Compound interest is a method used to calculate the interest on money that is either saved or borrowed. It is introduced as one of the two primary ways to compute interest, alongside simple interest.

Compound Interest

In algebraic finance applications, interest is the amount of money paid or earned on an initial deposit or loan, commonly denoted by the variable $$I$$. When money is deposited into a bank, the bank pays interest; conversely, when taking out a loan, interest must be paid on the amount borrowed.

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In algebraic finance problems, the initial amount of money deposited into a bank or borrowed as a loan is called the principal, denoted by the variable $$P$$. Whether earning interest on a savings account or paying interest on a loan, the amount of interest calculated is always based directly on this original principal sum.

Principal (Algebra)

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OpenStax Intermediate Algebra

Apply the problem-solving strategy to find the principal given the simple interest earned, rate of interest, and time.

**Problem:** Sean’s new car loan statement said he would pay $$\$4{,}866.25$$ in interest from a simple interest rate of $$8.5\%$$ over $$5$$ years. How much did he borrow to buy his new car?

1. **Read** the problem carefully.
2. **Identify** what to find: the amount borrowed (the principal).
3. **Name** the unknown: Let $$P$$ = principal borrowed.
4. **Translate** into an equation using the simple interest formula $$I = Prt$$. Given $$I = 4{,}866.25$$, $$r = 0.085$$, and $$t = 5$$:
$$4{,}866.25 = P(0.085)(5)$$
5. **Solve** the equation:
Multiply $$0.085$$ and $$5$$:
$$4{,}866.25 = 0.425P$$
Divide both sides by $$0.425$$:
$$11{,}450 = P$$
6. **Check** the result:
$$I = Prt$$
$$4{,}866.25 \stackrel{?}{=} (11{,}450)(0.085)(5)$$
$$4{,}866.25 = 4{,}866.25 \checkmark$$
7. **Answer**: The principal was $$\$11{,}450$$.

Solving a Word Problem for the Principal of Sean's Car Loan

Apply the problem-solving strategy to find the principal given the simple interest earned, rate of interest, and time.

**Problem:** Eduardo noticed that his new car loan papers stated that with a $$7.5\%$$ simple interest rate, he would pay $$\$6{,}596.25$$ in interest over $$5$$ years. How much did he borrow to pay for his car?

1. **Read** the problem carefully.
2. **Identify** what to find: the amount borrowed (the principal).
3. **Name** the unknown: Let $$P$$ = principal borrowed.
4. **Translate** into an equation using the simple interest formula $$I = Prt$$. Given $$I = 6{,}596.25$$, $$r = 0.075$$, and $$t = 5$$:
$$6{,}596.25 = P(0.075)(5)$$
5. **Solve** the equation:
Multiply $$0.075$$ and $$5$$:
$$6{,}596.25 = 0.375P$$
Divide both sides by $$0.375$$:
$$17{,}590 = P$$
6. **Check** the result:
$$I = Prt$$
$$6{,}596.25 \stackrel{?}{=} (17{,}590)(0.075)(5)$$
$$6{,}596.25 = 6{,}596.25 \checkmark$$
7. **Answer**: The principal was $$\$17{,}590$$.

Solving a Word Problem for the Principal of Eduardo's Car Loan

Apply the problem-solving strategy to find the principal given the simple interest earned, rate of interest, and time.

**Problem:** In $$5$$ years, Gloria’s bank account earned $$\$2{,}400$$ interest at $$5\%$$ simple interest. How much had she deposited in the account?

1. **Read** the problem carefully.
2. **Identify** what to find: the amount deposited (the principal).
3. **Name** the unknown: Let $$P$$ = principal deposited.
4. **Translate** into an equation using the simple interest formula $$I = Prt$$. Given $$I = 2400$$, $$r = 0.05$$, and $$t = 5$$:
$$2400 = P(0.05)(5)$$
5. **Solve** the equation:
Multiply $$0.05$$ and $$5$$:
$$2400 = 0.25P$$
Divide both sides by $$0.25$$:
$$9{,}600 = P$$
6. **Check** the result:
$$I = Prt$$
$$2400 \stackrel{?}{=} (9{,}600)(0.05)(5)$$
$$2400 = 2400 \checkmark$$
7. **Answer**: The principal was $$\$9{,}600$$.

Solving a Word Problem for the Principal of Gloria's Bank Account

Interest (Algebra)

In algebraic contexts such as calculating interest, time is represented by the variable $$t$$. It specifies the duration, typically measured in the number of years, for which a principal amount of money is either saved or borrowed.

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