# High School: Algebra

### Creating Equations HSA-CED.A.1

1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

Students should be able to interpret word problems and form equations and inequalities in order to solve the problem. That means translating a word problem to an algebraic equation.

Let's be real, here. Math is another language, just like Spanish, Japanese, or Icelandic. When you start learning a language, you don't start by translating words like "absquatulate" or "loquacious" or "pneumonoultramicroscopicsilicovolcanoconiosis" (and yes, that is a real word).

It's better to start easier, with words like "cat" and "girl" and slowly work your way up. Just the same, if you use simple linear equations that are familiar to students, they can focus on the translation process and it'll all go a lot smoother.

Translation is a useful analogy in and of itself because it emphasizes that the algebraic equation is the same as the word problem, just presented in a different way. In addition to helping students to understand the process, the translation analogy can also help reassure struggling learners and encourage practice.

After they've gotten a hang of the basics, students can start learning quadratic, rational, and exponential functions to address all aspects of this standard. Once students are familiar with these operations individually, they should be asked to distinguish them from each another.

As students gain experience, there are additional strategies that should be introduced. One experienced problem solver strategy is to read the question twice before beginning. It's a useful piece of advice in general, actually.

Writing a list of what is known and a list of what needs to be calculated is also an excellent strategy. Such lists are especially useful when sorting out unnecessary information, identifying an appropriate formula to utilize, or constructing a proof. These strategies should be suggested and shown to students after they are proficient with the basic translation process.

To start off, the chart below may be presented as a dictionary to support word to symbol translation. Students can also add to the chart as they find other key words or phrases.

 Algebra Symbol Key Words = equals allequalsgivesis, are, was, were, will beresultssameyields < is less than belowless than ≤ is less than or equal to maximum ofnot more than > is greater than greater thanmore thanover ≥ is greater than or equal to at leastminimum ofnot less than + addition addandcombineincreasemoreplusraisesumtogethertotal – subtraction decreasedifferencefewerlessloseminusreduce × multiplication directly proportionaldouble(× 2), triple(× 3), etc.group oflinearmultipliedproducttimes / division averagecutdivided by/intoeachinversely proportionalout ofperpiecesquotientratiosharesplit xn power powersquare (n = 2), cube (n = 3), etc. nx exponential decaysdoubles (n = 2), triples (n = 3), quadruples (n = 4), etc.growsrate of n per x

If you needed another word problem example or video to show your students, here is one such example:

#### Drills

1. There are 60 students going on a field trip to the chocolate factory. The students are from three different classes. Mrs. Hooper's class has 24 students and Mr. Gomez's class has 18 students. Which of the equalities correctly describes the students and could be used to solve for how many students are from Mr. Anderson's class? (Let A = the number of students in Mr. Anderson's class.)

24 + 18 + A = 60

The relation Hooper's class + Gomez's class + Anderson's class = 60 students going on a field trip becomes an equation by changing the written descriptions into numbers and variables. Mrs. Hooper's class has 24 students and Mr. Gomez's class has 18 students giving 24 + 18 + Anderson's class = 60 students going on a field trip. The number of students in Anderson's class is the unknown and must be represented by a variable like A for Anderson. That means 24 + 18 + A = 60.

2. There are 60 students going on a field trip to the chocolate factory. The students are from three different classes. Mrs. Hooper's class has 24 students and Mr. Gomez's class has 18 students. How many students are from Mr. Anderson's class?

18 students

The equation 24 + 18 + A = 60 (where A = students in Anderson's class) needs to be solved to isolate A. First, simplify 24 + 18 to 42 to get 42 + A = 60. Then, subtract 42 from both sides to get A = 60 – 42. This gives A = 18. There are 18 students from Anderson's class.

3. There are six chaperones going on a field trip. There are two buses for the trip. The chaperones divide so there is the same number of chaperones on each bus. Which of the equations could be utilized to find the number of chaperones on the first bus? (Let c = number of chaperones on first bus.)

The relation becomes an equation by changing the written descriptions into numbers and variables. There are six chaperones on the trip giving . There are two buses, giving . The number of chaperones on the first bus is an unknown represented by the variable c. This results in the equation .

4. There are six chaperones going on a field trip. There are two buses for the trip. The chaperones divide so there is the same number of chaperones on each bus. How many chaperones are there on the first bus?

3

The equation  = c (where c = number of chaperones on first bus) needs to be solved for the variable c. This gives 3 = c or c = 3. There are 3 chaperones on the first bus.

5. A total of 66 people attended a field trip to a chocolate factory for a tour. A maximum of 15 people are allowed to tour at one time. Which equation correctly describes how many tour groups to organize? (Let g = the number of groups.)

The relation  becomes an equation by changing the written descriptions into numbers and variables. The 66 and 15 become numbers directly giving . The number of tour groups is an unknown so we need to make a variable g = number of tour groups. The resulting equation is .

6. A total of 66 people attended a field trip to a chocolate factory for a tour. A maximum of 15 people are allowed to tour at one time. What is the minimum number of tour groups that can be formed?

