Transcript 4.1 Systems of Linear Equations in two variables

Learning Objectives for Section 4.1
Review: Systems of Linear
Equations in Two Variables
After this lesson, you should be able to
 solve systems of linear equations in two variables by graphing
 solve these systems using substitution
 solve these systems using elimination by addition
 solve applications of linear systems.
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System of Linear Equations
System of Linear Equations: refers to more than one
equation being graphed on the same set of coordinate axes.
Solution(s) of a System: the point(s) (ordered pair) at which
the lines ________________.
The solution of a system is an ordered pair(s) that will satisfy
all of the equations in the system.
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Solving: Graphically
1. On a single set of coordinate axes, graph each equation.
Label the equation of each line.
2. Find the coordinates of the point where the graphs
intersect. These coordinates give the solution of the
system. Label this point.
3. If the graphs have no point in common, the system has
no solution.
4. Check the solution in both of the original equations.
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Solve by Graphing
Example 1: Solve the following system by graphing:
y
x  y  4

2 x  y  5
x
Check your solution!
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Checking…
x  y  4

2 x  y  5
Solution:
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Consistent Systems
Consistent system- a system of equations that has a solution.
(the system has at least one point of intersection)
one solution exists
infinitely many solutions exist
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Inconsistent Systems
Inconsistent system- a system of equations that has no solutions.
(the lines are parallel)
no solution exists
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Independent and Dependent Equations
Independent Equations- the equations graph different lines
one solution exists
no solution exists
Dependent Equations- the equations graph the same line
infinitely many solutions exist
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Solving a System on the Calculator
Step 1: Graph each equation. (Use your calculator.)
4x – y = 9
x –3y = 16
*Make sure the equations are
in slope-intercept form!
Step 2: Find the coordinates of the point of intersection.
2nd
CALC
5: intersect
First curve?
Second curve?
(make sure cursor is
(make sure cursor on
the line for y2.)
on the line for y1.)
ENTER
ENTER
Guess?
(cursor to
where you feel the
intersection is)
ENTER
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Calculator Example Continued…
4 x  y  9

 x  3 y  16
Solution:
What type of system is
this?
What type of equations
are these?
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Example
Example: Solve the system by graphing on your calculator.
3x  5  2 y

3x  2 y  7
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Example with Dependent Equations
Example: Solve the system by graphing on your calculator.
5
 2 x  3 y  6

y   5 x  2

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Solving: Method of Substitution
1) Solve one of the equations for either x or y.
2) Substitute that result into the other equation to obtain an equation
in a single variable (either x or y).
3) Solve the equation for that variable.
4) Substitute this value into any convenient equation to obtain the
value of the remaining variable.
5) Check solution in BOTH ORIGINAL equations.
6) Write your solution as an ordered pair. If there is no solution, state
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that the system is inconsistent.
Example
Example: Solve the system using substitution and check.
x  3y  9

2 x  y  10
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Another Example
Example: Solve the system using substitution and check.
3x  2 y  7

 y  2x  3
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Solving: Method of Elimination
by Addition
1)
Write both equations in the general form Ax + By = C.
2)
Multiply the terms of one or both of the equations by
nonzero constants to make the coefficients of one variable
differ only in sign.
3)
Add the equations and solve for the variable.
4)
Substitute the value into one of the ORIGINAL equations to
find the value of the other variable.
5)
Check the solution in BOTH ORIGINAL equations.
6)
Write your solution as an ordered pair. If there is no solution,
state that the system is inconsistent.
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Addition Method Example
Example: Solve the system using addition method and check.
2 y  3x

2 x  7  y
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Another Addition Method Example
Example: Solve the system using addition method and check.
5 x  16  7 y

2 x  8 y  26
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Another Example
Example: Solve the system using addition method.
2 x  5 y  6

4 x  10 y  1
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Examples
Example: Solve the system using ANY method.
2 x  13 y  120

14 x  91y  840
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Examples
Example: Solve the system using ANY method.
x  2 y  8

x  4  2 y
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Special Cases Summary
When solving a system of two linear equations in two variables
algebraically:
If an identity is obtained, such as 0 = 0, then
the system has an infinite # of solutions.
The equations are dependent
 The system is consistent.

