=?

2) what is the sum of the solutions, x+y, in the following system of equations?
x/y=1/2
(x+2)/(y-9)=-4/5

The Attempt at a Solution



1) I started by separating the numerator of the initial big fractionk trying to simplify but failing. However, i did get to the point where I had to separated fractions with the same denominator, but it was a fail because I was just doing a round trip, lol.

2) I defined x in terms of y and plugged it in into the (x+2)/(y-9)=-4/5 equation, being as careful as I could with having all adding/subtracting terms with the same denominator, careful about signs etc. But I'm just doing something wrong.

BTW, I'm already a college freshman, I already passed algebra back in 9th and 10th, but I am just loosing the rhythm for this stuff and I don't want that, so any help is appreciated, thanks (just tell me which is the best path to start with if you can, thanks).

For #1, multiply the numerator & denominator by x2y2.
 
stonecoldgen said:
2) what is the sum of the solutions, x+y, in the following system of equations?
x/y=1/2
(x+2)/(y-9)=-4/5

2) I defined x in terms of y and plugged it in into the (x+2)/(y-9)=-4/5 equation, being as careful as I could with having all addingg/subtracting terms with the same denominator, careful about signs etc. But I'm just doing something wrong.



BTW, I'm already a college freshman, ...

If you're in college, have you learned Linear Algebra? You know, using matrices to solve linear equations? They're not necessary here, but it would be good practice to use them to solve #2 if you should already know them.

To start off, simplify both given equations into a linear combination of x and y, for example, \frac{x}{y}=\frac{1}{2}
2x=y
2x-y=0

Do the same for the second equation.

Now, we have a 3rd equation to solve, x+y=? but rather than a question mark, we'll introduce another variable, namely z for consistency.
You'll now have 3 linear equations with 3 unknowns. So either by using matrices or the conventional plugging and solving, try find the value of z.
 
Mentallic said:
If you're in college, have you learned Linear Algebra? You know, using matrices to solve linear equations? They're not necessary here, but it would be good practice to use them to solve #2 if you should already know them.

To start off, simplify both given equations into a linear combination of x and y, for example, \frac{x}{y}=\frac{1}{2}
2x=y
2x-y=0

Do the same for the second equation.

Now, we have a 3rd equation to solve, x+y=? but rather than a question mark, we'll introduce another variable, namely z for consistency.
You'll now have 3 linear equations with 3 unknowns. So either by using matrices or the conventional plugging and solving, try find the value of z.

hey this is pretty interesting! I am starting college today actually! I will see linear algebra next semester, but thanks a lot
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top