Transcript CH2 Section 2.1

Chapter 2
Linear Systems in Two Variables and
Inequalities with Applications
Section 2.2
Section 2.2
Substitution and Elimination
• Solving a Linear System in Two Variables by Substitution
• Solving a Linear System in Two Variables by Elimination
• Algebraic Modeling with Linear Systems
• Identifying Inconsistent and Dependent Systems Algebraically
Solving a Linear System with Two Variables by Substitution
(1) Choose one of the equations and solve it for one of the
variables in terms of the other.
(2) Substitute the expression from step (1) into the other
equation. This will result in an equation with just one
variable. Solve for the variable.
(3) Substitute the value found in step (2) into either of the
original equations to find the remaining variable.
(4) Write the answer as an ordered pair and check the
solution to the system in both original equations.
Note: When solving an application problem, state the
answer in the context of the problem.
Solve the system by substitution:
5 + y = –4x
3y – 6x = 21
(1) Solve “5 + y = –4x” for y.
y = –4x – 5
(2) Substitute y = –4x – 5 into 3y – 6x = 21 and solve for x:
3(–4x – 5) – 6x = 21
–12x – 15 – 6x = 21
–18x = 36
x = –2
(continued on the next slide)
(Contd.)
Solve the system by substitution:
5 + y = –4x
3y – 6x = 21
(3) Back-substitute x = –2 into any of the original equations
to find the value of y.
Hint: Using “y = –4x – 5” is a good choice because “y” is
already isolated.
y = –4x – 5
y = –4(–2) – 5
y=3
(4) The solution to this system is (–2, 3).
Check the solution:
5 + y = –4x
5 + 3 = –4(–2)
3y – 6x = 21
3(3) – 6(–2) = 21
True
True
Vlad does not manage his finances well, and during the last
years the bank has been increasing the interest rates on his 2
credit cards. His credit cards interest rates are 18% and 26.5%.
Vlad has a total balance of $5400 on both cards and he carries
$1167.50 in interest, excluding other fees. How much balance
does he have on each card?
Let x = amount owed on the 18% interest rate card
y = amount owed on the 26.5% interest rate card
Equation for total balance: x + y = 5400
Equation for interest charged: 0.18x + 0.265y = 1167.50
(1) Solve the first equation for x. (Note: Can solve for y instead)
x + y = 5400
x = 5400 – y
(continued on next slide)
(Contd.)
(2) Substitute “x = 5400 – y " into the second equation and
solve for y.
0.18x + 0.265y = 1167.50
0.18(5400 – y) + 0.265y = 1167.50
972 – 0.18y + 0.265y = 1167.50
972 + 0.085y = 1167.50
0.085y = 195.5
y = 2300
(3) Back-substitute y = 2300 into any of the original equations
to find x.
x + 2300 = 5400
x = 3100
(4) The solution to the system is (3100, 2300).Vlad has a
balance of $3,100 on the 18% interest rate card and $2,300
on the 26.5% interest rate card.
Solving a Linear System with Two Variables by Elimination
(1) Eliminate one of the variables by using the concept of
"opposites." If necessary, multiply one or both equations by a
nonzero constant that will produce coefficients of either x or y
that are opposites of each other.
(2) Add the two equations, making sure that one of the variables
drops out, leaving one equation and only one unknown.
(3) Solve the resulting equation for the variable.
(4) Back-substitute the value found in (3) into one of the original
equations to find the value of the remaining variable.
(5) Write the answer as an ordered pair and check the solution to
the system in both original equations.
Note: When solving an application problem, state the answer
in the context of the problem.
Solve the system by elimination:
3y – 6x = 15
2y + 4x = –14
Step 1: Eliminate one of the variables by using the concept of
"opposites.”
Multiply both sides of the first equation by 2.
Multiply both sides of the second equation by –3.
(Note: Can choose to multiply first equation by 4 and second
equation by 6 instead.)
2(3y – 6x = 15)  6y – 12x = 30
–3(2y + 4x = –14)  –6y – 12x = 42
Make sure you multiply all the terms throughout the equations!
(continued on next slide)
(Contd.)
Solve the system by elimination:
3y – 6x = 15
2y + 4x = –14
Steps 2-3: Add the equivalent equations to eliminate one of the
variables; solve for the remaining variable.
6y – 12x = 30
–6y – 12x = 42
0y – 24x = 72
x = –3
Step 4: Back-substitute x = –3 into any of the original equations
to find the value of y.
3y – 6(–3) = 15
3y + 18 = 15
y = –1
Step 5: The solution to the system is (–3, –1).
You can check the answer.
A group of fitness club members felt like "breaking the rules" for one
day, and enjoyed delicious desserts after their workout. Two had
chocolate cake and the other four ordered special lemon pie. Two
chocolate cakes and four special lemon pies cost $27.90. If four
club members had ordered chocolate cake and two had chosen the
special lemon pie, the cost would have been $29.70. Establish a
system of linear equations and solve it by elimination to find the
price of each of these desserts.
Let c = chocolate cake
p = special lemon pie
Equation for 2 cakes and 4 pies:
Equation for 4 cakes and 2 pies:
2c + 4p = 27.90
4c + 2p = 29.70
(continued on next slide)
(Contd.)
2c + 4p = 27.90
4c + 2p = 29.70
Step 1: Multiply both sides of the first equation by –2.
– 2(2c + 4p = 27.90)

–4c – 8p = –55.80
Make sure you multiply all the terms throughout the equation!
Steps 2-3: Add the equivalent equation from (1) to the second
equation of the system, and solve for p.
–4c – 8p = –55.80
4c + 2p = 29.70
0c – 6p = –26.10
p = 4.35
(continued on next slide)
(Contd.)
2c + 4p = 27.90
4c + 2p = 29.70
Step 4: Back-substitute p = 4.35 into any of the original equations
to find c.
2c + 4(4.35) = 27.90
2c + 17.4 = 27.90
2c = 10.5
c = 5.25
Step 5: The solution to the system is (5.25, 4.35). This means
that each chocolate cake was $5.25 and each special lemon pie
was $4.35.
Analyze the following system:
4x + 5y = 97
8x – 5y = –35
Which of the following statements is correct?
a. Solving by substitution is more convenient.
b. Solving by elimination is more convenient.
Since 5y and -5y are opposites, these y-terms would be
eliminated immediately, allowing us to solve for x.
Answer: b
Note: Solving by substitution requires to leave a variable by
itself on one side of the equation; in this case it would result in
fractions, making the solution process more laborious.
Identifying Inconsistent and Dependent Systems
Algebraically
If both variables of a 2 x 2 linear system are eliminated when
solving algebraically, this implies one of the following:
Case 1:
 If the resulting equation represents a false statement, the
system is inconsistent; the graph will display parallel lines, thus
the system has no real solution.
Example
2x – 6y = 1
–2x + 6y = 8
0 =9
False statement!
No x and y values will satisfy both equations simultaneously.
Identifying Inconsistent and Dependent Systems
Algebraically
Case 2:
 If the resulting equation represents an identity, the system is
dependent (coincident lines, infinitely many solutions).
Example
3x – 5y = –8
–3x + 5y = 8
0= 0
Identity!
All x and y values that satisfy one equation will also satisfy the
other equation.
Using your textbook, practice the
problems assigned by your instructor to
review the concepts from Section 2.2.