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Unit 3, Cluster 1, Learning Cycle A: Reasoning with Equations
I can explain or justify the steps used in solving linear equations using algebraic properties.
I understand and can use the notion of equality to construct an argument for solving an
equation.
EQUALITY
To help understand the properties of equality, we will build, rather than decompose an equation.
Example:
Start:
x=5
Multiply by 3
x=
Add 2
x=
Add x (to both sides
x=
of the equation)
Add x = 5
x=
(to 3x + 2 = 17)
The models show that ____________ the same quantity or ___________ the same number to
both sides of an equation ______ _____ _________ the original relationship. In a sense, we are
_______________ the equation while maintaining the value of the original relationship, x = 5.
This idea can be connected to transformations of ___________.
If a ___________ equality is introduced, the original equality will ______ be maintained.
Start:
x=5
Add a false equality
x=
The equation 3x + 2
x=
= 17 maintained x = 5
Add a new value:
x=
x=2
What is the difference between solving an equation and simplifying an algebraic expression?
_________________________
Can you solve an expression?_______ What’s the difference between an expression and an
equation? ____________________The properties of equality apply only to equations.
Skill-based Task
Justify the equation solution by writing the property
or reason why each solution step works.
3x + 7 = 12
3x + 7 – 7 = 12 – 7
3x + 0 = 5
3x = 5
(3x)(1/3) = (5)(1/3)
1x = 5/3
x = 5/3
Problem Task
When Sally picks any number between 1 and 20,
doubles it, adds 6, divides by 2 and subtracts 3,
she always gets the number she started with.
Why? Evaluate and use algebraic evidence to
support your conclusion.
Unit 3, Cluster 2, Learning Cycle A: Reasoning with Equations
I can solve linear equations and inequalities in one variable, including
absolute value equations/inequalities and compound inequalities.
SOLVING LINEAR EQUATIONS AND INEQUALITITES IN ONE VARIABLE
(absolute value and compound inequalities)
Absolute Value / Compound Inequalities:
Solving an __________ __________ inequality is similar to solving ____________ inequalities.
Start by understanding ________ ________ _________ and then link the meaning to the
algorithmic ___________.
|x| = 5 means, some value “x” is __________ 5 units from the origin. There are two such values
for x, ____ and _____.
The __________ is in the middle and the values for x are each ____ units left or right of the
origin. The “gap” between the solutions is ____ or twice ___ because we are looking for the
values of x that are exactly ___ _______ from the origin.
Now suppose we want to find |x| > 5. This means all values of 5 that are _______ ______ 5
units from the origin.
All values to the right of 5 are _______ ______ ___ units from the origin. For example, 6, 7,
8…and all values in between are more than 5 units from the origin. To express these values, we
can write ________________.
All values to the left of -5 are ______ _______ ___ units from the origin. Here, values such as 6,-7,-8…are more than 5 units from the origin. To express them mathematically we say
_________________.
Thus, the solution to |x| > 5 is __________ or ___________. Any value for x that meets either of
those conditions, satisfies |x| > 5. Again notice that the “gap” is 10 units.
Now suppose we want to find |x| < 5. Again, starting with the meaning is essential. The equation
|x| < 5 means we are looking for all values that are ______ _____ 5 units from the origin.
The solution includes all values for x between ____ and ___, all of the values are both to the
______ of -5 and to the ______ of 5, written two ways: a. ____________ or b. ___________
and the “gap” is ____units.
Notice from these two examples, there are always ______ parts.
For |x| ≥ a or |x| > a the distance is greater than a, _________ OR ___________
For |x| ≤ a or |x| < a the distance is less than a, ______________________________
-a
a
x______
x_______
_____ OR _____
Example 1: (solving with “greater than”)
Solve: |x – 20| > 5
x_____ OR x______
15 and 25 are 2x5 or ___ units apart
Example 3: (isolating the absolute value
first)
Solve: |3 + x| – 4 < 0
x____ AND x______
________________
Example 5: (no values work)
Solve: |x + 1| < -6
x_____ AND x_____???
______
-a
a
x______
x_______
_____________
Example 2: (solving with “less than or equal to”)
Solve: |x – 3| ≤ 4
x_____ AND x______
________________
7 and -1 are 2x4 or ___ units apart
Example 4: (all values work)
Solve: |x + 4| > -3
x____ OR x_____
______
Example 6: (word problem)
At Brooks Graphic Company, the average starting salary
for a new graphic designer is
$37, 600, but the actual salary could differ from the average
by as much as $2590.
a. Write an absolute value inequality to describe this
situation. |x - _________| < ______
|the __________ between _______ and ______|<$2590
b. Solve the inequality to find the range of the starting
salaries.
______________________
Unit 3, Cluster 2: Learning Cycle B
I can solve an equation for a specific variable and reason about the new equation.
Why would we ever need to solve a formula or equation for a specific value?________________
Example: Solve the formula d = rt for t or for r.
𝑟 =
𝑑
if d is held constant (i.e. the distance does not change) then we can see that as t
increases, r becomes smaller. Explanation: as time increases, the rate of speed _____________.
𝑡
Solve: 2(x + 4) – 3 > 4x – 2
The perimeter of a rectangle is given by P = 2W + 2L.
Solve for W and restate in words the meaning of this new formula in
terms of the meaning of the other variables.
