High School: Algebra

High School: Algebra

Creating Equations HSA-CED.A.3

3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional cost constraints on combinations of different foods.

Students have already translated words into algebraic equations (sometimes with more than one variable!) and have actually taken the time to solve the problem. We hate to make you the villain here, but you have to tell them their work isn't over. Students now have to interpret the results. This standard is about one thing: analysis. Well… actually it's about three things: 

  1. Creating equations/inequalities or systems of equations/inequalities
  2. Solving these equations/inequalities or system of equations
  3. Interpreting the answer properly

To analyze problems in which multiple relationships affect multiple variables, students must be able to create systems of equations, solve them, and interpret the results appropriately.

Creating Systems of Equations

In order to create systems of equations from word problems or other contexts, students need to be able to differentiate the relations and create equations for each. To support this, should already be able to create equations from word problems.

Creating equations from a word problem or similar context is a three-step translation process:

  1. Translate the equality or inequality (=, <, >, ≤, or ≥)
  2. Translate the operations (+, –, ×, ÷, xn, nx)
  3. Translate the numbers and variables

Systems of equations are identified during step one of this process. Students need to be able to read a problem and identify how many equality and inequality relations are described. Then, they should write each down separately.

Once these are written down, students perform steps two and three (translating the operations, numbers, and variables) independently for each equation. They should also simplify each equation individually before working to solve the system of equations.

Solving Systems of Equations

Solving systems of equations can be done through substitution or adding the two equations together to cancel out one of the variables. The goal is to eliminate one variable so that we can find the solution for the other and then substitute that answer back in to find the value of the second variable.

Hopefully, students already know how to solve systems of equations. After all, it's necessary in order to interpret results from a set of equations.

Interpreting Results from Systems of Equations

In many situations, students struggle to understand what an algebraic result means in the context of a word problem. This is especially true when systems of equations are involved and when they arrive at solutions that are correct algebraically, but incorrect in context.

For instance, a question about how many tops hats a giraffe can wear might produce the number 6.25 as the answer. This might make sense algebraically, but in the context of giraffes wearing top hats—how can a giraffe wear one fourth of a hat? The logical answer would then have to be 6 (although logic might not be our biggest concern if we're talking about giraffes in top hats).

Such algebraic solutions also present a challenge when multiple roots are encountered or when cancelling rational expressions. Describing what values of a variable are allowed when recording the variable information is one strategy for dealing with this problem. (For instance, we could write that giraffes only wear top hats in whole numbers.)

A common error is to report an answer based on a different variable in the problem. Recording the variable information helps to prevent such errors. If this is a common issue, students should try highlighting the quantity of interest in the problem, the matching variable name, and the eventual result. They can then check all highlighted items to ensure that they match.

Multiple variables present even more of a challenge, as students often need to find information for more than one variable in a problem. Highlighting the different information requested in different colors is one strategy. Another possibility is to have students solve for all variables before interpreting a solution; however, this becomes troublesome in more complex problems.

Interpreting results should be performed throughout the algebra curriculum. With enough practice, students will perform whatever strategies work best for them naturally.

Drills

  1. Small boxes of chocolates contain 12 pieces. Large boxes of chocolates contain 45 pieces. One chaperone buys 8 boxes of chocolate containing a combined total of 195 pieces. Which set of equalities correctly describes the chocolate purchase?

    Correct Answer:

    Small boxes and large boxes = 8 boxes of chocolate; pieces in small boxes and pieces in large boxes = combined total of 195 pieces

    Answer Explanation:

    There are two different equalities described in the problem. These need to be separated out to create two different equations. The first equality tells how many boxes of chocolate were purchased. This will be the number of small boxes and large boxes. The first equality is small boxes and large boxes = 8 boxes of chocolate. The second equality tells how many pieces of chocolate are purchased. This includes all pieces in the small boxes and all pieces in the large boxes. The second equality is pieces in small boxes and pieces in large boxes = combined total of 195 pieces.


  2. Small boxes of chocolates contain 12 pieces. Large boxes of chocolates contain 45 pieces. One chaperone buys 8 boxes of chocolate containing a combined total of 195 pieces. Which set of equations correctly describes the chocolate purchase?

    Correct Answer:

    s + g = 8; 12s + 45g = 195

    Answer Explanation:

    The relation small boxes + large boxes = 8 boxes of chocolate translates directly to s + g = 8, where s is the number of small boxes purchased and g is the number of large boxes purchased. Translating the relation pieces in small boxes and pieces in large boxes = combined total of 195 pieces is more complicated. If there are 12 pieces in each small box, there would be 12s total pieces in the small boxes. Since there are 45 pieces in each large box, there would be 45g total pieces in the large boxes. The relation pieces in small boxes and pieces in large boxes = combined total of 195 pieces creates the equation 12s + 45g = 195.


  3. Small boxes of chocolates contain 12 pieces. Large boxes of chocolates contain 45 pieces. One chaperone buys 8 boxes of chocolate containing a combined total of 195 pieces. How many small boxes and how many large boxes of chocolate did the chaperone purchase?

    Correct Answer:

    5 small boxes and 3 large boxes

    Answer Explanation:

    The system of equations s + g = 8 and 12s + 45g = 195 needs to be solved for s and for g. We can substitute g = 8 – s in the second equation to get 12s + 45(8 – s) = 195. Then, the second equation can be simplified to solve for s. The second equation expands to 12s + 360 – 45s = 195, which eventually simplifies to s = 5. Five small boxes of chocolates were purchased. The first equation g = 8 – s can then be utilized to find the number of large boxes g = 8 – 5 = 3. Three large boxes of chocolates were purchased.


