## Saturday, December 22, 2012

Many problems in algebra are in the form of words. In order to solve them, we must know how to translate the words into an equation. The key steps in solving a problem are analyzing the problem, forming the equation, solving the equation and checking the result.

Words and phrases can represent different mathematical operations. It's important to be able to translate words and phrases into equations when solving word problems. Here are some common words and phrases and their associated mathematical operation.
• minus, decreased, less than, reduced by: subtraction
• twice, product, multiplied by, times, of: multiplication
• quotient, divided by, ratio, into: division
It's best to try and write the equation as you are reading the problem. Here are some examples and steps to their solutions.

John and Steve have been collecting baseball cards since 2003. They now have accumulated a total of 40,000 cards. If John has 7,500 more cards than Steve, how many cards do each of them have?
First, assign a variable for the number of cards that either John or Steve has.

Let x = number of cards that Steve has. Since John has 7,500 more cards than Steve, he has 7,500 + x cards. Therefore, the number of cards that Steve has + Number of cards that John has = total number of cards. The equation that must be solved then is x + 7,500 + x = 40,000.

Here's another example which can be applied to real life. Suppose you wish to decide between two long distance phone plans. The first plan is \$0.20 for the first minute, \$0.07 for each additional minute. The second plan is \$0.15 for the first minute and \$0.10 for each additional minute. How many minutes must you talk for the first plan to be cheaper than the second plan?

For the first plan, let m = total minutes. Therefore, the cost for first minute plus cost for each additional minute is 0.20 + 0.07(m - 1).

For the second plan, the cost for first minute plus cost for each additional minute is 0.15 + 0.10(m - 1). We set equations equal to each other and solve for m to see how many minutes will the cost be equal.

Suppose a school wants to have an enclosed rectangular playground and has enough money in the budget to afford 550 feet of fencing. If the length is to be 125 feet longer than the width, what are the dimensions of the fence that encloses the playground?

Let W = width of the fence. The length is W + 125, therefore 2 times the length + 2 times the width = 550. The equation we must solve is 2(W + 125) + 2W = 550. Once we get a value for W, substitute back into the equation to solve for the length.

The important thing to remember is to analyze the problem, assign variable, set up equations which make sense in the context of the problem and check solutions. Be sure to try and write the equation as you are reading the problem.