Disassembling a product to it's factors

[Karol]

Homework Statement

In a physics problem where V is the volume i have $\displaystyle~3V-\frac{3}{4}V~$. i get 2 different answers when i calculate.

Homework Equations

$$a(b-c)=ab-ac$$

The Attempt at a Solution

I can:
$$3V-\frac{3}{4}V=3\left( 1-\frac{1}{4} \right)V=3\frac{3}{4}V$$
And if i solve it simply i get $~\displaystyle \left( 3-\frac{3}{4} \right)V=2\frac{1}{4}V$

 

[StoneTemplePython]

This is a notational bust.

the first one reads $3\big(\frac{3}{4}\big) = \frac{9}{4}$

the second one reads 2 and $\frac{1}{4}$ which just so happens to be equal to $\frac{9}{4}$

 

[kuruman]

$3 \left( 1-\frac{1}{4} \right)V=3 \left( \frac{3}{4} \right) V=\frac{9}{4}V=2.25V$

 

[tnich]

Homework Statement

In a physics problem where V is the volume i have $\displaystyle~3V-\frac{3}{4}V~$. i get 2 different answers when i calculate.

Homework Equations

$$a(b-c)=ab-ac$$

The Attempt at a Solution

I can:
$$3V-\frac{3}{4}V=3\left( 1-\frac{1}{4} \right)V=3\frac{3}{4}V$$
And if i solve it simply i get $~\displaystyle \left( 3-\frac{3}{4} \right)V=2\frac{1}{4}V$

Your problem is one of notation. In the first solution you use $3\frac{3}{4}$ to mean $(3)\left(\frac{3}{4}\right)$ (multiplication).
In the second solution you use $2\frac{1}{4}$ to mean $2+\frac{1}{4}$ (addition).

 

[Karol]

What is a bust, in slang?

 

[StoneTemplePython]

What is a bust, in slang?

Basically a break in logic, or fatal error.
- - - -
You need to find a way to make the notation work for you, not against you.

 

[Karol]

Thank you:
StoneTemplePython
Kuruman
And tnich