Problem 3.1 Solving Equations Using Tables and Graphs
A. Use the equation A = 5 + 0.5d.
1. Suppose Alana walks 23 kilometers. Show how you can use a table and a graph to find the amount of money Alana gets from each sponsor.
In the table, you can find 23 in the “d” column (independent variable) and read across to the right to find out the amount of money in the “A” column (dependent variable).
In the graph, you find 23 on the x-axis and follow the grid up to find the point on the line. Then you follow the grid across to see what the amount of money is on the y-axis.
2. Suppose Alana receives $60 from a sponsor. Show how you can use a table and a graph to find the number of kilometers she walks.
In the table you can find 60 in the “A” column (dependent variable), then read across to the left to find out how may km she walked in the “d” column(independent variable).
In the graph, you find $60 on the y-axis and follow the grid across to find the point on the line. Then you follow the grid down to see how many km Alana walked on the x-axis.
B. For each equation:
• Tell what information Alana is looking for.
• Describe how you can find the information.
1. A = 5 + 0.5(15)
Alana is trying to figure out how much money she would earn from a sponsor if she walked 15 kilometers. You can figure this out by multiplying 0.5*15, then adding 5 to that product. You could also use the graph or table as described in the previous problem to find the missing “A” value. In this case, she would earn $12.50.
2. 50 = 5 + 0.5d
Alana is trying to figure out how many kilometers she needs to walk in order to earn $50. You can figure this out by using the table or graph (find 50 in the “A” column), or by solving the equation to find “d”. You could do 50-5=45, then divide 45÷0.5=90. So, Alana would need to walk 90 kilometers to earn $50.
C. The following equations are related to situations that you have explored. Find the solution (the value of the variable) for each equation. Then, describe another way you can find the solution.
1. D = 25 + 2.5(7)
D=25+17.5 You can solve the equation as shown or use the
2. 70 = 25 + 2.5t
-25 -25 You can solve the equation as shown, use the table or
45 = 2.5t graph, or use guess and check to see what number for t
2.5 2.5 makes the equation true.
18 = t
Problem 3.2 Exploring Equality
A. Sarah draws the following picture. Each pouch contains the same number of $1 gold coins. How many gold coins are in each pouch? Explain your reasoning.
I cancelled 4 coins from each, then divided the remaining coins into 3 equal groups to figure out that there are 2 coins in each pouch.
B. For each situation, find the number of gold coins in the pouch. Write down your steps so that someone else could follow your steps to find the same number of coins in a pouch.
I cancelled out 3 coins from each side, then I divided the remaining coins into 3 equal groups (green, yellow, blue) because there were 3 pouches. I figured out that there are 9 coins in each pouch.
I cancelled two groups of two coins from both sides, and then I divided the remaining coins into two equal groups because there were 2 pouches. I figured out that there are 4 coins in each pouch.
I cancelled two pouches from each side and then saw that there was one pouch left, so there are 12 coins in each pouch.
I cancelled two pouches from each side and then cancelled 3 coins from each side. All of the remaining coins are in the 1 remaining pouch, so I figured out that there are 9 coins in each pouch.
First, I cancelled out two pouches from each side, and then I cancelled out three coins from each side. I split the remaining coins into three equal groups because there are three bags remaining. I found out that there are 6 coins in each pouch.
C. Describe how you can check your answer. That is, how do you know you found the correct number of gold coins in each pouch?
You can use your final answer of the number of coins in each pouch to “plug in” to the original picture. Find the total number of coins on each side and make sure that both sides have the same number of coins.
D. Describe how you maintained equality at each step of your solutions in Questions A and B.
To maintain equality, you have to do the same operation on both sides of the equation. (Add, Subtract, Multiply, or Divide by the same number.)
Problem 3.3 Writing Equations
A. For each situation:
• Represent the situation with an equation. Use an x to represent the number of gold coins in each pouch and a number to represent the number of coins on each side.
• Use the equation to find the number of gold coins in each pouch.
4x = 16
x = 4
3x = 6 see left hand column solving
x = 2
3x = 2x +6
1x = 6 or x = 6
6x = 4 + 4x
2x = 4
x = 2
B. For each equation:
• Use your ideas from Question A to solve the equation.
• Check your answer.
C. Describe a general method for solving equations using what you know about equality.
When solving equations (in order to maintain equality) whatever you do to one side of the equation, you must do to the other side of the equation.
F. 1. Describe how you could use a graph or table to solve the equation
5x + 10 = -20.
Find the value -20 on the y-axis and follow it to the graphed line to see what x-value is associated with that coordinate.
2. Suppose you use a different letter or symbol to represent the value of the unknown variable. For example, 5n + 10 = 6n instead of 5x + 10 = 6x. Does this make a difference in solving the equation? Explain.
It does not make a difference what letter we use to represent the variable.
Problem 3.5 Finding the Point of Intersection
At Fabulous Fabian’s Bakery, the expenses E to make n cakes per month is
given by the equation E = 825 + 3.25n.
The income I for selling n cakes is given by the equation I = 8.20n.
A. In the equations for I and E, what information do the y-intercepts represent?
In the equation for income, the y-intercept is 0, which means that they have zero income if they sell zero cakes.
In the equation for Expenses, the y-intercept is 825, which means they have an expense of $825 if they sell zero cakes.
What about the coefficients of n?
In the equation for income, the coefficient is 8.20, which means that they make $820 for each cake they sell.
In the equation for expenses, their expense increases by $3.25 for each cake they make.
B. Fabian sells 100 cakes in January.
1. What are his expenses and his income?
2. Does he make a profit? Describe how you found your answer.
No, he does not make a profit because his expenses were more than his income.
C. In April, Fabian’s expenses are $5,700.
1. How many cakes does he sell?
1500 cakes. and solve for x.
2. What is the income for producing this number of cakes?
$12,300 is the income. To find the income, substitute 1,500 into the equation.
3. Does he make a profit? Explain.
Yes, he makes $6,600 since $12,300—$5,700=$6,600.
D. The break-even point is when expenses equal income (E = I). Fabian thinks that
this information is useful.
1. Describe how you can find Fabian’s break-even point symbolically.
See work on board
Find the break-even point.
Between 166 and 167 cakes the bakery will break even.
2. Describe another method for finding the break-even point.
You could use the graph of the two equations and see where the graphs intersect to get the break-even point.