GED Mathematical Reasoning: Evaluating Algebraic Expressions

  • To evaluate means to “find the value of.” To evaluate an expression, we will replace the variables with the numbers given and then perform the operation to find the value.

 

Example 1

Evaluate \frac{a}{b}, given that a = 27 and b = 3

 

In this example, the expression is:   “a” over “b”   ~OR~   “a” divided by “b” since, as you may recall, a fraction is another way to denote division. And we’re given that “a” is equal to 27 and “b” is equal to 3. So we’ll replace “a” with the number 27 and “b” with the number 3, which makes the expression “a” over “b” become 27 over 3  or  27 divided by 3.

\frac{a}{b} \rightarrow \frac{27}{3}

As a general rule, when substituting a number for a variable, I recommend always including parenthesis around the number. Developing this habit now will help you down the road when problems get more complicated.

Okay! To find the value, of “27 over 3” you may think of this as a fraction and reduce to lowest terms by dividing the numerator and denominator by a common factor of 3.

\frac{27}{3} = \frac{27 \div 3}{3 \div 3} = \frac{9}{1} = 9

Or, you may simply divide 27 by 3. Either way, the final answer is 9.

\frac{27}{3} = 27 \div 3 = 9

 

Example 2

Evaluate 2y - 3, given that y = 7

 

Recall that when a number and a letter or two letters are written right up next to each other with nothing in between, the operation is assumed to be multiplication. So the operation between “2” and “y” is multiplication.

Also note that there is a minus sign in this expression, to denote subtraction. So this expression involves two operations: multiplication and subtraction.

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We will talk more about the order of operations later on, but for now it is enough for you to know that when there is more than one operation, do any multiplying and dividing before adding and subtracting.

With that in mind, let’s finish up this example. When we replace “y” with 7, the expression “two times ‘y’ minus 3” becomes “two times 7 minus 3”

2y - 3 \rightarrow 2(7) - 3

So we’ll multiply 2 times 7 FIRST, which gives us 14. And the problem then becomes: 14 – 3. Subtracting 3 from 14 yields a final result of:  11

2y - 3 \rightarrow 2(7) - 3 = 14 - 3 = 11

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