Quadratic equations...Help with 2/1-5/3z-1=2/(3z-1)^2

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Help! I just don't even know where to start with this one.

. . .Solve for 2/1-5/3z-1=2/(3z-1)^2

:? Can anybody help?
 
Hmm do you mean this: \(\displaystyle \L \;\frac{2}{1}\,-\,\frac{5}{3z-1}\,=\,\frac{2}{3z-1^2}\)

or maybe..........

\(\displaystyle \L \;\frac{2}{1}\,-\,\frac{5}{3z-1}\,=\,\frac{2}{(3z-1)^2}\)

Notice the former has (3z-1^2) and the latter has (3z-1)^2. Now which one is it or is it something totally different?
 
Hi jonboy, I's the second (3z-1) squared. I' m not sure how to show a raised power on my keyboard. Can you help?
 
Sure type it like: (2/1) - (5/3z-1) = 2/(3z-1)^2

Now I will see what I can do w/this problem.
 
Caryn said:
[It]'s the second [option,] (3z-1) squared.
Click to expand...
Your equation appears to be as follows:

. . . . .\(\displaystyle \L 2\,- \,\frac{5}{3z\,-\,1}\,= \,\frac{2}{(3z\,-\,1)^2}\)

I will assume that the instructions were to "Solve for the variable". A good first step would be to note that z cannot equal 1/3, and then multiply through by (3z - 1)<sup>2</sup> to get:

. . . . .\(\displaystyle \L 2(3z\,-\,1)^2\,- \,5(3z\,-\,1)\,= \,2\)

Multiply out to get:

. . . . .\(\displaystyle \L 2(9z^2\,-\,6z\,+\,1)\,- \,(15z\,-\,5)\,= \,2\)

Continue the simplification, and then solve the resulting quadratic equation by the usual methods.

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.
 
\(\displaystyle \L \;\frac{2}{1}\,-\,\frac{5}{3z-1}\,=\,\frac{2}{(3z-1)^2}\)


Get like terms: \(\displaystyle \L \;\frac{2(3z-1)^2}{(3z-1)^2}\,-\,\frac{5(3z-1)}{(3z-1)2}\,=\,\frac{2}{(3z-1)^2}\)


Now we can just drop the numerators and simplify: \(\displaystyle \L \;{2(9z^2-6z+1)\,-\,15z+5\,=\,2\)


.............Simplify: \(\displaystyle \L \;18z^2\,-\,12z\,+2\,-\,15z\,+\,5=2\,\to\,18z^2\,-\,27z\,+\,5\,=\,0\)


Solve by Quadratic Formula: \(\displaystyle \L \;z = \frac{{27 \pm \sqrt {( - 27)^2 - 4(18)(5)}}}{{2(18)}}\,\to\,z\,=\,\frac{{27 \pm \sqrt {369} }}{{36}}\,\to\,z = \frac{3}{4} \pm \frac{1}{{12}}\sqrt {41}\)

I hope this helps. Be free to ask how I got the steps.
 
jonboy said:
Get like terms: \(\displaystyle \L \;\frac{2(3z-1)^2}{(3z-1)^2}\,-\,\frac{5(3z-1)}{(3z-1)2}\,=\,\frac{2}{(3z-1)^2}\)


Now we can just drop the numerators and simplify: \(\displaystyle \L \;{2(9z^2-6z+1)\,-\,15z+5\,=\,2\)
Click to expand...
After dropping denominators, easier to let x = 3z-1; then you have:
2x^2 - 5x - 2 = 0
solve for x, then substitute back in...
 
Denis said:
jonboy said:
Get like terms: \(\displaystyle \L \;\frac{2(3z-1)^2}{(3z-1)^2}\,-\,\frac{5(3z-1)}{(3z-1)2}\,=\,\frac{2}{(3z-1)^2}\)


Now we can just drop the numerators and simplify: \(\displaystyle \L \;{2(9z^2-6z+1)\,-\,15z+5\,=\,2\)
Click to expand...
After dropping denominators, easier to let x = 3z-1; then you have:
2x^2 - 5x - 2 = 0
solve for x, then substitute back in...
Click to expand...

Where are you gettting an x?
 
jonboy, I'm temporarily replacing 3z-1 by x.

Simple example:
if 6(3z+1) = 60 ; let x = 3z-1, then :
6x = 60
x = 10

substitute back in:
3z-1 = 10
3z = 9
z = 3 : capish?
 
Denis said:
jonboy, I'm temporarily replacing 3z-1 by x.

Simple example:
if 6(3z+1) = 60 ; let x = 3z-1, then :
6x = 60
x = 10

substitute back in:
3z-1 = 10
3z = 9
z = 3 : capish?
Click to expand...

Ok Denis thank you for your explanation. Wow that does look much better I am going to experiment w/that. Yes it is capish! :)
 
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