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

ObjectiveThe 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

Manipulating

6x + 4y = 104x + y = 21

Steps for solving word problems:

1. Indicate what each variable stands for.

2. Write a system of 2 equations that can be used to solve the problem.

3. Solve the system- substitution or column elimination

4. Check your work in both original equations.

Substitution:

1. Pick the simpler of the two equations

2. Pick a variable and solve for it3. Go to “the other” equation and kick out that

variable you solved for. 4. Sub in the expression to which the variable is

equal (step 2)5. Solve for the single variable equation

Consider

y – x = 4y – 2x = -2

Elimination using multiplication.

Step 1: Put the equations in Standard Form.

Step 2: Determine which variable to eliminate based off of LCM.

Step 3: Multiply each equation fully to get opposite LCM coefficients

Step 5: Solve the one variable result equation

Standard Form: Ax + By = C

Look for variables that have the

same coefficient.

Solve for the variable.

Step 4: Add both equations column by column

1) Solve the system using elimination.

2x + 2y = 63x – y = 5

Step 1: Put the equations in Standard Form.

Step 2: Determine which variable to eliminate.

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 multiplythe bottom equation by 2)

1) Solve the system using elimination.

Step 4: Plug back in to find the other variable.

2(2) + 2y = 6

4 + 2y = 6

2y = 2

y = 1

2x + 2y = 6

3x – y = 5

Step 3: Multiply the equations and solve.

Multiply the bottom equation by 2

2x + 2y = 6

(2)(3x – y = 5)

8x = 16

x = 2

2x + 2y = 6(+) 6x – 2y = 10

(2, 1)

1) Solve the system using elimination.

Step 5: Check your solution.

(2, 1)

2(2) + 2(1) = 6

3(2) - (1) = 5

2x + 2y = 6

3x – y = 5(2, 1)

 

There are three possible outcomes that you may encounter when working with these systems:1.) one solution2.) no solution1.infinite solutions

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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.

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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

 

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

Example 6 Pg. 491

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7.1 Pg. 483 #1;5-7; 16; 67; 68

7.2 Pg. 495 # 2-6(even); 12-14; 48; 49

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