Transcript Sub and Elimination

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
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
Manipulating
6x + 4y = 104
x + 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 it
3. 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=4
y – 2x = -2
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
Step 4: Add both equations
column by column
Step 5: Solve the one
variable result equation
Solve for the variable.
1) Solve the system using elimination.
2x + 2y = 6
3x – 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 multiply
the bottom equation by 2)
1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
Multiply the bottom equation by 2
2x + 2y = 6
2x + 2y = 6
(2)(3x – y = 5) (+) 6x – 2y = 10
8x
= 16
Step 3: Multiply the
equations and solve.
x=2
Step 4: Plug back in to find
the other variable.
(2, 1)
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y=1
1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
(2, 1)
Step 5: Check your
solution.
(2, 1)
2(2) + 2(1) = 6
3(2) - (1) = 5
There are three possible outcomes that you may
encounter when working with these systems:
1.) one solution
2.) no solution
1.infinite solutions
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.
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
7.1 Pg. 483 #1;5-7; 16; 67; 68
7.2 Pg. 495 # 2-6(even); 12-14; 48; 49