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Study Guides > College Algebra CoRequisite Course

Evaluate and Simplify Algebraic Expressions

Learning Outcomes

  • List the constants and variables in an algebraic expression.
  • Evaluate an algebraic expression.
  • Use an algebraic formula.
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,\frac{4}{3}\pi {r}^{3}, or 2m3n2\sqrt{2{m}^{3}{n}^{2}}. In the expression x+5,5x+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.
\begin{align}&\left(-3\right)^{5}=\left(-3\right)\cdot\left(-3\right)\cdot\left(-3\right)\cdot\left(-3\right)\cdot\left(-3\right) && x^{5}=x\cdot x\cdot x\cdot x\cdot x \\ &\left(2\cdot7\right)^{3}=\left(2\cdot7\right)\cdot\left(2\cdot7\right)\cdot\left(2\cdot7\right) && \left(yz\right)^{3}=\left(yz\right)\cdot\left(yz\right)\cdot\left(yz\right)\\ \text{ }\end{align}
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: Describing Algebraic Expressions

List the constants and variables for each algebraic expression.
  1. x + 5
  2. 43πr3\frac{4}{3}\pi {r}^{3}
  3. 2m3n2\sqrt{2{m}^{3}{n}^{2}}

Answer:

Constants Variables
1. x + 5 5 x
2. 43πr3\frac{4}{3}\pi {r}^{3} 43,π\frac{4}{3},\pi rr
3. 2m3n2\sqrt{2{m}^{3}{n}^{2}} 2 m,nm,n

Try It

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Example: Evaluating an Algebraic Expression at Different Values

Evaluate the expression 2x−72x - 7 for each value for x.
  1. x=0x=0
  2. x=1x=1
  3. x=12x=\dfrac{1}{2}
  4. x=−4x=-4

Answer:

  1. Substitute 0 for xx.
    \begin{align}2x-7 & = 2\left(0\right)-7 \\ & =0-7 \\ & =-7\end{align}
  2. Substitute 1 for xx.
    \begin{align}2x-7 & = 2\left(1\right)-7 \\ & =2-7 \\ & =-5\end{align}
  3. Substitute 12\dfrac{1}{2} for xx.
    \begin{align}2x-7 & = 2\left(\frac{1}{2}\right)-7 \\ & =1-7 \\ & =-6\end{align}
  4. Substitute −4-4 for xx.
    \begin{align}2x-7 & = 2\left(-4\right)-7 \\ & =-8-7 \\ & =-15\end{align}

Try It

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Example: Evaluating Algebraic Expressions

Evaluate each expression for the given values.
  1. x+5x+5 for x=−5x=-5
  2. t2t−1\frac{t}{2t - 1} for t=10t=10
  3. 43πr3\dfrac{4}{3}\pi {r}^{3} for r=5r=5
  4. a+ab+ba+ab+b for a=11,b=−8a=11,b=-8
  5. 2m3n2\sqrt{2{m}^{3}{n}^{2}} for m=2,n=3m=2,n=3

Answer:

  1. Substitute −5-5 for xx.
    \begin{align}x+5 &=\left(-5\right)+5 \\ &=0\end{align}
  2. Substitute 10 for tt.
    \begin{align}\frac{t}{2t-1} & =\frac{\left(10\right)}{2\left(10\right)-1} \\ & =\frac{10}{20-1} \\ & =\frac{10}{19}\end{align}
  3. Substitute 5 for rr.
    \begin{align}\frac{4}{3}\pi r^{3} & =\frac{4}{3}\pi\left(5\right)^{3} \\ & =\frac{4}{3}\pi\left(125\right) \\ & =\frac{500}{3}\pi\end{align}
  4. Substitute 11 for aa and –8 for bb.
    \begin{align}a+ab+b & =\left(11\right)+\left(11\right)\left(-8\right)+\left(-8\right) \\ & =11-8-8 \\ & =-85\end{align}
  5. Substitute 2 for mm and 3 for nn.
    \begin{align}\sqrt{2m^{3}n^{2}} & =\sqrt{2\left(2\right)^{3}\left(3\right)^{2}} \\ & =\sqrt{2\left(8\right)\left(9\right)} \\ & =\sqrt{144} \\ & =12\end{align}

 

Be Careful when simplifying fractions!

Why does the fraction (25)3(25)−1\dfrac{(25)}{3(25)-1} not simplify to (25)3(25)−1=13−1=12\dfrac{\cancel{(25)}}{3\cancel{(25)}-1}=\dfrac{1}{3-1}=\dfrac{1}{2}? Using the inverse property of multiplication, we are permitted to "cancel out" common factors in the numerator and denominator such that aa=1\dfrac{a}{a}=1. But be careful! We have no rule that allows us to cancel numbers in the top and bottom of a fractions that are contained in sums or differences. You'll see this idea reappear frequently throughout the course.
 

