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| 11 Theory slides |
| 8 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.
Maya is researching a toy factory that currently produces robot action figures and car toys.
Sometimes the equations of a system can be rewritten in a way that is easier to work with. As an example, consider the following system of linear equations. x + 2y = 3 & (I) 12x - y = - 12 & (II) In Equation (II), the coefficient of the variable x is a fraction. It would be easier to solve the system if all the coefficients were integers. It is possible to rewrite the system using equivalent systems.
In a system of equations, an equivalent system can be created by replacing one equation with the sum of two or more equations in the system or by replacing an equation with a multiple of itself. An equation can be also replaced by the sum of that equation and a multiple of another equation in the system.
The statement will be proved for linear equations. For non-linear equations, the proof is similar. Consider a system of linear equations. ax+by=c & (I) dx+ey=f & (II) Let ( x_1, y_1) be a solution to the system. Therefore, this ordered pair satisfies both equations simultaneously. a x_1+b y_1=c ✓ d x_1+e y_1=f ✓ Consider now the system formed by the following two equations.
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.
For each system of linear equations, verify whether the coordinate pair is a solution.
A vlogger that Diego likes to watch bought silver and gold to make Olympic-style medals. The vlogger will show the process of making the medals as a multi-video series.
The price of gold is about $2000 per ounce, while the price of of silver is about $25 per ounce. The vlogger spent a total of $7075 in both silver and gold. Let g and s be the ounces of gold and silver, respectively, that the vlogger bought. Then, the given information can be modeled by a linear equation. 25s + 2000g = 7075 Diego notes that if he subtracts 50 times the number of ounces of gold from 5 times the number of ounces of silver, there would still be 65 ounces left. This can also be written as an equation. 5s - 50g = 65 A system of equations is obtained by combining these two equations. 25s + 2000g = 7075 & (I) 5s - 50g = 65 & (II)
(I): .LHS /25.=.RHS /25.
(I): Write as a sum of fractions
(II): .LHS /5.=.RHS /5.
(II): Write as a difference of fractions
(I), (II): a* b/c=a/c* b
(I), (II): Calculate quotient
(I), (II): Identity Property of Multiplication
(I): LHS-80g=RHS-80g
(II): LHS+10g=RHS+10g
Looking at the graph, it can be seen that the solution is about 3 ounces of gold and a bit more than 40 ounces of silver. In this case, since the point of intersection is not a lattice point, finding the exact solution is not possible.
(II): .LHS /5.=.RHS /5.
(II): Write as a difference of fractions
(II): a* b/c=a/c* b
(II): Calculate quotient
(II): Identity Property of Multiplication
(II): LHS+10g=RHS+10g
(I): s= 10g + 13
(I): .LHS /5.=.RHS /5.
(I): Write as a sum of fractions
(I): a* b/c=a/c* b
(I): Calculate quotient
(I): Subtract (II)
(I): Distribute - 1
(I): Add and subtract terms
(I): .LHS /450.=.RHS /450.
Concept | Definition |
---|---|
Consistent System | A system of equations that has at least one solution. |
Inconsistent System | A system of equations that has no solution. |
Dependent System | A system of equations with infinitely many solutions. |
Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, it is both consistent and independent.
As part of his student council duties, Tadeo is in charge of getting equipment for the school gym. At the moment, the gym needs new basketballs and volleyballs. Each basketball costs $75 and each volleyball costs $50. The gym's budget is $1400.
Tadeo can only go to the store once to buy the equipment. Each basketball weighs 22 ounces and each volleyball weighs 10 ounces. The total weight of the purchased balls is 364 ounces. For Tadeo to be able to spend all the budget in one go, the number of basketballs b and volleyballs v need to satisfy the following system of linear equations. 75b + 50v = 1400 & (I) 22b + 10v = 364 & (II)
(I): .LHS /(- 5).=.RHS /(- 5).
(I): Write as a sum of fractions
(I): a* b/c=a/c* b
(I): Put minus sign in front of fraction
a+(- b)=a-b
(I): Calculate quotient
(I): Add (II)
(I): Remove parentheses
(I): Add and subtract terms
(I): .LHS /7.=.RHS /7.
Equation (I) | Equation (II) | |
---|---|---|
Equation | 75b + 50v = 1400 | 22b + 10v = 364 |
Substitute | 75( 12) + 50( 10) = 1400 | 22( 12) + 10( 10) = 364 |
Simplify | 1400 = 1400 ✓ | 364 = 364 ✓ |
The values verify both equations of the system. Therefore, the solution is correct.
Concept | Definition |
---|---|
Consistent System | A system of equations that has at least one solution. |
Inconsistent System | A system of equations that has no solution. |
Dependent System | A system of equations with infinitely many solutions. |
Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, the system is both consistent and independent.