  4. On the day of the field trip, each teacher must call the parents of any student who has not returned a permission slip. All of Mr. Gomez's students returned their permission slips, so he did not have to make any calls. Mrs. Hooper and Mr. Anderson had to call a total of eight parents. Mrs. Hooper needed to call two more students than Mr. Anderson. Which set of relations correctly describes the phone calls made?

    Correct Answer:

    Mrs. Hooper + Mr. Anderson = total of eight students; Mrs. Hooper = Mr. Anderson + two more

    Answer Explanation:

    The key word in the equality Mrs. Hooper and Mr. Anderson had to call = total of eight students is "and," indicating addition. This gives the relation Mrs. Hooper + Mr. Anderson = total of eight students. The key phrase in the equality Mrs. Hooper needed to call = two more students than Mr. Anderson is "more than," indicating adding an amount to one term. This gives Mrs. Hooper = Mr. Anderson + two more.


  5. On the day of the field trip, each teacher must call the parents of any student who has not returned a permission slip. All of Mr. Gomez's students returned their permission slips, so he did not have to make any calls. Mrs. Hooper and Mr. Anderson had to call a total of eight parents. Mrs. Hooper needed to call two more students than Mr. Anderson. Which set of equations correctly describes the phone calls made? (Let H = Mrs. Hooper's calls and A = Mr. Anderson's calls.)

    Correct Answer:

    H + A = 8; H = A + 2

    Answer Explanation:

    The relation Mrs. Hooper + Mr. Anderson = total of eight students translates to H + A = 8 with H = Mrs. Hooper's calls and A = Mr. Anderson's calls. The relation Mrs. Hooper = Mr. Anderson + two more becomes H = A + 2.


  6. On the day of the field trip, each teacher must call the parents of any student who has not returned a permission slip. All of Mr. Gomez's students returned their permission slips, so he did not have to make any calls. Mrs. Hooper and Mr. Anderson had to call a total of eight parents. Mrs. Hooper needed to call two more students than Mr. Anderson. How many phone calls did Mrs. Hooper make to parents?

    Correct Answer:

    5

    Answer Explanation:

    The system of equations H + A = 8 and H = A + 2 needs to be solved for H. One of the equations must be rearranged to solve for A and then substituted back into the other equation. For instance, we can say that A = H – 2 and substitute it so that H + H – 2 = 8. This will allow us to get H = 5. That means that Mrs. Hooper made 5 phone calls.


  7. Tanya and Elaine have $20 to spend making model cars. Since wheel-axle pairs cost $0.50 each, they decide they will be able to make more cars if they use three wheels instead of four. The car body and other materials cost $1.25 per car. Tanya and Elaine use the equation 3 × 0.50c + 1.25c ≤ 20 to determine how many cars (c) they will be able to make. They calculate c ≤ 7.27. How many cars will they be able to make?

    Correct Answer:

    7

    Answer Explanation:

    Tanya and Elaine can only make whole numbers of cars. Their calculation of c ≤ 7.27 means that they can only make 7 cars since they do not have enough funds to make an entire eighth car. (Of course, they could try to make 0.27 of a car, but would they do with it?)


  8. Model cars are judged based on distance traveled and how straight a path is followed. A top rating is given if the distance traveled is at least 20 m and the deviation angle is less than 5 degrees. The distance and deviation for 12 cars are shown on the graph below. How many of the cars received a top rating?

    Correct Answer:

    3

    Answer Explanation:

    The shaded region of the chart shows the area where cars receive a top rating. In this region, distance ≥ 20 and angle < 5. There are three data points in the shaded region representing the three cars that received a top rating.


  9. Claudia and Nick design a model rocket. They run a test flight and find the height of the rocket in meters is given by the formula h = (-5t + 30) × (t + 2), where t is the flight time after fuel burn out in seconds. The height is h = 0 when the rocket hits the ground. How long into the flight does the rocket hit the ground?

    Correct Answer:

    6 seconds

    Answer Explanation:

    The height of h = 0 gives (-5t + 30) × (t + 2) = 0. This is possible if -5t + 30 = 0 or if t + 2 = 0. The equation -5t + 30 = 0 gives us t = 6. The equation t + 2 = 0 gives us t = -2. Both t = 6 and t = -2 are correct solutions to the equation. However, a negative flight time does not make sense. Only the t = 6 solution for a time of 6 seconds makes sense. The rocket hits the ground 6 seconds into the flight.


  10. Rockets are judged based on the maximum height flown. Claudia's and Nick's rocket rose 60 m while fuel burned during launch and then followed a parabolic path described by the equation h = (-5t + 30) × (t + 2), where t is the flight time after fuel burn out in seconds. Rockets receive a top rating if they fly higher than 85 m. Does Claudia and Nick's rocket receive top rating?

    Correct Answer:

    No

    Answer Explanation:

    In order to determine the maximum value of h for the equation, we can either graph or solve it algebraically. If we graph if, we will see that all points on the graph are below h = 85. We can also solve 85 = (-5t + 30) × (t + 2). This gives -5t2 + 20t + 60 = 85 which reduces to t2 – 4t + 5 = 0. There are no real solutions for t. Since there is no time when the rocket reaches 85 m, it does not earn top rating.


More standards from High School: Algebra - Creating Equations

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