Chapter 3 Notes Algebra II
3.2 Day 1 Substitution
You can use the substitution method to solve a system of equations when it is easy to isolate one of the variables.
1.) Isolate the variable that it’s easiest to isolate.
2.)After isolating the variable, substitute for that variable in the other equation.
3.)Then solve for the other variable.
Example 1: What is the solution of the system of equations?
3x + 4y = 12
2x + y = 10
Example 2: A music store offers piano lessons at a discount for customers buying new pianos. The costs for lessons and a one-time fee for materials (including music books, CDs, software, etc.) are shown in the advertisement. 6 lessons and the one-time fee cost $300. 12 lessons and the one-time fee cost $480. What is the cost of each lesson and the one-time fee for materials?
3.2 Day 2 Elimination
Another way to solve a system of equations is by using elimination method. If you can add a pair of additive inverses or subtract identical terms, you can eliminate a variable.
Example 1: What is the solution of the system of equations?
4x + 2y = 9
-4x + 3y = 16
If the system doesn’t already have additive inverses, you can create them by multiplying one or both equations in a system by a nonzero number. The new system will be an equivalent system to the original problem and will therefore have the same answer.
Example 2: What is the solution of the system of equations?
2x + 7y = 4
3x + 5y = -5
Sometimes systems of equations don’t have one unique solution. They can have infinitely many solutions, and they can also have no solutions.
Example 3: What is the solution of the system of equations?
-3x + y = -5
3x – y = 5
Example 4: What is the solution of the system of equations?
4x – 6y = 6
-4x + 6y = 10
3.5 Solving Systems with Three Variables
We’ve learned about systems of 2 equations with 2 variables that may have a solution (x,y) that makes both equations true.
In this lesson, we will learn about systems of 3 equations and 3 variables that may have a solution (x,y,z) that makes all three equations true.
Solve each system of equations. Check your answers.
1.) X – y + z = -1
X + y + 3z = -3
2x – y + 2z = 0
2.) X + y + 2z = -7
3x + y – 2z = 7
-x – 3y + z = -9
3.) X – 2y + 3z = 12
2x – y – 2z = 5
2x + 2y – z = 4
3.6 Using Matrices to Solve Systems of Equations
We will be using a graphing calculator to do the major computation in this section. On the test, you will be required to solve problems with the graphing calculator on your own, so be sure to pay close attention today!
About Matrices
A matrix is a rectangular array of numbers, symbols, or expressions. The individual items in the matrix are called elements or entries.
The matrix at the right has _____ elements.
The size of the matrix is given by the number of rows first and then the number of columns second.
The matrix at the right has _____ rows and _____ columns.
We say the size of this matrix is _____.
The elements in a matrix are named where i = row position and j = column position.
Name the elements: = _____= _____
Augmented Matrices
An augmented matrix can be used to represent a system of equations. It is important that each equation in the system be in standard form (Ax + By + … = D), that is, all the variable terms are on the left side of the equals sign in alphabetic order and the constant term isolated on the right side of the equals sign.
System of Equations in Standard FormAugmented Matrix
Write the following system of equations as augmented matrices:
1.)
2.)
Reduced Row Echelon Form
We are going to use a graphing calculator to change the augmented matrix for a system of equations into the reduced row echelon form (RREF) of the matrix. From RREF, we will be able to read the solutions of the system of equations that it represents.
Example of RREF:Solutions:
Read the solutions off the following RREF matrices:
1.)
2.)
TI-83 Matrix Menu
Note: The TI-83 may
have a matrix button
instead of a matrix
you have to shift to
use. This is actually
easier because it is
one less step to make.
Solve the system:
INSTRUCTIONS
1. Write your system as an augmented matrix and determine the size.
2. Go to MATRX and EDIT.
ENTER to select [A]
3. Define [A] to be a 3 x 4 matrix
Use the cursor and ENTER to move through the spaces.
4. Enter the coefficients and constants from the system (your augmented matrix)by entering the value and pressing ENTER.
5. Return to the home screen
2nd MODE to “QUIT”
6. Go to MATRX and MATH.
Scroll down and hit ENTER to select B: rref(
7. Go to MATRX and NAMES. Hit ENTER to paste [A] to your home screen.
Close the parenthesis.
8. Press ENTER and read the solution from the matrix.
Solution: x:_____ y:_____ z:_____
Use a graphing calculator to solve the following systems.
1.)
2.)
3.6 Day 2 Real World Problems
Set up a system of equations and solve using a graphing calculator. Does your answer make sense?
1.) The tickets for a school play are $7.00 for adults and $5.00 for students. There were 100 tickets sold. If the total profit was $550.00, how many of each were sold?
2.) You work at a fruit stand that sells apples for $2 per pound, oranges for $5 per pound, and bananas for $3 per pound.
Yesterday you sold 60 pounds of fruit and made $180. You sold 10 more pounds of apples than bananas.
How many pounds of each kind of fruit did you sell?
3.) Suppose you want to fill nine 1-lb tins with a snack mix.
You have $15 and plan to buy almonds for $2.45 per pound, hazelnuts for $1.85 per pound, and raisins for $0.80 per pound.
You want the mix to contain an equal amount of almonds and hazelnuts and twice as much of the nuts as raisin by weight.
How many of each ingredient should you buy?
4.) A hardware store mixes paints in a ratio of two quarts red to six quarts yellow to make two gallons of pumpkin orange. A ratio of five quarts red to three quarts yellow makes two gallons of pepper red. A gallon of pumpkin orange sells for $25 and a gallon of pepper red sells for $28. Find the cost of a quart of red and a quart of yellow.