Algebra 135 Seminar – Lesson 83
Part 1 – Bell Work
What is the slope of a line that is perpendicular to the given equation?
1. y = -3x +1
2. y = 7x - 2
4
3. y = x + 6
5
2
1
4. y = - x 7
3
Part 2 – Mini-lesson – Solutions to Systems
In today’s lesson, students will learn how to check if an ordered pair is a
solution to a system of equations.
Part 3 – Independent Work
Part 4 – Exit Card
SOLUTIONS TO SYSTEMS – LESSON 83
Algebra 135 Seminar – Lesson 84
Part 1 – Bell Work
Is the ordered pair a solution to the given system of equations?
1. (4, -2)
x – 2y = 8
2x + 2y = 4
2. (-1, 3)
4x + y = -1
3y – 2x = 1
3. (4, -3)
y = 3x - 15
x + 2y = -2
Part 2 – Mini-Lesson – Solving Equations
Today, we will review solving equations in which students will need to
distribute and then combine like terms. This is to prepare students for
solving systems of equations by substitution.
Part 3 – Independent Work
Part 4 – Exit Card
Solve for x.
2(2x – 4) – 2x = 10
SOLVING EQUATIONS REVIEW – LESSON 84
Solve for x.
2(3x – 4) + 4x = 42
-2(4x + 2) + 2x = 32
4(x + 5) – 2x = 16
5x – 2(3x + 7) = 25
-2x + 6(2x + 3) = 42
-0.5(3x + 4) + 3x = 8
3(-3x + 5) + 6x = 81
-5(4x – 13) – 6x = 101
2(x – 3) – 9x = 22
2(2x + 3) + 4x = 22
5(2x + 1) -5x = -40
1
2( - x + 5) + 5x = 40
4
-(x + 4) + 3x = 56
1
5( x + 1) + 2x = 12
3
3(x + 1) + 2x = 28
Algebra 135 Seminar – Lesson 85
Part 1 – Bell Work – Solving
Solve the following equations for x.
a. 2(2x + 3) + 3x = 13
b. -2(4x – 5) + 4x = 2
c. 3(0.5x + 3) - 4x = 5
Part 2 – Mini-Lesson – Substituting
This lesson will focus on one part of the substitution method for solving
systems of equations. Today, we will give students one equation already
solved for a variable
(y = 3x – 2), and they will substitute it into another equation (-5x + 3y = 10)
and we will solve for one variable (in this case, we should find that x=4).
We will spend two days on this.
Part 3 – Independent Work
Part 4 – Exit Card
Solve for x by substituting.
y = 2x – 10
2y + 2x = 10
SUBSTITUTION WORKSHEET - LESSON 85
Solve for x by
substituting.
y = 2x + 3
2y + 4x = 22
Solve for x by
substituting.
y = 4x – 10
2y – 2x = 10
Solve for y by
substituting.
x=y+3
4y + x = 28
Solve for y by
substituting.
x = 5y + 1
5y + 2x = 47
Solve for x by
substituting.
y = 0.5x – 2
-3y + 2x = 10
Solve for x by
substituting.
y = -1/4x + 4
2y – 2x = 14
Solve for y by
substituting.
x = 10y – 14
2x – 13y = 0
Solve for x by
substituting.
y = 4 – 3x
2y – 6x = 18
Solve for y by
substituting.
x = .4y + 1
-2y + 3x = 6
Solve for y by
substituting.
x = 2y
8y + x = 15
Solve for x by
substituting.
y=x–4
3y + 2x = 20
Solve for x by
substituting.
y = -1/3x + 1
2y + 2x = 6
Solve for y by
substituting.
x=y+1
4y + 2x = 32
Solve for x by
substituting.
y = -1.5x + 2
3y + 2x = 0
Solve for x by
substituting.
y = -4x + 4
2y – 2x = 14
Algebra 135 Seminar – Lesson 86
Part 1 – Bell Work
You may need to use this time to grade, or have students complete, the
worksheet
Part 2 – Mini-Lesson – Substituting
We will spend a second day on yesterday’s lesson. If the students finish the
worksheet, consider having them check and correct them.
Part 3 – Independent Work
Part 4 – Exit Card
Which variable will you solve for in this problem?
y = 2x + 3
-4y + 5x = 10
Algebra 135 Seminar – Lesson 87
Part 1 – Bell Work
Solve for x by substituting.
a.
y = 4x – 4
2y – 2x = 10
b.
y = 0.5x + 2
3y + 2x = 10
c.
y = 2 – 3x
2y – 6x = 20
Part 2 – Mini-Lesson – Review solving for y
Review solving an equation in standard form for y.
Part 3 – Independent Work
Part 4 – Exit Card
SOLVING FOR Y - LESSON 87
Solve the equation for y.
1. 6x + 6y = 42
2. -4x + y = -21
3. x + 3y = -4
4. 15x + 21y = 22
5. 5x – 2y = 18
6. 3x – 8y = -7
7. -3x – 4y = -2
8. 3x – y = -1
9. x + 2y = 3
10. -5x – 6y = 3
11. x – 6y = 6
12. 5x + 3y = 5
13. 2y + 7x = 22
14. -4y + 5x = 21