(2x-5)/x + (5x-10)/3x - (8x-30)/2x = 2

nadeem

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Nov 13, 2005
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7
Hello everyone :)

For the past hours, i've been trying to figure out how to solve this equation:

2x-5 + 5x-10 - 8x-30 = 2
..x.........3x.........2x

Since i made the problem, i know that x=5. I made the problem since i wanted to find out a way to solve such equations. But i must admit to defeat.

Please help, and feel welcome to write out the solution in a step-by-step manner.

Thanks :)

Nadeem.
 
One method is to find the common denomiantor?.

x(2x)(3x)

Multiply through by that:

\(\displaystyle x(2x)(3x)\frac{2x-5}{x}+x(2x)(3x)\frac{5x-10}{3x}-x(2x)(3x)\frac{8x-30}{2x}=2(x(2x)(3x))\)

Do the appropriate cancellations and it'll fall into place. You'll end up with a cubic equation to solve. I believe it will not be too bad to factor and find the solutions.
 
Hi dear Galactus,

thank you very much for your suggestion. However i don't have a clue on how i'll multiply x(2x)(3x) trough the equation.

This is how i tried to figure it out:

First, multiply the whole equation with 6. Then i got:

\(\displaystyle (12x-30)/x + (10x-20)/x - (24x -90)/x = 12\)

Then i multiplied it all with X, getting:

\(\displaystyle 12x^2 - 30x + 10x^2 - 20x - 24x^2 - 90x = 12x\)

And then.... Okey, i probably dont need to say more, since i've probably made the fatal mistake already.

Can you please tell, step by step what to do. Don't want to sound lazy, but i've been trying to get this right for hours and i'm frustrated.

Thanks for your time,

Nadeem.
 
You're close. Missing an \(\displaystyle x^{3}\). You should've multiplied by \(\displaystyle 6x\) and cancelled

the appropriate denomiantors.

Do you understand cancellation?.

The corresponding factor in the common denominator and the one in the denominator will cancel one another.

So you have:

\(\displaystyle (2x)(3x)(2x-5)+(2x)(x)(5x-10)-(3x)(x)(8x-30)=2(x(2x)(3x))\)

Now multipy out:

\(\displaystyle 6x^{2}(2x-5)+2x^{2}(5x-10)-3x^{2}(8x-30)=12x^{3}\)

You can finish now, can't you?.
 
Hi Galactus,

I'm feeling kind of stupid. :oops:

Yes, i believe i do understand cancellation. As in (2*3)/2 = 3. Since we cancelled the factor 2 with the denomintor 2, leaving 3 alone. Am i right?

I tried to figure out your last equation and got:

\(\displaystyle 12x^3 - 30x^2 + 10x^3 - 20x^2 - 24x^3 - 90x^2 = 12x^3\)

I've no clue what to do next.

This is not my homework. Im using my free-time on this since i really want to learn it. And i appreciate that you're helping me out. Thanks :)

Nadeem,.
 
You've got it except that's \(\displaystyle 90x^{2}\). Remember that negative times negative thing?.

Anyway, all you have to do is add like terms now.

\(\displaystyle 12x^{3}-30x^{2}+10x^{3}-20x^{2}-24x^{3}+90x^{2}=12x^{3}\)


You'll end up with a cubic which is easy to factor.
 
nadeem said:
Hello everyone :)
For the past hours, i've been trying to figure out how to solve this equation:
2x-5 + 5x-10 - 8x-30 = 2
..x.........3x.........2x
Since i made the problem, i know that x=5. I made the problem since i wanted to find out a way to solve such equations. But i must admit to defeat.
Please help, and feel welcome to write out the solution in a step-by-step manner.
Thanks :)
Nadeem.
To start with, if x=5, then (2x-5)/x + (5x-10)/3x - (8x-30)/2x = 1 , NOT 2

To solve, multiply by 6x to get:
12x-30 + 10x-20 - 24x+90 = 6x
x = 5

I have no idea why both of you are dealing with a cubic :?:

If you leave the equation the way you have it:
(2x-5)/x + (5x-10)/3x - (8x-30)/2x = 2
you can solve it the same way (multiply by 6x), and you'll get x = 20/7
 
galactus said:
You've got it except that's \(\displaystyle 90x^{2}\). Remember that negative times negative thing?.
Anyway, all you have to do is add like terms now.
\(\displaystyle 12x^{3}-30x^{2}+10x^{3}-20x^{2}-24x^{3}+90x^{2}=12x^{3}\)
You'll end up with a cubic which is easy to factor.
Ahhh, I see what you guys are doing; BUT that's NOT a cubic; only looks like one;
divide each term by x^2 to get:
12x - 30 + 10x - 20 - 24x + 90 = 12x
so: 14x = 40 ; x = 20/7
 
I know Denis. It only has two solutions(20/7 and 0). 'A pseudo-cubic'. I was just trying to impart some cancellation and factoring by 'going around the horn' so to speak. Should've just stuck to the Occam's razor approach.
 
Hello everybody,

I apologize for the slow response from my side, as i do appreciate the fast and precise help from your side. Makes one feel welcome, thanks :)

About the problem, what was i thinking?

You guys are absolutely correct, i see that now, thanks :-=

Regards,

Nadeem.
 
Hello, nadeem!

You were on the right track . . . two simple errors, though.

This is how i tried to figure it out:

First, multiply the whole equation with 6.

Then i got: .\(\displaystyle \frac{12x-30}{x}\,+\,\frac{10x-20}{x}\,-\,\frac{24x -90}{x}\:=\:12\) . . . . correct!

Then i multiplied it all with \(\displaystyle x\), getting:

\(\displaystyle 12x^2\,-\,30x\,+\,10x^2\,-\,20x\,-\,24x^2\,-\,90x\:=\:12x\) . . . . sorry, no
Too many x's . . . and you dropped part of the subtraction.
Did you write out the step . . . or do it mentally?

. . \(\displaystyle \L x\cdot\frac{12x-20}{x}\,+\,x\cdot\frac{10x-20}{x}\,-\,x\cdot\frac{24x-90}{x}\;=\;x\cdot12\) . . . . and cancel . . .

and we have: .\(\displaystyle \L(12x-20)\.+\.(10x-20)\,-\,(24x-90)\:=\:12x\)
 
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