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| 14 Theory slides |
| 11 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.
Vincenzo is fascinated by all things related to space and astronauts. He spends a lot of his free time reading books and watching movies about space travel, distant galaxies, and rocket science.
There are several methods for solving a system of equations. One of the most popular methods is the Substitution Method.
Solution: m=9, p=6
LHS+5m=RHS+5m
.LHS /8.=.RHS /8.
Write as a sum of fractions
a* b/c=a/c* b
Commutative Property of Addition
Draw a line through the two plotted points to get the graph of the first equation.
The lines intersect at (9,6). Therefore, m=9 and p=6, which indicates that Vincenzo spent 9 minutes spacewalking and installed 6 parts on the spaceship.
(I): LHS-4m=RHS-4m
(II): p= 42-4m
(II): Distribute 8
(II): Subtract term
(II): LHS-336=RHS-336
(II): .LHS /(- 37).=.RHS /(- 37).
(I): m= 9
(I): Multiply
(I): Subtract term
(I): l= 13
(I): Multiply
(I): Subtract term
l= 13, e= 28
Multiply
Add terms
The point of intersection lies on a lattice line where e=28. However, it can be difficult to determine the exact value of l just by looking at the graph. It can have values from 11 to 14. In Part A it was found that l is 13. The graph does support that value, so the solution is (28,13).
Multiply one of the equations by a constant so that one of the variable terms of the resulting equation is equal to or is the opposite of the corresponding variable term in the other equation. In this case, multiplying Equation (II) by - 2 will produce opposite coefficients for the y-variable. - 2 (- 2x+y)= - 2(- 11) ⇕ 4x-2y= 22 Both the original and the resulting equations have the same solutions because they are equivalent equations.
Remove parentheses
Commutative Property of Addition
Add and subtract terms
Substitute the value for the solved equation in one variable for any of the equations of the system. This produces an equivalent system of equations. In this case, Equation (I) will be replaced. & 3x+2y=6 & (I) - 2x+y=- 11 & (II) & ⇕ & x= 4 - 2x+y=- 11 Note that the first equation is the solution value of x.
Looking at the graph, the solution appears to be s=11 and n=24.
(I): LHS * 2=RHS* 2
(I): Add (II)
(I): a+(- b)=a-b
(I): Add and subtract terms
(I): .LHS /5.=.RHS /5.
Graphing Method:& s=11, n=24 Elimination Method:& s=11, n=24 Both methods resulted in the same solution, which means that they are both correct. Comparing the methods, using the Elimination Method might be a little easier and quicker than graphing the equations. This method also always results in finding the exact solution, while graphing sometimes results in finding just an estimation of the solution.
After refueling and repairing the spaceship, Vincenzo continued his way across space. His destination is a new galaxy called the Stellar Nebula.
Vincenzo used a laser measuring device on the spaceship to determine the dimensions of the galaxy. Its width w and height h are related by the following system of equations. 3w-4h=6 5h=78-w
(II): LHS+w=RHS+w
(II): LHS * 3=RHS* 3
(II): Subtract (I)
(II): Distribute - 1
(II): a-(- b)=a+b
(II): Commutative Property of Addition
(II): Add and subtract terms
(II): .LHS /19.=.RHS /19.
Equation (I) | Equation (II) | |
---|---|---|
Equation | 3w-4h=6 | 5h=78-w |
Substitute | 3( 18)-4( 12)? =6 | 5( 12)? =78- 18 |
Simplify | 6 = 6 ✓ | 60 = 60 ✓ |
The values verify both equations of the system. Therefore, the solution is correct!
Consider the given system of linear equations. Check whether the values of x and y correspond to a solution to the system.
Solve the system of linear equations to find the values of x and y.
Solving a system of equations can result in three different scenarios. One possible scenario is when a system of equations has exactly one solution. y=4x+5 7x-y=4 ⇓ x=3 y=17 The graph of this system of equations consists of two intersecting lines. The coordinates of the intersection point correspond to the solution of the system of equations.
Another possible scenario is when solving a system of equations results in an identity. y=4x+5 2y-8x=10 ⇔ y=4x+5 10=10 ✓ In this case, the system of equations has infinitely many solutions and the graph of the system is two coincidental lines.
The last possible scenario is when solving a system of equations results in a false statement. y=4x+5 2x-0.5y=3 ⇔ y=4x+5 - 2.5=3 * This means that the system of equations has no solution. The graph of this type of system of equations is two parallel lines.
These three scenarios are summarized in a table.
Number of Solutions | Graph |
---|---|
One solution | Intersecting lines |
Infinitely many solutions | Coincidental lines |
No solution | Parallel lines |
(II): LHS * 3=RHS* 3
(II): Rearrange equation
(I): Subtract ( II)
(I): Subtract term
(I): LHS+15p=RHS+15p
(I): .LHS /6.=.RHS /6.
(II): Rearrange equation
(II): .LHS /2.=.RHS /2.
(II): Commutative Property of Addition
Since the lines have the same equation, their graphs are coincidental lines. This piece of information highlights the fact that the lines have infinitely many common points. This means the system of equations has infinitely many solutions.
While exploring the new galaxy, Vincenzo and his team noticed a black hole on the edge of the galaxy. Curious, they flew closer to the black hole to register some of its characteristics.