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Special Cases Summary
When solving a system of two linear equations in two variables
algebraically:
If a contradiction is obtained, such as 0 = 7,
then the system has no solution.
The system is inconsistent.
 The equations are independent.

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Application
A man walks at a rate of 3 miles per hour and jogs at a rate of 5
miles per hour. He walks and jogs a total distance of
3.5 miles in 0.9 hours. For how many minutes does the man
jog?
Solution:
Let x represent the _____________________________________
Let y represent the _____________________________________
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Application
(continued)
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Supply and Demand
The quantity of a product that people are willing to buy
during some period of time is related to its price.
Generally, the higher the price, the less the demand; the
lower the price, the greater the demand.
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Supply and Demand
(continued)
Similarly, the quantity of a product that a supplier is
willing to sell during some period of time is also related
to the price.
Generally, a supplier will be willing to supply more of a
product at higher prices and less of a product at lower
prices.
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Supply and Demand
(continued)
The simplest supply and demand model is a linear
model where the graphs of a demand equation and a
supply equation are straight lines.
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Supply and Demand
(continued)
In supply and demand problems we are often interested
in finding the price at which supply will equal demand.
This is called the equilibrium price, and the quantity
sold at that price is called the equilibrium quantity.
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Supply and Demand
(continued)
If we graph the the supply equation and the demand
equation on the same axis, the point where the two lines
intersect is called the EQUILIBRIUM POINT.
•Its horizontal coordinate is the value of the equilibrium
quantity (q).
•Its vertical coordinate is the value of the equilibrium
price (p).
(q, p)
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Supply and Demand
Example
Example: Suppose that the supply equation for long-life
light bulbs is given by
(supply)
p = 1.04 q - 7.03,
and that the demand equation for the bulbs is
(demand)
p = -0.81q + 7.5
where q is in thousands of cases and p represents the price
per bulb in dollars.
Find the equilibrium price and quantity.
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Supply and Demand Example
Method I: Solve the system algebraically
We want to find the price at which the supply is equal to the
demand.
We can do this by __________________________________
________________________________________________.
Supply: p = 1.04 q - 7.03
Demand: p = -0.81q + 7.5
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Supply and Demand Example
(continued)
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Supply and Demand
Example
Method II: Solve the system graphically
Graph the two equations in the same coordinate system using
a graphing calculator and find the ______________________
_________________________________________________.
p = 1.04 q - 7.03
y1= 1.04 x - 7.03
p = -0.81q + 7.5
y2= -0.81x + 7.5
Notice x represents the quantity and y represents the price.
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Supply and Demand
(Example continued)
If we graph the two equations on a graphing calculator and
find the intersection point, we see the graph below.
Demand Curve
Supply Curve
Thus the equilibrium point is (7.854, 1.14).
The equilibrium quantity is ____________ cases and the
equilibrium price is $____________ per bulb.
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Another Example of a
Linear System
 A restaurant serves two types of fish dinners- small for $5.99
each and large for $8.99. One day, there were 134 total orders
of fish, and the total receipts for these 134 orders was
$1024.66. How many small dinners and how many large
dinners were ordered?
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Break-Even Analysis
P 187 #56
A small plant manufactures riding lawn mowers. The plant has
fixed costs (leases, insurance, etc.) of $65,000 per day and variable
costs (labor, materials, etc.) of $1100 per mower produced. The
mowers are sold for $1,600 each.
A) Find the cost and revenue functions.
Let x = ____________________________________________
Cost function:
Revenue function:
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Break-Even Analysis
(continued)
B) How many units must be manufactured and sold each day for
the company to break even?
C) Write the profit function, P(x).
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Break-Even Analysis
(continued)
D) How many units must be manufactured and sold each day for
the company to make a profit?
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