- Last updated

- Save as PDF

- Page ID
- 22474

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vectorC}[1]{\textbf{#1}}\)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

In this section we will evaluate *algebraic expressions* for given values of the variables contained in the expressions. Here are some simple tips to help you be successful.

## Tips for Evaluating Algebraic Expressions

- Replace all occurrences of variables in the expression with open parentheses. Leave room between the parentheses to substitute the given value of the variable.
- Substitute the given values of variables in the open parentheses prepared in the first step.
- Evaluate the resulting expression according to the Rules Guiding Order of Operations.

Let's begin with an example.

Example 1

Evaluate the expression \(x^2 − 2xy + y^2\) at \(x = −3\) and \(y = 2\).

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression *x*^{2} − 2*xy* + *y*^{2} with open parentheses.

\[ x^2 -2xy + y^2 = ( ~ )^2 -2(~)(~) + ( ~ )^2 \nonumber\nonumber \]

Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.

\[ \begin{aligned} x^2 -2xy + y^2 ~ & \textcolor{red}{ \text{ Original expression.}} \\ =( \textcolor{red}{-3} )^2 -2 ( \textcolor{red}{-3})( \textcolor{red}{2}) + (\textcolor{red}{2})^2 ~ & \textcolor{red}{ \text{ Substitute } -3 \text{ for } x \text{and 2 for }y.} \\ =9-2(-3)(2)+4 ~ & \textcolor{red}{ \text{ Evaluate exponents first.}} \\ = 9-(-6)(2)+4 ~ & \textcolor{red}{ \text{ Left to right, multiply } 2(-3)=-6.} \\ =9-(-12)+4 ~ & \textcolor{red}{ \text{ Left to right, multiply: } (-6)(2) = -12.} \\ = 9 + 12 + 4 ~ & \textcolor{red}{ \text{ Add the opposite.}} \\ = 25 ~ & \textcolor{red}{ \text{ Add.}} \end{aligned}\nonumber \]

Exercise

If *x* = −2 and *y* = −1, evaluate *x*^{3} − *y*^{3}.

**Answer**-
−7

Example 2

Evaluate the expression (*a* − *b*)^{2} If *a* = 3 and *b* = −5, at *a* = 3 and *b* = −5.

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression (*a* − *b*)^{2} with open parentheses.

\[ (a-b)^2 = (()-())^2\nonumber \]

Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.

\[ \begin{aligned} (a-b)^2 = (( \textcolor{red}{3})-( \textcolor{red}{-5}))^2 ~ & \textcolor{red}{ \text{ Substitute 3 for } a \text{ and } -5 \text{ for } b.} \\ = (3+5)^2 ~ & \textcolor{red}{ \text{ Add the opposite: } (3)-(-5)=3+5} \\ = 8^2 ~ & \textcolor{red}{ \text{ Simplify inside parentheses: } 3+5 = 8} \\ =64 ~ & \textcolor{red}{ \text{ Evaluate exponent: } 8^2 = 64} \end{aligned}\nonumber \]

Exercise

If *a* = 3 and *b* = −5, evaluate *a*^{2} − *b*^{2}.

**Answer**-
−16

Example 3

Evaluate the expression |a|−|b| at a = 5 and b = −7.

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression |a|−|b| with open parentheses.

\[ |a| - |b| = |( ~ )| - |( ~ )|\nonumber \]

Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.

\[ \begin{aligned} |a| - |b| = |( \textcolor{red}{5} )| = |( \textcolor{red}{-7})| ~ & \textcolor{red}{ \text{ Substitute 5 for } a \text{ and } -7 \text{ for } b.} \\ = 5 - 7 ~ & \textcolor{red}{ \text{ Absolute values first: } |(5)| = 5 \text{ and } |(-7)|=7|} \\ =5+(-7) ~ & \textcolor{red}{ \text{ Add the opposites: } 5 - 7 = 5+(-7).} \\ =-2 ~ & \textcolor{red}{ \text{ Add: } 5+(-7)=-2.} \end{aligned}\nonumber \]

Exercise

If *a* = 5 and *b* = −7, evaluate 2|*a*| − 3|*b*|.

**Answer**-
−11

Example 4

Evaluate the expression |*a* − *b*| at *a* = 5 and *b* = −7.

