n

2=n+1

can you help me solve this?

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- Oct 8th 2008, 06:29 AMgedprepAssistance is needed.
n

2=n+1

can you help me solve this? - Oct 8th 2008, 06:45 AMmasters
This is an addition equation because a value has been added to the variable n.

To isolate the variable (get the variable by itself on one side of the equation),

you must perform the inverse of addition (subtraction) on the value.

$\displaystyle 2=n+1$

Subtract 1 from both sides

$\displaystyle 2-1=n+1-1$

$\displaystyle 1=n$

$\displaystyle \boxed{n=1}$ - Oct 8th 2008, 06:51 AMgedprepI made a mistake.
I am sorry but tis was 2 over n equals n plus 1

- Oct 8th 2008, 07:03 AMmasters
$\displaystyle \frac{2}{n}=n+1$

First, multiply each term by n

$\displaystyle 2=n^2+n$

Now, set the equation = 0

$\displaystyle n^2+n-2=0$

Now, factor the trinomial

$\displaystyle (n+2)(n-1)=0$

Using the zero product property that says "If ab=0, then a=0 or b=0",

$\displaystyle n+2=0 \ \ or \ \ n-1=0$

$\displaystyle n=-2 \ \ or \ \ n=1$ - Oct 8th 2008, 08:36 AMgedprepThat error is my fault and I corrected it already. Check it out
I am sorry but it is n over 2 equals n plus 1

not 2 over n - Oct 8th 2008, 08:47 AMmasters
- Oct 8th 2008, 08:56 AMgedprepRe: Assistance is needed
why must you multiply the 1 as well as the number n?

- Oct 8th 2008, 09:46 AMmasters
One of the 10 commandants of math is:

Thou shalt do unto one side of an equation what thou doest to the other.

If I multiply 2 times one term in an equation, I must multiply 2 times every term in the equation.

Example:

$\displaystyle \frac{x}{2}+3=2x-1$

Multiple all terms by 2.

$\displaystyle {\color{red}2}\left(\frac{x}{2}\right)+{\color{red }2}(3)={\color{red}2}(2x)-{\color{red}2}(1)$