Category Archives: Applications of Linear Equations

Application of Linear Equation (Simple Interest)

Example:  Larry invested part of his $31,000 advance at 6% annual simple interest and the rest at 5% annual simple interest.  If his total yearly interest from both accounts was $1,760, find the amount invested at each rate.

Solution:  This question involved simple interest.  We will use the simple interest formula.


The I stands for interest, the P stands for principal (initial investment), r stands for the interest rate and t stands for time in years.

When starting an application problem like this it can be helpful to organize the information in a table using the formula.

I P r t
6% Account
5% Account

Begin filling in the information from the problem.  The interest rate is the most obvious and you should use the decimal version of the percent by moving the decimal place left two places.  Since the problem state that $1760 is the yearly interest for both accounts, we will use a time of 1 year.

I P r t
6% Account .06 1
5% Account .05 1

The goal in this question is to find the amount invested in each account.  I will let x be the amount invested in the 6% account.  The problem states that the rest of the $31,000 will be invested in the 5% account.  We can represent the rest of the money as 31000-x.

I P r t
6% Account x .06 1
5% Account  31000-x .05 1

You can fill in the interest column using the simple interest formula.

For the 6% account: I=Prt=(0.06)(x)(1)=0.06x

For the 5% account: I=Prt=(0.05)(31000-x)(1)=0.05(31000-x)

I P r t
6% Account 0 .06x x 0.06 1
5% Account 0.05(31000-x) 31000-x 0.05 1

You can now create an equation with this information.  The interest from the first account plus the interest from the second account should equal the total interest of $1760.


Solve the equation.
Use the distributive property and combine like terms to simplify each side of the equation.
Solve for x by subtracting 1550 on both sides and simplifying.
Solve for x by dividing both sides by 0.01 and simplifying.

Since x represents the amount invested in the 6% account, $21000 is invested in the 6% account and the rest is invested in the 5% account.  The rest is $31,000-$21,000=$10,000.  $10,000 is invested in the 5% account.

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