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression |*a* − *b*| with open parentheses.

\[ |a-b| = |(~)-(~)|\nonumber \]

\[ \begin{aligned} |a-b| = |( \textcolor{red}{5})-( \textcolor{red}{-7})| ~ & \textcolor{red}{ \text{ Substitute 5 for } a \text{ and } -7 \text{ for } b.} \\ = |5+7| ~ & \textcolor{red}{ \text{ Add the opposite: } 5-(-7)=5+7.} \\ =|12| ~ & \textcolor{red}{ \text{ Add: } 5+7=12.} \\ =12 ~ & \textcolor{red}{ \text{ Take the absolute value: } |12| = 12.} \end{aligned}\nonumber \]

Exercise

If *a* = 5 and *b* = −7, evaluate |2*a* − 3*b*|.

**Answer**-
31

Example 5

Evaluate the expression

\[ \frac{ad-bc}{a+b}\nonumber \]

at *a* = 5, *b* = −3, *c* = 2, and *d* = −4.

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of variables in the expression with open parentheses.

\[ \frac{ad-bc}{a+b} = \frac{(~)(~)-(~)(~)}{(~)+(~)}\nonumber \]

\[ \begin{aligned} \frac{ad-bc}{a+b} = \frac{( \textcolor{red}{5}) -( \textcolor{red}{-3}) ( \textcolor{red}{2})}{( \textcolor{red}{5}) + ( \textcolor{red}{-3})} ~ & \textcolor{red}{ \text{ Substitute: } 5 \text{ for } a,~ -3 \text{ for } b,~ 2 \text{ for } c,~ -4 \text{ for } d.} \\ = \frac{-20-(-6)}{2} ~ & \begin{aligned} \textcolor{red}{ \text{ Numerator: } (5)(=4)=-20,~ (-3)(2) = -6.} \\ \textcolor{red}{ \text{ Denominator: } 5+(-3)=2.} \end{aligned} \\ = \frac{-20+6}{2} ~ & \textcolor{red}{ \text{ Numerator: Add the opposite.}} \\ = \frac{-14}{2} ~ & \textcolor{red}{ \text{ Numerator: } -20+6=-14.} \\ = -7 ~ & \textcolor{red}{ \text{Divide.}} \end{aligned}\nonumber \]

Exercise

If a = −7, b = −3, c = −15, 15, and *d* = −14, evaluate:

\[\frac{a^2+b^2}{c+d}\nonumber \]

**Answer**-
−2

Example 6

Pictured below is a rectangular prism.

The volume of the rectangular prism is given by the formula

\[V=LWH,\nonumber \]

where *L* is the length, *W* is the width, and *H* is the height of the rectangular prism. Find the volume of a rectangular prism having length 12 feet, width 4 feet, and height 6 feet.

**Solution**

Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of of L, W, and H in the formula

\[ V = LWH\nonumber \]

with open parentheses.

\[V = (~)(~)(~)\nonumber \]

Next, substitute 12 ft for *L*, 4 ft for *W*, and 6 ft for H and simplify.

\[ \begin{aligned} V = (12 \text{ft})(4 \text{ft})(6 \text{ft}) \\ = 288 \text{ft}^3 \end{aligned}\nonumber \]

Hence, the volume of the rectangular prism is 288 cubic feet.

Exercise

The surface area of the prism pictured in this example is given by the following formula:

\[S = 2(W H + LH + LW) \nonumber \]

If *L* = 12, *W* = 4, and *H* = 6 feet, respectively, calculate the surface area.

**Answer**-
288 square feet

## Exercises

In Exercises 1-12, evaluate the expression at the given value of x.

1. −3x^{2} − 6x + 3 at x = 7

2. 7x^{2} − 7x + 1 at x = −8

3. −6x − 6 at x = 3

4. 6x − 1 at x = −10

5. 5x^{2} + 2x + 4 at x = −1

6. 4x^{2} − 9x + 4 at x = −3

7. −9x − 5 at x = −2

8. −9x + 12 at x = 5

9. 4x^{2} + 2x + 6 at x = −6

10. −3x^{2} + 7x + 4 at x = −7

11. 12x + 10 at x = −12

12. −6x + 7 at x = 11

In Exercises 13-28, evaluate the expression at the given values of x and y.

