# Ed bought $2,000 in stock shares one week before the stock price dropped$10.00. If he had waited for the price setback, he could have bought 10 more shares for the same amount. How many shares did he buy?

Sep 29, 2017

See a solution process below:

#### Explanation:

We can write a formula for the shares Ed bought as:

n * p = $2000 Where: • $n$is the number of shares Ed bought • $p$is the the price Ed paid per share •$2000 is the total Ed paid for all the shares

We can write an equation for what would of happened if Ed had waited one week for the price per share to drop $10 per share when he could of bought 10 more shares as: (n + 10)(p -$10) = $2000 Or np -$10n + 10p - $100 =$2000

Step 1" We can solve the first equation for $p$:

(n * p)/color(red)(n) = ($2000)/color(red)(n) (color(red)(cancel(color(black)(n))) * p)/cancel(color(red)(n)) = ($2000)/n

p = ($2000)/n Step 2: Substitute ($2000)/n for $p$ in the second equation and solve for $n$:

np - $10n + 10p -$100 = $2000 becomes: (n * ($2000)/n) - $10n + (10 * ($2000)/n) - $100 =$2000

(color(red)(cancel(color(black)(n))) * ($2000)/color(red)(cancel(color(black)(n)))) -$10n + ($20000)/n -$100 = $2000 $2000 - $10n + ($20000)/n - $100 =$2000

$2000 -$100 - $10n + ($20000)/n = $2000 $2000 - $100 -$10n + ($20000)/n =$2000

$1900 -$10n + ($20000)/n =$2000

$1900 - color(red)($1900) - $10n + ($20000)/n = $2000 - color(red)($1900)

0 - $10n + ($20000)/n = $100 -$10n + ($20000)/n =$100

$10(-n + (2000)/n) =$100

($10(-n + (2000)/n))/color(red)($10) = ($100)/color(red)($10)

(color(red)(cancel(color(black)($10)))(-n + (2000)/n))/cancel(color(red)($10)) = 10

$- n + \frac{2000}{n} = 10$

$\textcolor{red}{n} \left(- n + \frac{2000}{n}\right) = \textcolor{red}{n} \times 10$

$\left(\textcolor{red}{n} \times - n\right) + \left(\textcolor{red}{n} \times \frac{2000}{n}\right) = 10 n$

$- {n}^{2} + 2000 = 10 n$

$- {n}^{2} - \textcolor{red}{10 n} + 2000 = 10 n - \textcolor{red}{10 n}$

$- {n}^{2} - 10 n + 2000 = 0$

$\textcolor{red}{- 1} \left(- {n}^{2} - 10 n + 2000\right) = \textcolor{red}{- 1} \times 0$

${n}^{2} + 10 n - 2000 = 0$

$\left(n + 50\right) \left(n - 40\right) = 0$

Solution 1:

$n + 50 = 0$

$n + 50 - \textcolor{red}{50} = 0 - \textcolor{red}{50}$

$n + 0 = - 50$

$n = - 50$

Solution 2:

$n - 40 = 0$

$n - 40 + \textcolor{red}{40} = 0 + \textcolor{red}{40}$

$n - 0 = 40$

$n = 40$

The solution is: Ed bought 40 shares.

Solution 1 is an extraneous solution because Ed could not have bought a negative 50 (or -50) shares.