They tried to measure the density d and mass m of the black hole and got the following system of equations. 16d=- 8m+20 m=4-2d
(I): m= 4-2d
(I): Distribute - 8
(I): Add terms
(I): LHS-16d=RHS-16d
(II): Commutative Property of Addition
(I): Rearrange equation
(I): LHS-20=RHS-20
(I): .LHS /(- 8).=.RHS /(- 8).
The lines are parallel. Since they do not intersect, there is no solution to the system of equations. Vincenzo's team was getting closer and closer to the dark hole when, suddenly, he woke up. Wow, what a cool dream he had tonight!
Vincenzo spends a lot of his free time reading books and watching movies about space travel, distant galaxies, and rocket science.
Write two equations that describe the total number of movies and books about space that Vincenzo has watched or read. Then solve the system of equations by using the Substitution Method.
(I): m= b+9
(I): Add terms
(I): LHS-9=RHS-9
(I): .LHS /2.=.RHS /2.
(II): b= 9
(II): Add terms
We are asked to find the numbers of children and adults that visited the HowItWorks Museum on a certain summer day. Let's first try to create a system of equations using the given information. Here, c is the number of children and a is the number of adults. &c→ Number of children at the museum &a→ Number of adults at the museum We are told that 590 people visited the museum that day. This means that the sum of the numbers of adults a and children c is equal to 590. a+c=590 We also know that the tickets cost $4.75 for children and $6.25 for adults. This means that the total cost of the tickets for the children is 4.75* c and the total cost of the tickets for the adults is 6.25* a. 4.75 * c = 4.75c 6.25 * a = 6.25a The cost of admission totaled $3276.50, so we can say that the sum of 4.75c and 6.25a should be equal to 3276.5. 4.75c+ 6.25a= 3276.5 We have created two equations for a and c. Together they form a system of equations. a+c=590 4.75c+6.25a=3276.5 We can solve this system using the Substitution Method. Let's begin by isolating a in the first equation. Then we can substitute the corresponding expression into the second equation.
Great! We found that c=274. Now we can substitute 274 for c into the first equation to find the value of a.
We found that a=316 and c=274. This means that 316 adults and 274 children were at the HowItWorks Museum that day!
The number of lollipops l made in m minutes by each of two machines is given by equations below.
We are given a system of equations that represents the relations between the number of lollipops l and the time m, in minutes, in which they are made by each of two machines. 180m+4l=60 l+45m=60 We will use the substitution to determine if there is a point in time when the machines will have made the same number of lollipops. Let's begin by isolating l in the second equation.
Now we can substitute l=60-45m into the first equation.
The first equation is a false statement because 240 can never be equal to 60. This means that this system of equations has no solutions. In other words, the machines will never produce the same number of lollipops in the same amount of time.
At a basketball game, a team made 53 successful shots. The shots were a combination of 1- and 2-point baskets. The team scored 97 points in total.
We want to solve the given system of equations using the Elimination Method to find the number of 1- and 2-point shots the team made. x+y = 53 x+2y=97 Recall that the Elimination Method consists of eliminating one of the variables by adding or subtracting the equations. This means that either the x- or the y-terms must cancel each other out. x+ y = 53 & (I) x+2 y=97 & (II) We can see that the x-terms will eliminate each other if we subtract Equation (I) from Equation (II).
Now we can solve for x by substituting the value of y into either equation and simplifying.
The solution to the system of equations is x=9 and y=44. This solution means that the team made 9 1-point shots and 44 2-point shots.
Train A and Train B together weigh a total of 356 tons. Train A is heavier than Train B. The difference between their weights is 42 tons.
Let a be the weight of Train A and b the weight of Train B. We know that the sum of their weights is 356 tons. Let's use this to write our first equation. a+b=356 We are also told that Train A is heavier than Train B by 42 tons. This tells us that a-b is equal to 42. Let's write our second equation! a-b=42 Next, we can gather these two equations in a system of equations to find the values of a and b. a+ b = 356 & (I) a- b=42 & (II) We can see that the b-terms will eliminate each other if we add Equation (II) to Equation (I). Let's do it!
Now we can solve for b by substituting the value of a into either equation and simplifying.
The solution to the system of equations is a=199 and b=157. Therefore, Train A weighs 199 tons and Train B weighs 157 tons.
Davontay and Emily are siblings. The sum of their ages is 42. Also, adding Davontay's age to 3 times Emily's age results in 76.
Let d be Davontay's age and e be Emily's age. We know that the sum of their ages is 42. Let's use this information to write an equation. d+e=42 We are also told that Davontay's age plus 3 times Emily's age is equal to 76. Let's use this information to write a second equation. d+3e=76 Together, the equations form the following system of equations. d+e=42 d+3e=76 Let's solve this system of equations using the Substitution Method. First, we will isolate d on one side of the equation, then substitute the corresponding expression into the other equation.
Now we can substitute 17 for e into either equation and solve for d.
The solution is d=25 and e=17. This means that Davontay is 25 years old and Emily is 17 years old.
Let's solve the same system of equations using the Elimination Method this time. d+e=42 & (I) d+3e=76 & (II) Notice that subtracting Equation (I) from Equation (II) will eliminate the variable d.
Next, substitute 17 for e into Equation (I) and solve for d.
The solution is d=25 and e=17.
Let's compare the solutions we found by using two different solution methods. rcc Substitution Method:& d=25, & e=17 Elimination Method:& d=25, & e=17 We can see that the solutions are the same. This means that both methods are correct, despite being different.