Creating equivalent forms of an expression is a high emphasis topic on the TASC Test Assessing Secondary Completion™ Mathematics subtest. Expressions, like people, are multifaceted. You can't know someone's life story by simply glancing at them. You can’t understand what expressions are representing by a quick look; they require you to work through them for understanding.
A mathematical expression can be considered a mathematical thought or idea communicated by the language of mathematics. Find a good review of interpreting expressions that represent a quantity in a previous TASC test math blog post.
You may be asked to choose and produce an equivalent form of an expression to explain the properties of the quantity represented by the expression.
Let’s look at a few examples from Khan Academy:
Example 1: Which expressions are equivalent to 2(4f+2g)?
- 8f + 4g
- 2f (4+2g)
- 8f + 2g
- 4 (2f+g)
The goal of this example is to manipulate 2(4f+2g) in order to get one more of the answers.
First, distribute the two. By doing this you get:
2·4f + 2·2g = 8f + 4g
So we know that (a) is an equivalent form of the original expression. But what about the other options? By solving the same way, we can distribute the two:
b. 2f (4+2g) = 8f + 4fg (not an equivalent form of the original expression)
- 8f + 2g = 10fg (not an equivalent form of the original expression)
- 4 (2f+g) = 8f + 4g (an equivalent form of the original expression)
Example 2: Find an equivalent expression to 6l + 5m - 3n.
- 3 (2l – n) + 5m à 6l – 3n + 5m = equivalent form.
- 3n + 6l – 5m à This is not equivalent to the original. The 3n is negative and 5m is positive in the original, and in this form, the 3 is positive and the 5 is now negative.
- 5m + (6l – 3n) à You can remove the parenthesis and it will not change the outcome. It is an equivalent form.
The equivalent expressions are (a) and (c).
Writing a function that’s defined by an expression in different but equivalent forms is another important skill that the TASC Mathematics subtest will test you on.
Example 1: The quantity of a car C after time t can be written C(t) = 1.02t . Is the quantity of the car growing or decaying? Identify the percent rate of change from the equation for C(t).
Answer: C is growing at a rate of 2% for each time interval t.
Example 2: Solve x2 − 5x + 6 = 0.
Answer: x = 3 and x = 2.
Find more example function problems at Shmoop.com.