Q 4.57 - Intermediate Algebra

Question

Translate to a system of equations and solve: 

Laura owes \($18,000$ on her student loans. The interest rate on the bank loan is $2.5$% and the interest rate on the federal loan is $6.9$%. The total amount of interest she paid last year was \)$1066$. What was the principal for each loan? 

Step-by-Step Solution

Answer

The principal for bank loan is $$4000$

and the principal for federal loan is $$14000$

Step 1. Given Information

Laura has $$18000$ student loan.

The interest rate on the bank loan is $2.5$% and the interest rate on the federal loan is $6.9$%.

The total amount of interest she paid last year was $$1066$. 

Step 2. Identify and name what we are looking for

We need to find the principal for each loan.

Let $x$ be the principal of bank loan and $y$ be the principal of federal loan.

Step 3. Form the equations

The total loan amount is $$18000$, so the equation is

$x+y=18000 ...\left(1\right)$

The federal loan has interest rate of $6.9$% and the for the bank loan it is $2.5$%, and the amount paid last year is $$1066$, so the second equation is given as

$\begin{array}{rcl}(2.5\%)x+(6.9\%)y& =& 1066\\ 0.025x+0.069y& =& 1066 ...\left(2\right)\end{array}$

Step 4. Solve using substitution

Solve the first equation for $y$

$\begin{array}{rcl}x+y& =& 18000\\ x+y-x& =& 18000-x\\ y& =& 18000-x ...\left(3\right)\end{array}$

Using the third equation substitute $18000-x$ for $y$  in the second equation and solve for $x$

$\begin{array}{rcl}0.025x+0.069y& =& 1066\\ 0.025x+0.069(18000-x)& =& 1066\\ 0.025x+1242-0.069x& =& 1066\\ 0.025x-0.069x+1242-1242& =& 1066-1242\\ -0.044x& =& -176\\ \frac{-0.044x}{-0.044}& =& \frac{-176}{-0.044}\\ x& =& 4000\end{array}$

 

Step 5. Find the value of y

Substitute $4000$ for $x$ in the third equation

$$\begin{gathered} y=18000-x \\ y=18000-4000 \\ y=14000 \end{gathered}$$

So amount invested in bank loan is $$4000$ and on the federal loan is$$14000$.

Step 6. Check the solution.

Substitute $4000$ for $x$ and $14000$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 18000\\ 4000+14000& =& 18000\\ 18000& =& 18000\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.025x+0.069y& =& 1066\\ 0.025·4000+0.069·14000& =& 1066\\ 100+966& =& 1066\\ 1066& =& 1066\end{array}$

This is also a true statement.

So the point satisfies both the equations.