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| 7 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Mark has just started working with his father at a car dealership. They sell cars and motorcycles.
Zain volunteered to work at the reception desk at the school concert.
At the end of the evening, they picked up the cash box and noticed a dollar lying on the floor next to it. Zain wonders whether the dollar belongs inside the cash box or not.
Without considering the extra dollar found on the floor, write and solve a system of equations. Does the solution make sense in this context?
(I): LHS−x=RHS−x
(II): y=47−x
(I): LHS−x=RHS−x
(II): y=47−x
(II): Distribute 4
(II): Subtract term
(II): LHS−188=RHS−188
(I): x=13
(I): Subtract term
Write and solve a system of equations.
(I): LHS⋅2.5=RHS⋅2.5
(II): Subtract (I)
Tiffaniqua is selling juice to make some money for a trip to the beach. To prepare a big jug, she used oranges and peaches.
Tiffaniqua's math teacher stopped by to buy some juice and told her the amount of carbohydrates that oranges and peaches have.
Write and solve a system of equations.
(I): LHS−x=RHS−x
(II): LHS−12x=RHS−12x
(II): LHS/8=RHS/8
(II): Write as a sum of fractions
(II): ca⋅b=ca⋅b
(II): Put minus sign in front of fraction
(II): ba=b/4a/4
(II): Calculate quotient
Finally, the point of intersection can be determined.
The lines intersect at the point with coordinates (6,7). Therefore, the solution to the system is x=6 and y=7. In the context of the situation, this means that Tiffaniqua used 6 oranges and 7 peaches to prepare a jug of juice.
Adventures on the Water is a company that organizes river safaris in Sri Lanka. They take tourists on boats along a river in the middle of the jungle for a full day.
Each boat can hold at most 8 people and can only carry 1200 pounds of weight, including passengers and gear, for safety reasons. The company assumes that, on average, an adult weighs 150 pounds and a child weighs 75 pounds. It is also assumed that each group will require 200 pounds of gear plus 10 pounds of gear per person. There are three groups who wish to take a river safari.
Write and solve a system of inequalities.
(I): LHS−x=RHS−x
(II): LHS−200=RHS−200
(II): LHS−160x=RHS−160x
(II): LHS/85=RHS/85
(II): Write as a sum of fractions
(II): Put minus sign in front of fraction
(II): ca⋅b=ca⋅b
(II): ba=b/5a/5
By following the same procedure, the region that corresponds to the second inequality can be determined.
Test Point: (1,2) | ||
---|---|---|
Inequality | Substitute | Simplify |
x+y≤8 | 1+2≤?8 | 3≤8 ✓ |
160x+85y+200≤1200 | 160(1)+85(2)+200≤?1200 | 230≤1200 ✓ |
The second inequality is also satisfied by (1,2). Therefore, the region that contains this point will be shaded.
To fully see the region that satisfies both inequalities, the unwanted regiones will be removed.
Finally, a point that represents each of the three groups will be plotted to see if they belong to the shaded area.
Group | Adults & Children | Point |
---|---|---|
A | 4 adults and 2 children | (4,2) |
B | 3 adults and 5 children | (3,5) |
C | 8 adults | (8,0) |
These points will be plotted on the coordinate plane.
The points that represent groups A and B are in the shaded area, and the point that represents group C is not in the shaded area. Therefore, A and B are the only groups that can safely take a river safari.
The challenge presented at the beginning of the lesson can also be modeled by a system of equations. It is known that Mark and his father sell cars and motorcycles.
Write and solve a system of equations.
(I): LHS−x=RHS−x
(II): y=-x+18
(II): Distribute 2
(II): Subtract term
(II): LHS−36=RHS−36
(II): LHS/2=RHS/2
(I): x=7
(I): Add terms
Let x represent the weight of the 40 % silver alloy in kilograms and y represent the weight of the 20 % silver alloy in kilograms. The sum of these two alloys should be equal to 5 kilograms. x+y=5 To create a second equation, we will first rewrite percentages as decimals. ccc Percent & → & Decimal [0.5em] 40 % &→ & 0.4 20 % &→ & 0.2 25 % &→ & 0.25 We want 5 kilograms of an alloy that contains 25 % silver. With this information in mind and using the above percentages, we can write the following equation. 0.4x+ 0.2y= 0.25(5) ⇕ 0.4x+0.2y=1.25 We can form our system of equations by combining the two equations. x+y=5 & (I) 0.4x+0.2y=1.25 & (II) Let's solve the system by using the Substitution Method. We will first isolate y in the first equation and substitute it into the second equation to find the value of x.
Next, we can substitute 54 for x in Equation (I), and solve for y.
