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by **MathDrudge** on Sun Jul 24, 2011 10:33 am

Hi, Im trying to learn algebra on the side as I am in the army so I have no tutor or teacher to help, and Im have no idea how to solve this next equation. Hope you can Help. Its question "b)"

This isnt my homework or anything that Im trying to put on you or anything, Im just really confused about this. IS the

--v-1 a mixed number (the second division from the left)? how about the squared numbers? how do I deal with them? would appreciate if someone could show a step by step solution. sry bad format
v ----
---4

**Sponsor**

by **stapel_eliz** on Sun Jul 24, 2011 10:53 am

Is the equation as follows?

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

Thank you. :wink:

by **MathDrudge** on Sun Jul 24, 2011 11:18 am

stapel_eliz wrote:Is the equation as follows?

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

Thank you.

Yes it is

by **stapel_eliz** on Sun Jul 24, 2011 5:55 pm

MathDrudge wrote:Im trying to learn algebra on the side...and Im have no idea how to solve this next equation....
IS the --v-1 a mixed number (the second division from the left)?

No; mixed numbers are whole numbers together with fractions, like "three and a half". They do not contain variables.

MathDrudge wrote:how about the squared numbers?

I'll guess that you're referring to the squared variables, for which you need to find the numerical value(s)?

MathDrudge wrote:how do I deal with them?

Assuming you know what variables are, how to solve linear equations, how to multiply polynomial expressions, how to factor quadratics, and how to convert to common denominators, a good start would be to multiply things out on the second term. Then multiply the entire equation by the lowest common denominator to clear the fractions. Then combine like terms, and solve the resulting quadratic equation, using whichever method you prefer. :wink:

by **MathDrudge** on Mon Jul 25, 2011 6:51 am

stapel_eliz wrote:
MathDrudge wrote:Im trying to learn algebra on the side...and Im have no idea how to solve this next equation....
IS the --v-1 a mixed number (the second division from the left)?

No; mixed numbers are whole numbers together with fractions, like "three and a half". They do not contain variables.

Ok, so its multiplied by v?

stapel_eliz wrote:
MathDrudge wrote:how do I deal with them?

Assuming you know what variables are, how to solve linear equations, how to multiply polynomial expressions, how to factor quadratics, and how to convert to common denominators, a good start would be to multiply things out on the second term. Then multiply the entire equation by the lowest common denominator to clear the fractions. Then combine like terms, and solve the resulting quadratic equation, using whichever method you prefer.

All these topics you put look familiar and I think I know them. The equation is given as a non quadratic equation, or in other words just a simple equation. Pardon me cause I dont know the right terms, Im working in finnish and cant find a right term.

So cause its not a quadratic equation. so I dont think you need this http://www.purplemath.com/modules/quadform.htm altough Im familiar with the equation.

The answer for it by the way is 4/5

What I tried to do was multiply both sides with 20.

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

$20v\, -\, 25v\, +\, 5 +\, 4v^2\, -\, 4 +\, v^2\, =\, 0$

$20v\, -\, 25v\, +\, 4v^2\, +\, v^2\, =\, 4 -\,5$

$-\, 5v\, +\, 5v^2\, =-\,1$

So, Im a bit stuck here, or I doubt that its right up to here, anyway. how do I get rid of squared? how do I deal with the negatives Units in the fractions, when the whole fraction is a negative itself and is multiplied with a positive. Maybe from here I would square root both sides to get rid of the squared? or break it down into

$v\, {(5v\, -\, 5\,)}\, = -\,1$

but I cant see how that helps

by **stapel_eliz** on Mon Jul 25, 2011 11:18 am

MathDrudge wrote:Ok, so its multiplied by v?

Um.... So what is multiplied by v?

MathDrudge wrote:The equation is given as a non quadratic equation....

Um.... No; the fact that the highest-degree terms are squares indicates that this is indeed a quadratic.

MathDrudge wrote:The answer for it by the way is 4/5

Um.... Do you mean that "the solution for the equation" is "4/5"?

MathDrudge wrote:What I tried to do was multiply both sides with 20.

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

$20v\, -\, 25v\, +\, 5 +\, 4v^2\, -\, 4 +\, v^2\, =\, 0$

Um.... How did you arrive at this equation? Since v(v - 1) = v² - v and since (20/4)(v² - v) does not equal 25v + 5 and (20/5)(v² - 5) does not equal 4v² - 4, I'm afraid I don't follow.

by **MathDrudge** on Mon Jul 25, 2011 2:29 pm

@stapel_eliz

this is all the stages Im trying to solve the equation. Can you help?

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

$20v\, -\, 20v\, 20(\frac{v\, -\, 1}{4}\), +\, 20(\frac{v^2\, -\, 5}{5}\), +\, 20(\frac{1}{20}v^2\), =\, 20*0$

$20v\, -\, 20v\frac{20v\, -\, 20}{4}\, +\, \frac{20v^2\, -\, 100}{5}\, +\, \frac{20}{20}v^2\, =\, 0$

$20v\, -\, 20v\, -\, 5v\, -\, 5\, +\, 4v^2\, +\, 20\, +\, 1v^2\, =\, 0$

$20v\, -\, 25v\, +\, 5v^2\, +\, =\, 5\, -\, 20\,$

$-\, 5v\, +\, 5v^2\, +\, =\,-\, 15\,$

by **stapel_eliz** on Mon Jul 25, 2011 5:21 pm

MathDrudge wrote:this is all the stages Im trying to solve the equation. Can you help?

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

$20v\, -\, 20v\, 20(\frac{v\, -\, 1}{4}\), +\, 20(\frac{v^2\, -\, 5}{5}\), +\, 20(\frac{1}{20}v^2\), =\, 20*0$

It looks like you multiplied the second term twice by 20...?

First, try multiplying things out. (Follow the link, provided earlier, for a lesson on how to do that.) Then try doing the multiplication. :wink:

by **MathDrudge** on Mon Jul 25, 2011 6:35 pm

What is the second term? a mixed number made up by variables? or is it v * (v-1)/4? or is it either?

by **stapel_eliz** on Mon Jul 25, 2011 11:11 pm

MathDrudge wrote:What is the second term? a mixed number made up by variables? or is it v * (v-1)/4? or is it either?

There are still no mixed numbers in this equation. To learn what "terms" are, please try this lesson.

To learn how to simplify v(v - 1), please try this lesson or this lesson. :wink:

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How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0

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