3.1 systems of linear equations using graphs and tables to solve systems using substitution and...
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3.1 Systems of Linear Equations
• Using graphs and tables to solve systems
• Using substitution and elimination to solve systems
• Using systems to model data
• Value, interests, and mixture problems
• Using linear inequalities in one variable to make predictions
Using Two Models to Make a Prediction
• When will the life expectancy of men and women be equal?– L = W(t) = 0.114t + 77.47– L = M(t) = 0.204t + 69.90
Years since 1980
Yea
rs o
f Li
fe
60
80
100
20 40 60 80 100 120
(84.11, 87.06)
Equal at approximately 87 years old in 2064.
System of Linear Equations in Two Variables (Linear System)
• Two or more linear equations containing two variables
y = 3x + 3
y = -x – 5
Solution of a System
• An ordered pair (a,b) is a solution of a linear system if it satisfies both equations.
• The solution sets of a system is the set of all solutions for that system.
• To solve a system is to find its solution set.
• The solution set can be found by finding the intersection of the graphs of the two equations.
• Graph both equations on the same coordinate plane– y = 3x + 3– y = -x – 5
• Verify– (-3) = 3(-2) + 3– -3 = -6 + 3– -3 = -3
– (-3) = -(-2) – 5 – -3 = 2 – 5 – -3 = -3
• Only one point satisfiesboth equations
• (-2,-3) is the solution set of the system
Find the Ordered Pairs that Satisfy Both Equations
Solutions for
y = 3x + 3
Solutions for
y = -x – 5
Solution for both (-2,-3)
Example• ¾x + ⅜y = ⅞• y = 3x – 5 • Solve first equation for y
– ¾x + ⅜y = ⅞– 8(¾x + ⅜y) = 8(⅞)– 24x + 24y = 56
4 8 8
– 6x -6x + 3y = 7 – 6x– 3y = -6x + 7
3 3 3
– y = -2x + 7/3
(1.45,-.6)
y = 3x – 5
-.6 = 3(1.45) – 5
-.6 = 4.35 – 5
-.6 ≈ -.65
y = -2x + 7/3
-.6 = -2(1.45) + 7/3
-.6 = -2.9 + 7/3
-.6 ≈ -.57
Inconsistent System
• A linear system whose solution set is empty– Example…Parallel lines never intersect
• no ordered pairs satisfy both systems
Dependent System
• A linear system that has an infinite number of solutions– Example….Two equations of the same line
• All solutions satisfy both lines
y = 2x – 2
-2x + y = -2• -2x +2x + y = -2 +2x• y = 2x – 2
One Solution System
• There is exactly one ordered pair that satisfies the linear system– Example…Two lines
that intersect in only
one point
Solving Systems with Tables
x 0 1 2 3 4
y = 4x – 6 -6 -2 2 6 10
y = -6x + 14 14 8 2 -4 -10
• Since (2,2) is a solution to both equations, it is a solution of the linear system.
• Isolate a variable on one side of either equation
• Substitute the expression for the variable into the other equation
• Solve the second equation
• Substitute the solution into one of the equations
3.2 Using Substitution
1. y = x – 1
2. 3x + 2y = 13• 3x + 2(x – 1) = 13• 3x + 2x – 2 = 13• 5x – 2 +2 = 13 +2• 5x = 15
5 5• x = 3
• y = 3 – 1 • y = 2
Solution set for the linear system is (3,2).
Example 1
Example 2
1. 2x – 6y = 4
2. 3x – 7y = 8• 2x – 6y +6y = 4 +6y• 2x = 6y + 4
2 2 2
• x = 3y + 2
• 3(3y + 2) – 7y = 8• 9y + 6 – 7y = 8• 2y + 6 -6 = 8 -6• 2y = 2
2 2
• y = 1• x = 3(1) + 2• x = 5
The solution set is (5, 1).
Using Elimination
• Adding left and right sides of equations
• If a=b and c=d, then a + c = b + d
– Substitute a for b and c for d• a + c = a + c• both sides are the same
Using Elimination
• Multiply both equations by a number so that the coefficients of one variable are equal in absolute value and opposite sign.
• Add the left and right sides of the equations to eliminate a variable.
• Solve the equation.
• Substitute the solution into one of the equations and solve.
