So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as x+5,43πr3x+5,43πr3, or √2m3n2√2m3n2. In the expression x+5x+5, 5 is called a constant because it does not vary and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.
We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way.
In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables.
Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before.
Example 8: Describing Algebraic Expressions
List the constants and variables for each algebraic expression.
- x + 5
- 43πr343πr3
- √2m3n2√2m3n2
Solution
Constants | Variables | |
---|---|---|
1. x + 5 | 5 | x |
2. 43πr343πr3 | 43,π43,π | rr |
3. √2m3n2√2m3n2 | 2 | m,nm,n |
Try It 8
List the constants and variables for each algebraic expression.
- 2πr(r+h)2πr(r+h)
- 2(L + W)
- 4y3+y4y3+y
Example 9: Evaluating an Algebraic Expression at Different Values
Evaluate the expression 2x−72x−7 for each value for x.
- x=0x=0
- x=1x=1
- x=12x=12
- x=−4x=−4
Solution
- Substitute 0 for xx.
2x−7=2(0)−7=0−7=−72x−7=2(0)−7=0−7=−7
- Substitute 1 for xx.
2x−7=2(1)−7=2−7=−52x−7=2(1)−7=2−7=−5
- Substitute 1212 for xx.
2x−7=2(12)−7=1−7=−62x−7=2(12)−7=1−7=−6
- Substitute −4−4 for xx.
2x−7=2(−4)−7=−8−7=−152x−7=2(−4)−7=−8−7=−15
Try It 9
Evaluate the expression 11−3y11−3y for each value for y.
a. y=2y=2
b. y=0y=0
c. y=23y=23
d. y=−5y=−5
Example 10: Evaluating Algebraic Expressions
Evaluate each expression for the given values.
- x+5x+5 for x=−5x=−5
- t2t−1t2t−1 for t=10t=10
- 43πr343πr3 for r=5r=5
- a+ab+ba+ab+b for a=11,b=−8a=11,b=−8
- √2m3n2√2m3n2 for m=2,n=3m=2,n=3
Solution
- Substitute −5−5 for xx.
x+5=(−5)+5=0x+5=(−5)+5=0
- Substitute 10 for tt.
t2t−1=(10)2(10)−1=1020−1=1019t2t−1=(10)2(10)−1=1020−1=1019
- Substitute 5 for rr.
43πr3=43π(5)3=43π(125)=5003π43πr3=43π(5)3=43π(125)=5003π
- Substitute 11 for aa and –8 for bb.
a+ab+b=(11)+(11)(−8)+(−8)=11−8−8=−85a+ab+b=(11)+(11)(−8)+(−8)=11−8−8=−85
- Substitute 2 for mm and 3 for nn.
√2m3n2=√2(2)3(3)2=√2(8)(9)=√144=12√2m3n2=√2(2)3(3)2=√2(8)(9)=√144=12
Try It 10
Evaluate each expression for the given values.
a. y+3y−3y+3y−3 for y=5y=5
b. 7−2t7−2t for t=−2t=−2
c. 13πr213πr2 for r=11r=11
d. (p2q)3(p2q)3 for p=−2,q=3p=−2,q=3
e. 4(m−n)−5(n−m)4(m−n)−5(n−m) for m=23,n=13m=23,n=13
Candela Citations
- College Algebra. Authored by: OpenStax College Algebra. Provided by: OpenStax. Located at: http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1/Preface. License: CC BY: Attribution