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

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### Re:

stapel_eliz wrote:
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.

is the solution for v(v-1) = v^2-v ?
MathDrudge

Posts: 11
Joined: Sun Jul 24, 2011 10:17 am

MathDrudge wrote:is the solution for v(v-1) = v^2-v ?

If, by "solution", you mean "the multiplied-out form", then yes.

stapel_eliz

Posts: 1785
Joined: Mon Dec 08, 2008 4:22 pm

### Re: How to solve this problem

i still dont know what to do with this

$20(v\, \frac{v\, -\, 1}{4}\) \,$
MathDrudge

Posts: 11
Joined: Sun Jul 24, 2011 10:17 am

MathDrudge wrote:i still dont know what to do with this

$20(v\, \frac{v\, -\, 1}{4}\) \,$

Multiply it out. Cancel the four with the twenty.

stapel_eliz

Posts: 1785
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

stapel_eliz wrote:
MathDrudge wrote:i still dont know what to do with this

$20(v\, \frac{v\, -\, 1}{4}\) \,$

Multiply it out. Cancel the four with the twenty.

$5(v^2\, -\, 1\,) \,$
MathDrudge

Posts: 11
Joined: Sun Jul 24, 2011 10:17 am

### Re: Re:

MathDrudge wrote:$5(v^2\, -\, 1\,) \,$

Almost. Use the correct multiplication from two posts before the above-quoted one.

stapel_eliz

Posts: 1785
Joined: Mon Dec 08, 2008 4:22 pm

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