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List the constants and variables for each algebraic expression.

  1. 2 π r ( r + h )
  2. 2( L + W )
  3. 4 y 3 + y
Constants Variables
a. 2 π r ( r + h ) 2 , π r , h
b. 2(L + W) 2 L, W
c. 4 y 3 + y 4 y
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Evaluating an algebraic expression at different values

Evaluate the expression 2 x 7 for each value for x.

  1. x = 0
  2. x = 1
  3. x = 1 2
  4. x = −4
  1. Substitute 0 for x .
    2 x 7 = 2 ( 0 ) 7 = 0 7 = −7
  2. Substitute 1 for x .
    2 x 7 = 2 ( 1 ) 7 = 2 7 = −5
  3. Substitute 1 2 for x .
    2 x 7 = 2 ( 1 2 ) 7 = 1 7 = −6
  4. Substitute −4 for x .
    2 x 7 = 2 ( 4 ) 7 = 8 7 = −15
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Evaluate the expression 11 3 y for each value for y.

  1. y = 2
  2. y = 0
  3. y = 2 3
  4. y = −5
  1. 5;
  2. 11;
  3. 9;
  4. 26
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Evaluating algebraic expressions

Evaluate each expression for the given values.

  1. x + 5 for x = −5
  2. t 2 t −1 for t = 10
  3. 4 3 π r 3 for r = 5
  4. a + a b + b for a = 11 , b = −8
  5. 2 m 3 n 2 for m = 2 , n = 3
  1. Substitute −5 for x .
    x + 5 = ( −5 ) + 5 = 0
  2. Substitute 10 for t .
    t 2 t 1 = ( 10 ) 2 ( 10 ) 1 = 10 20 1 = 10 19
  3. Substitute 5 for r .
    4 3 π r 3 = 4 3 π ( 5 ) 3 = 4 3 π ( 125 ) = 500 3 π
  4. Substitute 11 for a and –8 for b .
    a + a b + b = ( 11 ) + ( 11 ) ( −8 ) + ( −8 ) = 11 88 8 = −85
  5. Substitute 2 for m and 3 for n .
    2 m 3 n 2 = 2 ( 2 ) 3 ( 3 ) 2 = 2 ( 8 ) ( 9 ) = 144 = 12
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Evaluate each expression for the given values.

  1. y + 3 y 3 for y = 5
  2. 7 2 t for t = −2
  3. 1 3 π r 2 for r = 11
  4. ( p 2 q ) 3 for p = −2 , q = 3
  5. 4 ( m n ) 5 ( n m ) for m = 2 3 , n = 1 3
  1. 4;
  2. 11;
  3. 121 3 π ;
  4. 1728;
  5. 3
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Formulas

An equation    is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation 2 x + 1 = 7 has the unique solution of 3 because when we substitute 3 for x in the equation, we obtain the true statement 2 ( 3 ) + 1 = 7.

A formula    is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area A of a circle in terms of the radius r of the circle: A = π r 2 . For any value of r , the area A can be found by evaluating the expression π r 2 .

Using a formula

A right circular cylinder with radius r and height h has the surface area S (in square units) given by the formula S = 2 π r ( r + h ) . See [link] . Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of π .

A right circular cylinder with an arrow extending from the center of the top circle outward to the edge, labeled: r. Another arrow beside the image going from top to bottom, labeled: h.
Right circular cylinder

Evaluate the expression 2 π r ( r + h ) for r = 6 and h = 9.

S = 2 π r ( r + h ) = 2 π ( 6 ) [ ( 6 ) + ( 9 ) ] = 2 π ( 6 ) ( 15 ) = 180 π

The surface area is 180 π square inches.

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A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm 2 ) is found to be A = ( L + 16 ) ( W + 16 ) L W . See [link] . Find the area of a matte for a photograph with length 32 cm and width 24 cm.

/ An art frame with a piece of artwork in the center. The frame has a width of 8 centimeters. The artwork itself has a length of 32 centimeters and a width of 24 centimeters.

1,152 cm 2

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Simplifying algebraic expressions

Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.

Simplifying algebraic expressions

Simplify each algebraic expression.

  1. 3 x 2 y + x 3 y 7
  2. 2 r 5 ( 3 r ) + 4
  3. ( 4 t 5 4 s ) ( 2 3 t + 2 s )
  4. 2 m n 5 m + 3 m n + n

  1. 3 x 2 y + x 3 y 7 = 3 x + x 2 y 3 y 7 Commutative property of addition = 4 x 5 y 7 Simplify

  2. 2 r 5 ( 3 r ) + 4 = 2 r 15 + 5 r + 4 Distributive property = 2 r + 5 r 15 + 4 Commutative property of addition = 7 r 11 Simplify

  3. 4 t 4 ( t 5 4 s ) ( 2 3 t + 2 s ) = 4 t 5 4 s 2 3 t 2 s Distributive property = 4 t 2 3 t 5 4 s 2 s Commutative property of addition = 10 3 t 13 4 s Simplify

  4. m n 5 m + 3 m n + n = 2 m n + 3 m n 5 m + n Commutative property of addition =   5 m n 5 m + n Simplify
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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