Example 1
5x – 6y = 9
+ 2x + 6y = 12
7x + 0 = 21
7x = 21
7 7 x = 3
2(3) + 6y = 12
6 -6 + 6y = 12 -6
6y = 12
6 6
y = 2
Solution set for the system is (3,2)
Example 21. 3x + 7y = 29
2. 6x – 12y = 32
-2(3x + 7y) = -2(29)
-6x – 14y = -58
+ 6x – 12y = 32
0 – 26y = -26
-26 -26
y = 1
• 3x + 7(1) = 29• 3x + 7 -7 = 29 -7• 3x = 22
3 3
x = 22/3
Solution (22/3, 1)
Example 3
1. 2x + 5y = 14
2. 7x – 3y = 8
3(2x + 5y) = 3(14)
6x + 15y = 42
5(7x – 3y) = 5(8)
35x – 15y = 40
6x + 15y = 42
+ 35x – 15y = 40
41x + 0 = 82
41 41
x = 2
2(2) + 5y = 14
4 -4 + 5y = 14 -4
5y = 10
5 5
y = 2Solution (2,2)
Using Elimination with Fractions
2x – 5y -1 3 9 3
3x – 2y 715 3 5
9( ) = ( )9
15( ) = ( )15
-2(6x – 5y) = (-3)-2
-12x + 10y = 6+___________
-9x + 0 = 27
-9 -9x = -3
3x – 10y = 21
6x – 5y = -3
6(-3) – 5y = -3
-18 +18 – 5y = -3 +18
-5y = 15 -5 -5 y = -3
Inconsistent Systems
• y = 3x + 3• y = 3x – 2 • 3x + 3 = 3x – 2
3x -3x + 3 = 3x -3x – 2
3 ≠ - 2 False
• y = 3x + 3
-1(y) = -1(3x + 3)• -y = -3x – 3
+ y = 3x – 2
0 = 0 – 2
0 ≠ -2 False
•If the result of substitution or elimination of a linear system is a false statement, then the system is inconsistent.
Dependent Systems
1. y = 3x – 4
2. -9x + 3y = -12• -9x + 3(3x – 4) = -12• -9x + 9x -12 = -12 • -12 = -12• True
• y -3x = 3x -3x – 4 • -3(-3x + y) = (-4)-3• 9x + -3y = 12
+ -9x + 3y = -12
0 = 0
True
• If the result of applying substitution or elimination to a linear system is a true statement, then the system is dependent.
Solving systems in one variable
• To solve an equation A = B, in one variable, x, where A and B are expressions,
• Solve, graph or use a table of the system– y = A– y = B
• Where the x-coordinates of the solutions of the system are the solutions of the equation A = B
Systems in one variable
• y = 4x – 3 • y = -x + 2• 4x – 3 = -x + 2• 4x +x – 3 = -x +x + 2• 5x – 3 +3 = 2 +3• 5x = 5• 5 5• x = 1
• y = 4(1) – 3 • y = 1
(1,1)
Graphing to solve equations in one variable
• -2x + 6 = 5/4x – 3 – (3,1)
• -2x + 6 = -4– (6,-4)
(3,1)
(6,-4)
Using Tables to Solve Equations in One Variable
• -2x + 7 = 4x – 5
• Solution (2,3)
x y y
-2 11 -13
-1 9 -9
0 7 -5
1 5 -1
2 3 3
3 1 7
4 -1 11
3.3 Systems to Model Data
• Predict when the life expectance of men and women will be the same.
• L = W(t) = .114t + 77.47• L = M(t) = .204t + 69.90• .114t -.114t + 77.47 = .204t + 69.90 -.114t• 77.47 -69.90 = .09t + 69.90 -69.90• 7.57 = .09t• t ≈ 84.11
Solving to Make Predictions
• In 1950, there were 4 women nursing students at a private college and 84 men. If the number of women nursing students increases by 13 a year and the number of male nursing students by 6 a year. What year will the number of male and female students be the same?
• A = W(t) = 13t + 4• A = M(t) = 6t + 84
Using substitution
• 13t + 4 = 6t + 84
• 13t + 4 -4 = 6t + 84 -4
• 13t -6t = 6t -6t + 80
• 7t = 80
7 7
• t ≈ 11.43
• Equal in 1961~1962