Therefore, we need 154 kilograms of the alloy that is 40 % silver and 54 kilograms of the alloy that is 20 % silver. These values can also be expressed as decimal numbers. ccc Fraction & &Decimal [1em] 15/4 & → &3.75 [1em] 5/4 & → &1.25
To find the break even point for the scooter store, we will write a system of equations. Let s be the number of scooters sold and d be the amount of money from the sale of scooters. The store sells scooters for $100 each, so the income d can be represented by the following equation. d=100s Buying a scooter from the manufacturer costs the store $50. Additionally, the cost of operating the store each month is $2000. The expenses the store incurs each month is equal to the sum of these two costs. Since we are asked to find the break even point for the store, the sum must be equal to d. d=50s+2000 With these two equations, we can form a system of equations. d=100s & (I) d=50s+2000 & (II) If we solve this system of equations, we can find the monthly break even point for the store. To determine the number of scooters the store needs to sell, we solve for s. Let's use the Substitution Method.
The break even point is reached when 40 scooters are sold. Therefore, the store needs to sell at least 40 scooters per month to avoid loss.
Let p represent the number of pennies and q the number of quarters. We can write an equation knowing that Vincenzo wants to choose 12 coins. p+q=12 Vincenzo wants the coins to have a value of $ 2.28. Knowing that the value of a penny is $ 0.01 and that the value of a quarter is $ 0.25, we can write a second equation. 0.01p+ 0.25q= 2.28 Then we can combine these two equations we can form a system of equations. p+q=12 & (I) 0.01p+0.25q=2.28 & (II) Let's solve this system by using the Substitution Method. We will start by isolating p in Equation (I). Then, we will substitute its equivalent expression into Equation (II).
Now that we know the value of q, we can substitute it into Equation (I) to find the value of p.
Therefore, Vincenzo has to take 3 pennies and 9 quarters in order to make $ 2.28 from the 12 coins.
The Panthers football team scored 68 points during the game last night. They made one more field goal than twice the number of touchdowns. Touchdowns are worth 7 points after a one-point conversion, and field goals are worth 3 points.
We know that the Panthers scored 68 points. We also know that each touchdown t is worth 7 points, after a one-point conversion, which can be expressed as 7t. Furthermore, each field goal f is worth 3 points, which we can write as 3f. With this information, we can write our first equation. 7t+4f=68 We are also told that the number of field goals was 1 more than twice the number of touchdowns. Let's write an equation to represent this statement. f=2t+1 The two equations we wrote form the system of linear equations that corresponds to choice B. cc A. t=2f+1 7t+3f=68 & B. f=2t+1 7t+3f=68 [1.75em] C. t=2f+1 3t+7f=68 &D. f=2t+1 3t+7f=68
To find the number of touchdowns and field goals that the Panthers scored last night, we will solve the system we wrote in Part A. Since f is already isolated in the first equation, we can use the Substitution Method.
The number of touchdowns the Panthers scored is 5. Now we can substitute this number for t into the first equation and simplify to determine the value of f.
The Panthers scored 11 field goals.
LaShay is climbing down a ladder from her roof, while Tiffaniqua is climbing up a second identical ladder that is placed next to the first ladder. Each ladder has 30 rungs. Their friend Davontay recorded the following information.
Let R be the rung number from the base of the ladder and t be the time in seconds. We know that LaShay climbs down the 30-rung ladder at a speed of 2 rungs per second. At the same time, Tiffaniqua climbs up an identical ladder at a speed of 1 rung per second.
We will write an equation for each person, then form a system of equations. LaShay starts at the 30^(th) rung and steps down 2 rungs each second. With this information, we can write our first equation. R = 30 - 2t Tiffaniqua starts from the base of the ladder and climbs up 1 rung every second. Let's write a second equation. R= 1t+0 ⇔ R=t If we combine these equations, we can form a linear system that we can solve by using the Substitution Method.
Tiffaniqua and LaShay will meet on the 10^(th) rung from the base.
To write a system of equations that represents the situation, we need to first assign variables to each item. We can say that the number of hardcover books is h and the number of paperback books is p. We know that Ramsha bought 10 books in an unknown combination of hardcovers and paperbacks. This can be illustrated by the following equation. h+p=10 We also know that each hardcover book costs $ 5 and that each paperback book costs $ 3. Furthermore, Ramsha spent a total of $38. This can be shown by a second equation. 5h+3p=38 The two equations above form a system of equations. h+p=10 & (I) 5h+3p=38 & (II) We can most easily use the Substitution Method to solve this system, since we can easily isolate either variable in the first equation. Let's begin by isolating h.
Ramsha bought 6 paperback books. We can find the value of h by substituting p=6 in the first equation. However, since we are only asked how many paperback books Ramsha bought, we do not need the value of h.