Modeling activity sheet
The modeling activity sheet I used was adapted from a colleague's pre-algebra quiz about linear equations. Since she did not have a digital copy, I just made changes in class while going through the activity. On this page I have provided all of the questions and solution methods.
REASONING WITH TWO-STEP EQUATIONS
Directions: Working with your partner, write equations using variables to describe the situations in each problem.
1. There are 8 presents on a table, and each present weighs the same amount. Combined, the presents weigh 24 pounds. How much does one present weigh? (Even if you can solve this without an algebraic phrase, use a variable to write an equation).
2. 3 of the presents on the table are replaced by 3 books. Now the total weight is 42 pounds. How much does each book weigh?
3. Jackie and Danny went shopping. Jackie bought 5 pens for $15, and Danny bought 2 pens and 6 toy cars for $48.
a) If each pen costs the same amount, how much does each pen cost?
b) How much does one toy car cost if they are all the same price?
4. 7 pumpkins, each weighing the same, weigh 49 pounds total. 4 of these pumpkins, plus 8 starfish, weigh 52 pounds. Find the weight of 1 starfish.
5. Create your own situation that can be modeled by an equation (make sure you can answer it!). Swap problems with another group and solve each other's situations.
1. Let p represent the weight of one present. Then 8p=24. When solved by dividing both sides by 8, we get p=3, so one present weighs one pound
2. From the last question, we know that the 5 remaining presents weigh 3(5)=15 pounds. So if b represents the weight of one book, we now have the equation 3b+15=42. After subtracting 15 from both sides and dividing by 3, we get b=9, so one book weighs 9 pounds.
3. a) If p represents the cost of one pen, then 5p=15. Solving this results in p=3, so one pen costs $3.
b) Using the information from part (a), Danny's 2 pens cost $6. So if c represents the cost of a car, then 6c+6=48. Solving this using subtraction and division, we get c=7, so each toy car costs $7.