# Thread: [SOLVED] Pre-Calc applications of equations

1. ## [SOLVED] Pre-Calc applications of equations

Hi, I need help on this word problem:

You have already invested 550$in a stock with an annual return of 11%. How much of an additional 1100 should be invested at 12% and how much at 6% so that the total return on the entire 1650$ is 9%

I tried to set up my equations as (550 times .11) + (.12x times 1100) + (.06y times 1100) = (1650 times .09)
and x + y = 1100

I am wondering if i am going in the right direction. The answer in the book says 366.67 at 12% and 733.33 at 6% but i tried to work my math and i got nowhere close to that answer.

2. Hello, soccer49!

You're on the right track, but you put "too much" in your equations.

You have already invested $550 in a stock with an annual return of 11%. How much of an additional$1100 should be invested at 12% and how much at 6%
so that the total return on the entire $1650$ is 9% ?

Let $x$ = amounted invested at 12%.
Let $y$ = amount invested at 6%.

Then: . $x + y \:=\:1100$ .[1]

$550 at 11% returns$60.50 in interest.

$x$ dollars at 12% returns $0.12x$ dollars in interest.

$y$ dollars at 6% returns $0.06y$ dollars in interest.

The total interest is: . $60.50 + 0.12x + 0.06y$
. . and we want this to be equal to 9% of $1650 =$148.50

There is our equation: . $60.50 + 0.12x + 0.06y \:=\:148.50$

Multiply by 100: . $6050 + 12x + 6y \:=\:14850 \quad\Rightarrow\quad 12x + 6y \:=\:8800$ .[2]

Solve the system of equation.

$\begin{array}{cccccccc}\text{Multiply {\color{blue}[1]} by -6:} & \text{-}6x - 6y &=& \text{-}6600 \\ \text{Add {\color{blue}[2]}:} & 12x + 6y &=& 8800 \end{array}$

And we get: . $6x \:=\:2200 \quad\Rightarrow\quad x \:=\:\frac{2200}{6} \quad\Rightarrow\quad\boxed{ x\:\approx\:\366.67}$

Substitute into [1]: . $366.67 + y \:=\:1100\quad\Rightarrow\quad \boxed{y \:=\:\733.33}$

3. Thank you very much, yes that would make sense, i am just not very good at setting up the equations.