Q 4.58 - Intermediate Algebra

Question

Translate to a system of equations and solve: 

Jill’s Sandwich Shoppe owes \($65,200$ on two business loans, one at $4.5$% interest and the other at $7.2$% interest. The total amount of interest owed last year was \)$3,582$. What was the principal for each loan? 

Step-by-Step Solution

Answer

The amount of loan at a rate of $4.5\%$ is $$41,200$ and the amount of loan at a rate of $7.2\%$ is $$24,000$.

Step 1. Given Information

The total amount of the loan is $$65,200$.

One loan is at $4.5\%$ interest rate and the other loan at $7.2\%$.

The amount paid last year was $$3582$

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  the first business loan at $4.5\%$ interest and $y$ be the principal of the second business loan at $7.2\%$

Step 3. Form the equations

The total amount of loan taken is $$65,200$, so the equation can be written as

$x+y=65200 ...\left(1\right)$

One loan of $x$ dollars is at $4.5\%$ interest and the other loan of $y$ dollars is at $7.2\%$ interest.

The amount paid last year is $3582$, so equation can be written as

$\begin{array}{rcl}(4.5\%)x+(7.2\%)y& =& 3582\\ 0.045x+0.072y& =& 3582 ...\left(2\right)\end{array}$

Step 4. Solve using substitution

Solve the first equation for $y$

$\begin{array}{rcl}x+y& =& 65200\\ x+y-x& =& 65200-x\\ y& =& 65200-x ...\left(3\right)\end{array}$

So in the second equation substitute $65200-x$  for $y$and solve for $x$

$\begin{array}{rcl}0.045x+0.072y& =& 3582\\ 0.045x+0.072(65200-x)& =& 3582\\ 0.045x+4694.4-0.072x& =& 3582\\ 0.045x-0.072x+4694.4-4694.4& =& 3582-4694.4\\ -0.027x& =& -1112.4\\ \frac{-0.027x}{-0.027}& =& \frac{-1112.4}{0.027}\\ x& =& 41200\end{array}$

 

Step 5. Find the value of y

Substitute $41200$ for $x$ in the third equation

$$\begin{gathered} y=65200-x \\ y=65200-41200 \\ y=24000 \end{gathered}$$

Thus, the principal amount of the loan at $4.5\%$ interest is $$41,200$ and at $7.2\%$ interest is $$24,000$

Step 6. Check the solution.

Substitute $41200$ for $x$ and $24000$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 65200\\ 41200+24000& =& 65200\\ 65200& =& 65200\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}0.045x+0.072y& =& 3582\\ 0.045\times 41200+0.072\times 24000& =& 3582\\ 1854+1728& =& 3582\\ 3582& =& 3582\end{array}$

This is also a true statement.

So the point satisfies both the equations.