Class: Algebra 2
Instructional Date(s): 1/24/05
Text Chapter: 7.5 Special Types of Linear Systems
(EXAMPLE)
LESSON PLAN
1. Warm-Up Exercise:
Write Each Equation in slope-intercept form
1.) x + y = 12
2.) -x + 3y = 3
3.) 4x – 6y = 15
4.) x – 5y = 10
2. Interaction of Lines on Coordinate Plane Sub-Objective
Intersecting lines – One solution (include sketch)
Parallel lines – No solutions (include sketch)
Coincident lines – Infinite solutions (include sketch)
3. Linear Systems with No Solutions Sub-Objective
Example Problem
-6x + 2y = -8
-3x + y = 7
Graphing
Solve equations into y = mx + b form
-6x + 2y = -8 → y = 3x – 4 (sketch graph)
-3x + y = 7→ y = 3x + 7 (sketch graph)
Compare the slopes of the two equations
3=3
m1 = m2 → parallel lines
Same slopes, parallel lines
Substitution
Solve one equation for one variable, which one and why?
-3x + y = 7→ y = 3x + 7
Substitute y = 3x + 7 into -6x + 2y = -8
-6x + 2(3x + 7) = -8
-6x - 6x + 14 = -8
14 = -8 FALSE
When both variables are eliminated and result is a false statement, the system has no solutions or is
parallel lines
Linear Combinations
Multiply one (or both) equations by a factor, which one (ones) and why?
-3x + y = 7 by -2→ 6x - 2y = -14
Add -6x + 2y = -8 and 6x + 3y = 14
-6x + 2y = -8
6x +-32 = -14
0 = -22 FALSE
Variables eliminated, result is a false statement, and the system has no solutions or is parallel lines
Intermediate Closure:
State result of algebraic solution that indicates parallel lines. (False)
4. Linear Systems with Infinite Solutions Sub-Objective
Example Problem
-x + 2y = -2
3x - 6y = 6
Graphing
Solve equations into y = mx + b form
-x + 2y = -2 → y = ½ x – 1 (sketch graph)
3x - 6y = 6 → y = ½ x - 1 (sketch graph)
Compare the slopes of the two equations
1/2 =1/2
m1 = m2
Compare the y-intercepts of the two equations
-1 = -1
b1 = b2
Same slopes and y- intercepts, coincident lines
Compare the two equations, do you see a factor that if multiplied by one equation results in the other equation?
-x + 2y = -2 multiplied by -3 results in 3x - 6y = 6, therefore coincident lines.
Substitution
Solve one equation for one variable, which one and why?
-x + 2y = -2→ x = 2y + 2
Substitute x = 2y + 2 into 3x - 6y = 6
3(2y + 2) - 6y = 6
6y + 6 – 6y = 6
6 = 6 TRUE
When both variables are eliminated and result is a true statement, the system has infinite solutions or is
coincident lines
Linear Combinations
Multiply one (or both) equations by a factor, which one (ones) and why?
-x + 2y = -2 by 3→ -3x + 6y = -6
Add -3x + 6y = -6 and 3x - 6y = 6
-3x + 6y = -6
3x - 6y = 6
0 = 0 TRUE
Variables eliminated, result is a true statement, and the system has infinite solutions or is coincident
lines
Intermediate Closure:
State result of algebraic solution that indicates coincident lines. (True)
Independent Practice
Solve the following systems of linear equations using both substitution and linear combinations.
Interpret your results.
1)
6x - 2y = 10
-3x + y = 12
Solution: 0 = 34, parallel lines
1)
3x + y = 4
-6x – 2y = -8
Solution: 0 = 0, coincident lines
5. Real-life example problem of parallel lines.
Trudy bought 2 notebooks and 3 packs of paper for $6.25. Tran bought 6 packs of paper and 4
notebooks for $12.50. Write a system of linear equations to model the situation and interpret the results.
Price of Notebooks = x
Price of Packs of paper = y
Trudy 2x + 3y = 6.25 → 200x + 300y = 625
Tran 6x + 4y = 12.50 → 600x + 400y = 1250
Solve by Linear Combinations
-3(200x + 300y) = -3(625) → -600x - 400y = -1250
Add 600x + 400y = 1250 and -600x - 400y = -1250
600x + 400y = 1250
-600x - 400y = -1250
0 = 0 TRUE
True solution, coincident lines, not possible to determine x (price notebooks) and y (price of packs of
paper) from information given.
6. Final Closure
Review topics of lesson:
What is algebraic result of ║ lines?
What is algebraic result of coincident lines?
Homework Assignment:
Pg. 429: 12-17,20,22,26,28,31-33