# Solving inequality by multiplying across

• September 25th 2013, 01:41 AM
corbomite1
Solving inequality by multiplying across
Just looking for some guidance on the following problem:

Find the least value of n (Natural number) for which (5n + 3 / 7 - n) < 1, n <> 7

One approach:

5n + 3 < 7 - n
6n + 3 < 7
6n < 4
3n < 2
n < 2/3

Since n is a Natural number we get n = 0 as the least value

The 1st step above involves multiplying across by 7 - n, however 7 - n is negative for n > 7 so this would equate to multiplying an inequality by a negative number in which case the sign would have to be changed.

Is it valid to multiply across by 7 - n as described?

Thanks(Hi)
Corbomite1
• September 25th 2013, 02:01 AM
Prove It
Re: Solving inequality by multiplying across
It's valid but you will then need to consider two cases. When n > 7, the denominator is negative and so will change the sign of the inequality.

So Case 1: n < 7

\displaystyle \begin{align*} \frac{5n + 3}{7-n} &< 1 \\ 5n+3 &< 7-n \\ 6n+3 &> 7 \\ 6n &> 4 \\ n &> \frac{2}{3} \end{align*}

So the values of n which satisfy this case are \displaystyle \begin{align*} n \in \left( \frac{2}{3}, 7 \right) \end{align*}

Now consider Case 2, where n > 7...
• September 25th 2013, 02:20 AM
corbomite1
Re: Solving inequality by multiplying across
Thanks for the quick reply! I have 2 further questions:

In case 1 above we have n < 7 which means 7 - n > 0 so we should be able to multiply across by 7 - n without changing the sign, yet you have changed the sign when arriving at 6n + 3 > 7

In case 2, n > 7 which means 7 - n < 0 so we need to change the sign when multiplying across by 7 - n:

5n + 3 / 7 - n < 1

5n + 3 > 7 - n [change sign]

6n + 3 > 7

6n > 4

3n > 2

n > 2/3

So this case cannot yield the least value of n for which the inequality holds, so we revert to n < 2/3 which gives n = 0 as n has to be a Natural number.

Correct?

Thanks a lot
Corbomite1
• September 25th 2013, 02:36 AM
Plato
Re: Solving inequality by multiplying across
Quote:

Originally Posted by corbomite1
Just looking for some guidance on the following problem:
Find the least value of n (Natural number) for which (5n + 3 / 7 - n) < 1, n <> 7

I much prefer to avoid the problem cases of cross multiplying. So I write it as:
$\\\frac{5n+3}{7-n}-1<0\\\frac{6n-4}{7-n}<0\\n=0\text{ or }n>7$

Be careful. Some textbooks and/or authors do not count zero among the natural numbers.
• September 25th 2013, 03:05 AM
corbomite1
Re: Solving inequality by multiplying across
Nice solution, thanks!
• September 25th 2013, 11:47 PM
Prove It
Re: Solving inequality by multiplying across
Quote:

Originally Posted by corbomite1
Thanks for the quick reply! I have 2 further questions:

In case 1 above we have n < 7 which means 7 - n > 0 so we should be able to multiply across by 7 - n without changing the sign, yet you have changed the sign when arriving at 6n + 3 > 7

Yes that would be a typo. I'm sure you can fix it...