Ignacio is looking through his dad's toolbox to see if there are any duplicates that can be sold at a garage sale. He discovers that there are some extra hammers and wrenches.
In fact, the number of wrenches is equal to the difference between 13 and twice the number of hammers. He figures out that if he sells each hammer at $7 and each wrench at $3.50, he will make a profit of $45.50. If h and w are the number of hammers and wrenches, respectively, the information can be written as a system of equations. w = 13 - 2h & (I) 7h + 3.50w = 45.50 & (II)
(I): LHS * 3.5=RHS* 3.5
(I): Distribute 3.5
(II): Subtract (I)
(II): Distribute - 1
(II): Subtract terms
Concept | Definition |
---|---|
Consistent System | A system of equations that has at least one solution. |
Inconsistent System | A system of equations that has no solution. |
Dependent System | A system of equations with infinitely many solutions. |
Independent System | A system of equations that has exactly one solution. |
Since the system has infinitely many solutions, the system is consistent and dependent. In a system of linear equations, if the system is dependent, then the equations that make the system are equations of coincidental lines.
(I): Subtract (II)
Concept | Definition |
---|---|
Consistent System | A system of equations that has at least one solution. |
Inconsistent System | A system of equations that has no solution. |
Dependent System | A system of equations with infinitely many solutions. |
Independent System | A system of equations that has exactly one solution. |
Since the system has no solution, it is an inconsistent system.
Solve the system of linear equations to find the values of x and y.
Multiply or divide an equation by a number so that a variable has the same coefficient in both equations.
(I): LHS * 10=RHS* 10
(I): Distribute 10
(I): Commutative Property of Multiplication
(I): a/c* b = a* b/c
(I): Calculate quotient
(II): .LHS /5.=.RHS /5.
(II): Write as a sum of fractions
(II): a* b/c=a/c* b
(II): Calculate quotient
(I), (II): Identity Property of Multiplication
(I): Subtract (II)
Dominika is running on a treadmill that can be set to different speeds. On the display she can read the amount of calories she burns during a workout session on the treadmill.
Dominika usually sets the treadmill to a speed of 8 kilometers per hour, which is her low speed. After a while, she usually increases the speed to 12 kilometers per hour, her high speed. The table shows two different workout sessions.
Low speed | High speed | Calories Burnt | |
---|---|---|---|
Workout 1 | 20 min | 10 min | 300 |
Workout 2 | 10 min | 15 min | 280 |
We need to create two equations, one describing the first workout and the other describing the workout. Let x and y be the calories per minute that Dominika burns when she runs at low and high speed, respectively.
Dominika runs 20 minutes at low speed and 10 minutes at high speed. With this information, we can write expressions for the calories she burns at these speeds. Low Speed: 20x calories High Speed: 10y calories The sum of these expressions is 300, which is the amount of calories she burns in the first session. 20x+10y=300
Similar to the first workout, we can write an equation describing the amount of calories Dominika burns when running at high and low speed during her second workout session. 10x+15y=280
We can combine these equations to create a system of equations, which we can solve by using the Elimination Method.
We can substitute 13 for y in Equation (I) to find the value of x. However, knowing that y=13 is enough information to state that Dominika burns 13 calories per minute when running at a higher speed.
An equivalent system has the same solution as the original system. Therefore, let's start by finding the solution to the original system.
To solve this system, we will use the Elimination Method.
Next, to find the value of x, we will substitute y=1 in Equation (I).
The solution to the original system is (1,1).
We know that the coefficient of x is 13 and that the constant is 27. 13x+ay=27 The new and the original systems are equivalent. Therefore, (1,1) is also a solution to the new system. This means that the solution x=1, y=1, also satisfies the new equation. With this information in mind, we can find the value of a.
The coefficient of y is 14. 2x+3y=5 & (I) 13x+14y=27 & (II)
Speed is distance over time. speed=distance/time With the given information, we have enough to calculate Zosia's speed when she swims with and against the current. With Current:& s=100/80 [1em] Against Current:& s=150/200 When swimming with the current, Zosia's total speed is the sum of her swimming speed the speed of the current. Conversely, when she swims against the current, the speed of the current is subtracted from the her swimming speed to find the total speed. There are two unknown variables, the current's speed and Zosia's swimming speed. Let them be r_C and r_Z, respectively. With Current:& r_Z+r_C=100/80 [1em] Against Current:& r_Z-r_C=150/200 Now we have two equations that relate the speeds of Zosia and the current. If we combine them, we can write a system of equations which we can solve by using the Elimination Method.
The speed of the current is 0.25 meters per second.