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 x2 − 2xy + y2 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 x3 − y3.
- 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 a2 − b2.
- 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.
\[ |a-b| = |(~)-(~)|\nonumber \]
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
\[ \begin{aligned} |a-b| = |( \textcolor{red}{5})-( \textcolor{red}{-7})| ~ & \textcolor{red}{ \text{ Substitute 5 for } a \text{ and } -7 \text{ for } b.} \\ = |5+7| ~ & \textcolor{red}{ \text{ Add the opposite: } 5-(-7)=5+7.} \\ =|12| ~ & \textcolor{red}{ \text{ Add: } 5+7=12.} \\ =12 ~ & \textcolor{red}{ \text{ Take the absolute value: } |12| = 12.} \end{aligned}\nonumber \]
Exercise
If a = 5 and b = −7, evaluate |2a − 3b|.
- 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.
\[ \frac{ad-bc}{a+b} = \frac{(~)(~)-(~)(~)}{(~)+(~)}\nonumber \]
Secondly, replace each variable with its given value, and thirdly, follow the “Rules Guiding Order of Operations” to evaluate the resulting expression.
\[ \begin{aligned} \frac{ad-bc}{a+b} = \frac{( \textcolor{red}{5}) -( \textcolor{red}{-3}) ( \textcolor{red}{2})}{( \textcolor{red}{5}) + ( \textcolor{red}{-3})} ~ & \textcolor{red}{ \text{ Substitute: } 5 \text{ for } a,~ -3 \text{ for } b,~ 2 \text{ for } c,~ -4 \text{ for } d.} \\ = \frac{-20-(-6)}{2} ~ & \begin{aligned} \textcolor{red}{ \text{ Numerator: } (5)(=4)=-20,~ (-3)(2) = -6.} \\ \textcolor{red}{ \text{ Denominator: } 5+(-3)=2.} \end{aligned} \\ = \frac{-20+6}{2} ~ & \textcolor{red}{ \text{ Numerator: Add the opposite.}} \\ = \frac{-14}{2} ~ & \textcolor{red}{ \text{ Numerator: } -20+6=-14.} \\ = -7 ~ & \textcolor{red}{ \text{Divide.}} \end{aligned}\nonumber \]
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.
![Screen Shot 2019-08-19 at 6.09.16 PM.png](https://math.libretexts.org/@api/deki/files/27859/Screen_Shot_2019-08-19_at_6.09.16_PM.png?revision=1)
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. −3x2 − 6x + 3 at x = 7
2. 7x2 − 7x + 1 at x = −8
3. −6x − 6 at x = 3
4. 6x − 1 at x = −10
5. 5x2 + 2x + 4 at x = −1
6. 4x2 − 9x + 4 at x = −3
7. −9x − 5 at x = −2
8. −9x + 12 at x = 5
9. 4x2 + 2x + 6 at x = −6
10. −3x2 + 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. −5x2 + 2y2 at x = 4 and y = 2
16. −5x2 − 4y2 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. 5x2 − 4xy + 3y2 at x = 1 and y = −4
22. 3x2 + 5xy + 3y2 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. −5x2 − 3xy + 5y2 at x = −1 and y = −2
26. 3x2 − 2xy − 5y2 at x = 2 and y = 5
27. 5x2 + 4y2 at x = −2 and y = −2
28. −4x2 + 2y2 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 v0 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 v0 = 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 v0 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 v0 = 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.