Unit 3, Cluster 2, Learning Cycle C: Reasoning with Equations
I can solve simple exponential equations using laws of exponents.
LAWS OF EXPONENTS
Use knowledge of laws of exponents to solve simple exponential equations.
Examples: 1. Solve for x:
3. These equations could also be solved
graphically
3x= 243
x=___
2. Solve for x:
using technology. For example: Graph y = 3x
and y = 243, and use technology to find the point
of intersection. This connects nicely to solving
systems of equations.
5x + 2 = 127
5x = ____
x=___
Unit 3, Cluster 3, Learning Cycle A: Reasoning with Equations
I can explain how to solve a system of two equations and justify the method
used to find the solution.
SYSTEM OF TWO EQUATIONS
The focus in Secondary 1 should be on the ________ of each equation, values that do or do not
___________ each of the equations, the meaning of a ________ to the system, and
understanding, justifying and explaining the _________ used to solve the system. Explore
systems where both equations describe the ______ line and result in _________ _____ solutions,
and systems describing ________ lines with ____ solution.
Examples:
This system has no point of
This has lines that ___________ and an
______________ and thus ___ solution.
____________ number of solutions.
It is an _________________ system. The system is _________ and ___________.
Unit 3, Cluster 3, Learning Cycle B: Reasoning with Equations
I can explain why replacing one equation by the sum of that equation with a
multiple of the other produces a system with the same solution.
A SYSTEM WITH THE SAME SOLUTION
There are three possibilities regarding a system of two linear
functions:
1. the lines intersect at _________ ____ _________,
2. there is ____ ________________, or
3. the two lines intersect at ____ ___________.
Suppose (2, -3) is the point where two lines intersect. For this
point, we know that x = ___ and y = ____.
We can think of each of the equations as a point on a number
line, or as a line on the coordinate plane.
Using the properties of equal., we can operate on these equations
Add the equations: x = 2
Sum: ________
y = -3
This equation is based on the original equality, and can function
as our first equation in the system. (blue)
Multiply x = 2 by 3 and add it to the 2nd equation we get:
-3(x = 2)
-3x = -6
y = -3
y = -3
_______________
This equation is also based on the original equality and it will
function as the second equation in the system. (red)
The new lines also go through the point (__,__). They are
_______ of the original x = 2 and y = -3 around the point (2, -3).
For the line x + y = -1 the slope is -1, how is this related to the
lines x = 2 and y = -3? ____________________________
How was this line created? ________________________
For the line -3x + y = -9, the slope is 3, how is this line related to
each of x=2 and y=-3 and how was it generated? ____________
Add the equations: x + y = -1
-3x + y = -9
-2x + 2y = -10 -x + y = -5.
This is a rotation of the original lines, all rotated on the point
(__, ___). All combinations of equations render lines that go
through the same point of intersection. This point (2, -3) is the
solution of the system.
Every _____-________ or ____-_________ line that passes through the point (2, -3) has both
an x and a y that depend on each other. Each line that passes through (2, -3) is some
combination (with or without scalar multiplication) of the sum of x=2 and y=-3. Solving the
system in essence means that you are _________ ________ (through rotations) the lines x=2
and y=-3. The combinations are systematically separating and rotating the system back to x=2
and y=-3. In solving this system of equations, the goal is to rotate the line back to the original
x = 2 and y = -3. Multiply the first equation by 3 and then add to the second:
3(x + y = -1)
3x + 3y = -3
-3x + y = -9
-3x + y = -9
4y = -12
4(y = -3)
y = -3 is just a rotation of x + y = -1 and/or -3x + y = -9.
We can add y = -3 to either x + y = -1 or -3x + y = -9 and get another rotation, but the new
equation will just be a non-vertical or non-horizontal rotation on the point (2, -3) of either
equation where x and y are dependent on each other. Instead, because we know that y = -3 is a
true equation, we can ___________ -3 in for y in either of the equations and get x = 2.
x + y = -1
-3x + y = -9
x + -3 = -1
-3x + -3 = -9
x = ___
-3x = -6
x = ___
Skill-based Task
Verify that (-4, 13) is the solution to the system:
2x + y = 5
-5x – 2y = -6
Justify that the following is an equivalent system:
-3x – y = -1
-5x – 2y = -6
Verify that (-4, 13) is the solution to this new
system.
Problem Task
Create a new system using both the addition and
multiplication properties of equality. Then verify
that the new system has the same solution as the
original.
2x + y = 5
-5x – 2y = -6
Unit 3, Cluster 3, Learning Cycle C: Reasoning with Equations
I can create a system of equations to represent relationships between quantities.♦
REPRESENTING RELATIONSHIPS WITH A SYSTEM OF EQUATIONS
Example:
The admission fee at a county fair is $1.50 for children and $4.00 for adults. On the last day of
the fair, 2200 people enter the fair and $5,050 is collected. How many children and
how many adults attended?
Skill-based Task
Approximate the solution to the system of
equations graphically. Then verify the
solution algebraically.
3x – 5y = -15
2x - y = 2
Problem Task
The high school is putting on the musical Footloose. The
auditorium has 300 seats. Student tickets are $3 and
adult tickets are $5. The royalty for the musical is $1300.
What combination of student and adult tickets do you
need to fill the house and pay the royalty?
How could you change the price of tickets so more
students can go?