Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##

  • Thread starter Thread starter RChristenk
  • Start date Start date
  • Tags Tags
    Precalculus
AI Thread Summary
The discussion revolves around solving the equation 3(x-2)/(5-x) + (1/2)(11-2x)/(5-2x) = 3.5. The original poster found solutions x = 2 and x = 15/4 but noted that direct multiplication is tedious and suspected a simplification might exist. They highlighted the similarity of the denominators 5-x and 5-2x, indicating a potential for simplification that remains elusive. Participants emphasize a preference for methodical approaches over clever tricks, advocating for a solid understanding of basic principles in algebra. The conversation reflects a common challenge in algebraic manipulation and the search for efficient solving methods.
RChristenk
Messages
73
Reaction score
9
Homework Statement
Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
Relevant Equations
Algebraic manipulation
I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed.

##3\cdot\dfrac{x-2}{5-x}+\dfrac{1}{2}\cdot\dfrac{11-2x}{5-2x}=\dfrac{7}{2}##

Multiply both sides by ##2##:

##\dfrac{6(x-2)}{5-x}+\dfrac{11-2x}{5-2x}=7##

And then I'm stuck. But the denominators ##5-x, 5-2x## are tantalizingly close to each other, but I just can't figure out how to simplify/substitute/manipulate it to process this problem. Besides just multiplying it out of course.
 
Physics news on Phys.org
I don't see a way to simplify things. The best you could do is:
$$6(x-2)(5-2x) + (11-2x)(5-x) = 7(5-x)(5-2x)$$$$6(x-2)(5-2x) + (5-x)(11-2x - 35 + 14x) = 0$$$$6(x-2)(5-2x) + (5-x)(12x-24) = 0$$$$6(x-2)(5-2x) +6(x-2)(10 - 2x) = 0$$$$6(x-2)(15 - 4x) = 0$$
 
  • Like
Likes MatinSAR, chwala, SammyS and 2 others
<br /> -3-\frac{9}{x-5}+\frac{1}{2}-\frac{3}{2x-5}=3+\frac{1}{2}
\frac{3}{x-5}+\frac{1}{2x-5}=-2
4x^2-23x+30=0
(x-2)(4x-15)=0
 
Last edited:
  • Like
Likes MatinSAR, chwala, RChristenk and 1 other person
RChristenk said:
Homework Statement: Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
Relevant Equations: Algebraic manipulation

I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed.

##3\cdot\dfrac{x-2}{5-x}+\dfrac{1}{2}\cdot\dfrac{11-2x}{5-2x}=\dfrac{7}{2}##

Multiply both sides by ##2##:

##\dfrac{6(x-2)}{5-x}+\dfrac{11-2x}{5-2x}=7##

And then I'm stuck. But the denominators ##5-x, 5-2x## are tantalizingly close to each other, but I just can't figure out how to simplify/substitute/manipulate it to process this problem. Besides just multiplying it out of course.
Here is my two cents.
I much prefer methodical approaches that use basic principles to reliably solve the vast majority of problems. I don't really care if they require a little more work. If you look for cute and clever tricks, you might learn a million of them and still not have a basic understanding.
 
  • Like
Likes MatinSAR, chwala, docnet and 2 others
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
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top