5

The equation  (where g = number of tour groups) needs to be solved for the variable g. Multiplying each side by g to remove it from the denominator gives 66 ≤ 15 × g. Dividing each side by 15 then gives  which becomes 4.4 ≤ g. Reversing this gives g ≥ 4.4. The actual number of tour groups formed must be a whole number since there cannot be fractions of tours. The smallest whole number that is greater than or equal to 4.4 is 5. A minimum of 5 tour groups must be formed.

7. A heart shaped chocolate box is composed of one square and two half circles. The total number of chocolates in the box is calculated by adding the area of a square given by 4x2 and the area of a circle approximated by 3x2. The company plans to add a small additional box for a promotional campaign containing one row (2x) of chocolates. If the total combined heart shape and small box contain 69 chocolates, which of these equations could be utilized to solve for the number of chocolates in the small box (2x)?

4x2 + 3x2 + 2x = 69

Begin with the relation: combine heart shape and small box = 69 chocolates total. The key words "and, "adding," and "add" all indicate summing the three relations for numbers of chocolates. This gives 4x2 + 3x2 + 2x. Combining these gives the equation 4x2 + 3x2 + 2x = 69.

8. A heart shaped chocolate box is composed of one square and two half circles. The total number of chocolates in the box is calculated by adding the area of a square given by 4x2 and the area of a circle approximated by 3x2. The company plans to add a small additional box for a promotional campaign containing one row (2x) of chocolates. If the total combined heart shape and small box contain 69 chocolates, how many chocolates are in the small box (2x)?

6

The equation 4x2 + 3x2 + 2x = 69 can be solved for 2x, the number of chocolates in the small box. This gives 7x2 + 2x – 69 = 0 which is factored to (7x + 23)(x – 3) = 0. The solutions are and x = 3. These give 2x = -6.572 and 2x = 6, respectively. The negative solution does not make sense because you cannot have a negative number of chocolates in a box. So, the correct solution must be 2x = 6. There are 6 chocolates in the small box.

9. On the day of the class field trip, the chocolate factory produced three times as many plain chocolate bars as crispy bars. They produced 50 more nutty bars than crispy bars. The ratio of plain chocolate bars produced to nutty bars produced was 2 to 1. Which of the equations below could be utilized to solve for the number of crispy bars produced on the day of the field trip?

The key word "ratio" indicates division such that . Then, you need number of plain chocolate bars in terms of crispy bars, which is 3c from the key word "times." And, you need number of nutty bars in terms of crispy bars, which is c + 50 from the key word "more." Combining these gives the equation .

10. On the day of the class field trip, the chocolate factory produced three times as many plain chocolate bars as crispy bars. They produced 50 more nutty bars than crispy bars. The ratio of plain chocolate bars produced to nutty bars produced is 2 to 1. How many crispy bars were produced?

100

The equation  equates the 2 to 1 ratio with 3c (the number of plain chocolate bars) to c + 50 (the number of nutty bars). To solve for c, the number of crispy bars, first multiply both sides by c + 50 to get 3c = 2c + 100. Then, subtract 2c from both sides to get c = 100. So, 100 crispy bars were produced.

11. A large box of 144 chocolates has a width that is three times the height of the box and a length that is twice the width of the boxes. Each chocolate rests in a cube that is 1 in × 1 in × 1 in. Which equation could be utilized to calculate the height of the box in inches?

(2 × 3h) × 3h × h = 144

The initial relation for this problem requires knowing that the volume of the box will be length × width × height. Since each chocolate has a volume of 1 in3, the total volume of the box with 144 chocolates will be 144 in3. So, the basic relationship is length × width × height = 144. We do not know length, but we do know that it is twice the width. And, we know the width is three times the height. The key words "twice" and "times" both indicate multiplication so we have width = 3 × height = 3h and length = 2 × width = 2 × 3h. And the height is, of course, just h. So, the overall equation becomes (2 × 3h) × 3h × h = 144.

12. A large box of 144 chocolates has a width that is three times the height of the box and a length that is twice the width of the boxes. Each chocolate rests in a cube that is 1 in × 1 in × 1 in. What is the height of the box in inches?

2 in

Given the relation length × width × height = volume, the resulting equation is (2 × 3h) × 3h × h = 144. This simplifies to 18h3 = 144. First, each side is divided by 18 to get h3 = 8. Then, taking the cubed root of both sides gives h = 2. The height of the box is 2 in.

13. The candy company's total revenue this year was 1.1 times the revenue last year. If they experience the same growth every year, what equation describes how many years will it take before the revenue is more than double?

1.1y ≥ 2

The basic relation here is for year revenue to be at least initial revenue. That is yearly revenue ≥ 2 × initial revenue. But there is no information given about the total initial revenue or the total year revenue. Instead, we know that the first year the revenue will be 1.1 × initial revenue. The second year, it will be 1.1 × first year revenue, or 1.1 × 1.1 × initial revenue, or 1.12 × initial revenue. For any given year, the revenue is 1.1y × initial revenue, where y is the number of years. Going back into the inequality gives 1.1y × initial revenue ≥ 2 × initial revenue. Canceling the initial revenue from both sides gives the equation 1.1y ≥ 2.

14. The candy company's total revenue this year was 1.1 times the sales last year. If they experience the same growth every year, how many years will it take before the revenue is more than double?

8