Chapter 10, Problem 43RE

To determine

To find: the amount invested in each account

Answer to Problem 43RE

The amount invested on $6\%$ interest account is $\$3000$ and amount invested on $7\%$ interest account is $\$6000$

Explanation of Solution

Given:

Paying interest 6% and 7%

Annual interest income $600

Calculation:

Consider the data

Interest per year on first account is $6\%$

Interest per year on second account is $7\%$

Investment on second account twice than the first account and annual interest income is $\$600$ .

Let x = amount invested on first account.

y = amount invested on second account.

Convert each fact in the problem to an equation.

The investment on second account twice than the first account

  $y=2x$

And

  $0.06x+0.07y=600$

Therefore, the system of equations is

  $\begin{array}{l}2x-y=0 Equation1\\ 0.06x+0.07y=600 Equation\text{ 2}\end{array}$

Use these together to solve the system. Change the system to triangular form by eliminating the $x-term$ from the equation $3$ by adding $2$ times equation $2to-0.06$ times equation $1$ .

The work is shown as follows.

  $\frac{\begin{array}{l}-0.12x+0.06y=0\\ 0.12x+0.14y=1200\end{array}}{0.2y=1200}$

So, the equivalent system one step closer to triangular form is

  $\begin{array}{l}2x-y=0\\ 0.2y=1200 -0.6Equation 1+2.Equation 2=New equation 2\end{array}$

The system is in triangular form. Use back substitution to solve the system. From the second equation,

  $\begin{array}{l}0.2y=1200\\ y=\frac{1200}{0.2}\\ =6000\end{array}$

Back substitute $y$ in first equation and solve for $x$ .

  $\begin{array}{l}2x-y=0\\ 2x-6000=0\\ 2x=6000\\ x=\frac{6000}{2}Simplify\\ =3000\end{array}$

Therefore, the solution to the system of equations is $x=3000$ , $y=6000$ .

Thus, the amount invested on $6\%$ interest account is $\$3000$ and amount invested on $7\%$ interest account is $\$6000$ .

Conclusion: The amount invested on $6\%$ interest account is $\$3000$ and amount invested on $7\%$ interest account is $\$6000$