1. You sell tickets for admission to your school play and collect a total of $104. Admission prices are $6 for adults and $4 for children. You sold 21 tickets. How many adult tickets and how many children tickets did you sell?
11:30

2. Objective
The student will be able to: Set up a system of equations Use the “substitution” method and the “elimination” effectively to solve for all unknown variables. Manipulate- within reason

3. Manipulating
6x + 4y = 104 x + y = 21

4. Steps for solving word problems:
1. Indicate what each variable stands for.
Write a system of 2 equations that can be used to solve the problem.
Solve the system- substitution or column elimination
4. Check your work in both original equations.

5. Substitution:
1. Pick the simpler of the two equations
Pick a variable and solve for it
Go to “the other” equation and kick out that variable you solved for.
Sub in the expression to which the variable is equal (step 2)
Solve for the single variable equation

6. Consider
y – x = 4
y – 2x = -2

7. Elimination using multiplication.
Step 1: Put the equations in Standard Form.
Standard Form: Ax + By = C
Step 2: Determine which variable to eliminate based off of LCM.
Look for variables that have the
same coefficient.
Step 3: Multiply each equation fully to get opposite LCM coefficients
Solve for the variable.
Step 4: Add both equations column by column
Step 5: Solve the one variable result equation

8. 1) Solve the system using elimination.
2x + 2y = 6 3x – y = 5
Step 1: Put the equations in Standard Form.
They already are!
None of the coefficients are the same!
Find the least common multiple of each variable.
LCM = 6x, LCM = 2y
Which is easier to obtain?
2y (you only have to multiply the bottom equation by 2)
Step 2: Determine which variable to eliminate.

9. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
Multiply the bottom equation by 2
2x + 2y = 6
(2)(3x – y = 5)
8x = 16
x = 2
2x + 2y = 6
(+) 6x – 2y = 10
Step 3: Multiply the equations and solve.
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y = 1
Step 4: Plug back in to find the other variable.
(2, 1)

10. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
(2, 1)
(2, 1)
2(2) + 2(1) = 6
3(2) - (1) = 5
Step 5: Check your solution.

11. There are three possible outcomes that you may encounter when working with these systems:
1.) one solution
2.) no solution
infinite solutions

12. One Solution If the system has one solution, it is an ordered pair (x, y) that makes BOTH equations true.  In other words, when you plug in the values of the ordered pairs, it makes ALL equations TRUE. 

13. No Solution If the equations are parallel to each other, they will never intersect.  This means they do not have any points in common.  In this situation, no pair values will make both equations true. Example 5: pg 491

14. Infinite Solutions If the two equations are equivalent to each other, then there is an infinite number of solutions. 
Example 6 Pg. 491

15. 7.1 Pg. 483 #1;5-7; 16; 67; 68
7.2 Pg. 495 # 2-6(even); 12-14; 48; 49