13. |x|−|y| at x = −5 and y = 4

14. |x|−|y| at x = −1 and y = −2

15. −5x^{2} + 2y^{2} at x = 4 and y = 2

16. −5x^{2} − 4y^{2} at x = −2 and y = −5

17. |x|−|y| at x = 0 and y = 2

18. |x|−|y| at x = −2 and y = 0

19. |x − y| at x = 4 and y = 5

20. |x − y| at x = −1 and y = −4

21. 5x^{2} − 4xy + 3y^{2} at x = 1 and y = −4

22. 3x^{2} + 5xy + 3y^{2} at x = 2 and y = −1

23. |x − y| at x = 4 and y = 4

24. |x − y| at x = 3 and y = −5

25. −5x^{2} − 3xy + 5y^{2} at x = −1 and y = −2

26. 3x^{2} − 2xy − 5y^{2} at x = 2 and y = 5

27. 5x^{2} + 4y^{2} at x = −2 and y = −2

28. −4x^{2} + 2y^{2} at x = 4 and y = −5

In Exercises 29-40, evaluate the expression at the given value of x.

29. \( \frac{9+9x}{−x}\) at x = −3

30. \( \frac{9 − 2x}{−x}\) at x = −1

31. \(\frac{−8x + 9}{−9 + x}\) at x = 10

32. \(\frac{2x + 4}{1 + x}\) at x = 0

33. \(\frac{−4+9x}{7x}\) at x = 2

34. \(\frac{−1 − 9x}{x}\) at x = −1

35. \(\frac{−12 − 7x}{x}\) at x = −1

36. \(\frac{12 + 11x}{3x}\) at x = −6

37. \(\frac{6x − 10}{5}\) + x at x = −6

38. \(\frac{11x + 11}{−4}\) + x at x = 5

39. \(\frac{10x + 11}{5}\) + x at x = −4

40. \(\frac{6x + 12}{−3}\) + x at x = 2

41. The formula

\[d=16t^2\nonumber \]

gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 4 seconds.

42. The formula

\[d = 16t^2\nonumber \]

gives the distance (in feet) that an object falls from rest in terms of the time t that has elapsed since its release. Find the distance d (in feet) that an object falls in t = 24 seconds.

43. The formula

\[C = \frac{5(F − 32)}{9}\nonumber \]

gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature (◦ C) if the Fahrenheit temperature is F = 230◦ F.

44. The formula

\[C = \frac{5(F − 32)}{9}\nonumber \]

gives the Celcius temperature C in terms of the Fahrenheit temperature F. Use the formula to find the Celsius temperature (^{◦}C) if the Fahrenheit temperature is F = 95 ^{◦}F.

45. The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0^{◦} K, the temperature at which molecules have zero kinetic energy. Water freezes at 273^{◦} K and boils at K = 373^{◦} K. To change Kelvin temperature to Fahrenheit temperature, we use the formula

\[F = \frac{9(K − 273)}{5} + 32.\nonumber \]

Use the formula to change 28^{◦ }K to Fahrenheit.

46. The Kelvin scale of temperature is used in chemistry and physics. Absolute zero occurs at 0^{◦} K, the temperature at which molecules have zero kinetic energy. Water freezes at 273^{◦} K and boils at K = 373^{◦} K. To change Kelvin temperature to Fahrenheit temperature, we use the formula

\[F = \frac{9(K − 273)}{5} + 32.\nonumber \]

Use the formula to change 248^{◦} K to Fahrenheit.

47. A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula

\[v = v0 − gt,\nonumber \]

where v_{0} is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v_{0} = 272 feet per second, find the speed of the ball after t = 6 seconds.

48. A ball is thrown vertically upward. Its velocity t seconds after its release is given by the formula

\[v = v_0 − gt,\nonumber \]

where v_{0} is its initial velocity, g is the acceleration due to gravity, and v is the velocity of the ball at time t. The acceleration due to gravity is g = 32 feet per second per second. If the initial velocity of the ball is v_{0} = 470 feet per second, find the speed of the ball after t = 4 seconds.

49. **Even numbers**. Evaluate the expression 2n for the following values:

i) n = 1

ii) n = 2

iii) n = 3

iv) n = −4

v) n = −5

vi) Is the result always an even number? Explain.

50. **Odd numbers**. Evaluate the expression 2n + 1 for the following values:

i) n = 1

ii) n = 2

iii) n = 3

iv) n = −4

v) n = −5

vi) Is the result always an odd number? Explain.

## Answers

1. −186

3. −24

5. 7

7. 13

9. 138

11. −134

13. 1

15. −72

17. −2

19. 1

21. 69

23. 0

25. 9

27. 36

29. −6

31. −71

33. 1

35. 5

37. 46

39. −29

41. 256 feet

43. 110 degrees

45. −409^{◦} F

47. 80 feet per second

49.

i) 2

ii) 4

iii) 6

iv) −8

v) −10

vi) Yes, the result will always be an even number because 2 will always be a factor of the product 2n.