Category Archives: Applications

Example:  The product of some negative number and 5 less than twice that number is 273.  Find the number.

Solution:  Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows.  Twice that number refers to 2 times the negative number that was described before.  (2x-5)

“…is 273”

This translates to equals 273.

Solve the equation.

 The translated equation Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation) Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula)  I will demonstrate factoring so I need to move all terms to the same side to have zero on one side. Factor Use the zero product property and set each factor equal to zero and solve.

Since the statement says that the number is negative, the number is .

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:

For the 5% account:

 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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