Try It

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In the following video we present more examples of how to evaluate an expression for a given value. https://youtu.be/MkRdwV4n91g

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 2x+1=72x+1=7 has the unique solution x=3x=3 because when we substitute 3 for xx in the equation, we obtain the true statement 2(3)+1=72\left(3\right)+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 AA of a circle in terms of the radius rr of the circle: A=πr2A=\pi {r}^{2}. For any value of rr, the area AA can be found by evaluating the expression πr2\pi {r}^{2}.

Example: Using a Formula

A right circular cylinder with radius rr and height hh has the surface area SS (in square units) given by the formula S=2πr(r+h)S=2\pi r\left(r+h\right). Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of π\pi.
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

Answer: Evaluate the expression 2πr(r+h)2\pi r\left(r+h\right) for r=6r=6 and h=9h=9.

\begin{align}S&=2\pi r\left(r+h\right) \\ & =2\pi\left(6\right)[\left(6\right)+\left(9\right)] \\ & =2\pi\left(6\right)\left(15\right) \\ & =180\pi\end{align}
  The surface area is 180π180\pi square inches.

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/ 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. Figure 4
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 cm2) is found to be A=(L+16)(W+16)−L⋅WA=\left(L+16\right)\left(W+16\right)-L\cdot W. Find the area of a matte for a photograph with length 32 cm and width 24 cm.

Answer: 1,152 cm2

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Try it

Click on the black dot in the graph below to explore how changing length or width changes the area of a rectangle. https://www.desmos.com/calculator/eq1cow0lcj

Simplify 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.

Recall: operations on Fractions

When simplifying algebraic expressions, we may sometimes need to add, subtract, simplify, multiply, or divide fractions. It is important to be able to do these operations on the fractions without converting them to decimals. To multiply fractions, multiply the numerators and place them over the product of the denominators.

ab⋅cd=acbd\dfrac{a}{b}\cdot\dfrac{c}{d} = \dfrac {ac}{bd}

To divide fractions, multiply the first by the reciprocal of the second.

 ab÷cd=ab⋅dc=adbc\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}

To simplify fractions, find common factors in the numerator and denominator that cancel.

 2432=2⋅2⋅2⋅32⋅2⋅2⋅2⋅2=32⋅2=34\dfrac{24}{32}=\dfrac{2\cdot2\cdot2\cdot3}{2\cdot2\cdot2\cdot2\cdot2}=\dfrac{3}{2\cdot2}=\dfrac{3}{4}

To add or subtract fractions, first rewrite each fraction as an equivalent fraction such that each has a common denominator, then add or subtract the numerators and place the result over the common denominator.

 ab±cd=ad±bcbd\dfrac{a}{b}\pm\dfrac{c}{d} = \dfrac{ad \pm bc}{bd}

 

Example: Simplifying Algebraic Expressions

Simplify each algebraic expression.
  1. 3x−2y+x−3y−73x - 2y+x - 3y - 7
  2. 2r−5(3−r)+42r - 5\left(3-r\right)+4
  3. (4t−54s)−(23t+2s)\left(4t-\dfrac{5}{4}s\right)-\left(\dfrac{2}{3}t+2s\right)
  4. 2mn−5m+3mn+n2mn - 5m+3mn+n

Answer:

  1. \begin{align}3x-2y+x-3y-7 & =3x+x-2y-3y-7 && \text{Commutative property of addition} \\ & =4x-5y-7 && \text{Simplify} \\ \text{ }\end{align}
  2. \begin{align}2r-5\left(3-r\right)+4 & =2r-15+5r+4 && \text{Distributive property}\\&=2r+5r-15+4 && \text{Commutative property of addition} \\ & =7r-11 && \text{Simplify} \\ \text{ }\end{align}
  3. \begin{align} 4t-\frac{5}{4}s -\left(\frac{2}{3}t+2s\right) &=4t-\frac{5}{4}s-\frac{2}{3}t-2s &&\text{Distributive property}\\&=4t-\frac{2}{3}t-\frac{5}{4}s-2s && \text{Commutative property of addition}\\&=\frac{12}{3}t-\frac{2}{3}t-\frac{5}{4}s-\frac{8}{4}s && \text{Common Denominators}\\ & =\frac{10}{3}t-\frac{13}{4}s && \text{Simplify} \\ \text{ }\end{align}
  4. \begin{align}mn-5m+3mn+n & =2mn+3mn-5m+n && \text{Commutative property of addition} \\ & =5mn-5m+n && \text{Simplify}\end{align}

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Example: Simplifying a Formula

A rectangle with length LL and width WW has a perimeter PP given by P=L+W+L+WP=L+W+L+W. Simplify this expression.

Answer:

\begin{align}&P=L+W+L+W \\ &P=L+L+W+W && \text{Commutative property of addition} \\ &P=2L+2W && \text{Simplify} \\ &P=2\left(L+W\right) && \text{Distributive property}\end{align}

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If the amount PP is deposited into an account paying simple interest rr for time tt, the total value of the deposit AA is given by A=P+PrtA=P+Prt. Simplify the expression. (This formula will be explored in more detail later in the course.)

Answer: A=P(1+rt)A=P\left(1+rt\right)

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