Q 4.56 - Intermediate Algebra

Question

Translate to a system of equations and solve: Julius invested $$7000$ into two stock investments. One stock paid $11$% interest and the other stock paid data-custom-editor="chemistry" $13$% interest. He earned $12.5$% interest on the total investment. How much money did he put in each stock? 

Step-by-Step Solution

Answer

Julius put $$1750$ in the first stock and $$5250$ in the second stock.

Step 1. Given Information

Julia invested a sum of $$7000$ in two stocks.

The first stock gave $11$% interest while the second stock gave $13$% interest.

On the total investment, he earned $12.5$% interest.

Step 2. Identify and name what we are looking for

We need to find the amount of money invested in each stock.

Let $x$ dollars be invested in the first stock and $y$ dollars be invested in the second stock.

Step 3. Form the equations

The total amount invested is $$7000$. So an equation can be written as

$x+y=7000 ...\left(1\right)$

The first stock paid $11$% interest and the second stock paid $13$% interest and he earned $12.5$% interest on the total investment, so

$\begin{array}{rcl}(11\%)x+(13\%)y& =& (12.5\%)\left(7000\right)\\ 0.11x+0.13y& =& 0.125\times 7000\\ 0.11x+0.13y& =& 875 ...\left(2\right)\end{array}$

Step 4. Solve using substitution

Solve the first equation for $y$

$\begin{array}{rcl}x+y& =& 7000\\ x+y-x& =& 7000-x\\ y& =& 7000-x ...\left(3\right)\end{array}$

Using the third equation substitute $7000-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}0.11x+0.13y& =& 875\\ 0.11x+0.13(7000-x)& =& 875\\ 0.11x+910-0.13x& =& 875\\ 0.11x-0.13x+910-910& =& 875-910\\ -0.02x& =& -35\\ \frac{-0.02x}{-0.02}& =& \frac{-35}{-0.02}\\ x& =& 1750\end{array}$

Step 5. Solve for y

Substitute $1750$ for $x$ in the thrid equation

$$\begin{gathered} y=7000-x \\ y=7000-1750 \\ y=5250 \end{gathered}$$

So amount invested in the first stock is $$1750$ and the amount invested in the second stock is $$5250$

Step 6. Check the solutions

Substitute $1750$ for $x$ and $5250$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 7000\\ 1750+5250& =& 7000\\ 7000& =& 7000\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.11x+0.13y& =& 875\\ 0.11·1750+0.13·5250& =& 875\\ 192.5+682.5& =& 875\\ 875& =& 875\end{array}$

This is also a true statement.

So the point satisfies both the equations.