Systems of Linear Equations in Two Variables

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7.3
SYSTEMS OF LINEAR EQUATIONS IN
TWO VARIABLES
OBJECTIVES
1.
2.
3.
4.
5.
6.
Decide whether an ordered pair is a solution of a
linear system.
Solve linear systems by graphing.
Solve linear systems by substitution.
Solve linear systems by addition.
Identify systems that do not have exactly one
ordered-pair solution.
Solve problems using systems of linear equations.
SYSTEM OF LINEAR EQUATIONS IN 2
VARIABLES

System of Linear Equations:
-- 2 linear equations


Solution to a System of Equations:
-- an ordered pair that satisfy both equations


E.g., 2x – 3y = -4
2x + y = 4
E.g., (1, 2), or, x = 1 and y = 2 is a solution to the
system
Check
2x – 3y = -4
2(1) – 3(2) = -4
2 – 6 = -4 true
2x + y = 4
2(1) + 2 = 4
2 + 4 = 4 true
GRAPHICAL INTERPRETATION OF THE
SOLUTION

2x – 3y = -4
2x + 4 = 3y
y = (2/3)x + (4/3)

2x + y = 4
y = -2x + 4
(1, 2)
EXAMPLE

Solve by graphing the system of equations.
x + 2y = 2 an x – 2y = 6

Find x-intercept and y-intercept for both equations.
x + 2y = 2
x – 2y = 6
when x = 0,
0 + 2y = 2
y=1
when x = 0,
0 – 2y = 6
y = -3
when y = 0
x + 2(0) = 2
x=2
when y = 0,
x – 2(0) = 6
x=6
(0, 1) -- y-intercept
(2, 0) -- x-intercept
(0, -3) -- y-intercept
(6, 0) -- x-intercept
EXAMPLE (CONT.)
• Lines intersect at (4, -1).
• Check:
x + 2y = 2
4 + 2(-1) = 2
4–2=2
true
x – 2y = 6
4 – 2(-1) = 6
4+2=6
true
SOLVING EQUATION SYSTEM BY
SUBSTITUTION
Solve:
y = -x – 1
4x – 3y = 24
 Solution

Substitute the expression for y in the first
equation for y in the second equation.
4x – 3y = 24
4x – 3(-x – 1) = 24
4x + 3x + 3 = 24
7x = 21
x=3
y = -x - 1
y = -(3) - 1
y = -4
Solution: (3, -4)
SOLVING A LINEAR SYSTEM BY SUBSTITUTION
 Solve
y = -x – 1
4x – 3y = 24
 This
gives us: 4x – 3(−x – 1) = 24.
Solving for x, we get: x = 3
 Substitute x value back in the first equation.
y = -(3) – 1
 This gives us: y = -4
 Solution: (3, -4)
EXAMPLE
Solve the linear system.
-4x + y = -11
2x – 3y = 3
 Solution
1. Using one equation, express y in terms of x.
-4x + y = -11
y = 4x - 11
2. Substitute this in the second equation.
2x – 3y = 3
2x – 3(4x – 11) = 3
3. Solve for x
2x – 12x + 33 = 3
-10x = -30
x=3

EXAMPLE (CONT.)
4. Substitute this value of x in the first equation
and solve for y.
y = 4x – 11
y = 4(3) – 11
y = 12 – 11
y=1
 Solution: (3, 1)
YOUR TURN
Solve the linear system
y = 2x + 7
2x – y = -5
 Solution
2x – y = -5
2x – (2x + 7) = -5
2x – 2x + 7 = -5
7 = -5
 What does this mean?
 Check the slopes of the 2 lines.

SOLVING A LINEAR SYSTEM BY ADDITION
Solve:
3x + 2y = 48
9x – 8y = -24
1. The idea is to eliminate either the x column or
the y column and add the two equations.
4(3x + 2y) = 4(48)
9x – 8y = -24
2. 12x + 8y = 192
9x – 8y = -24
3. 21x = 168
4. x = 8

SOLVING A LINEAR SYSTEM BY ADDITION
Substitute this value of x in either equation and
solve for y.
3x + 2y = 48
3(8) + 2y = 48
24 + 2y = 48
2y = 24
y = 12
 Solution: (8, 12)

SPECIAL CASES
Number of Solutions
Graphically
One ordered-pair solution
Two lines intersect at one point.
No solution
Two lines are parallel.
Infinitely many solutions
Two lines are the same line.
A SYSTEM WITH NO SOLUTION
Solve:
4x + 6y = 12
6x + 9y = 12
 Using the addition method,
4x + 6y = 12 Multiply by 3
6x + 9y = 12 Multiply by -2

0 = 12
false.
There is no solution to the system.
A SYSTEM WITH INFINITELY MANY SOLUTIONS
Solve:
y = 3x – 2
15x – 5y = 10
 Using the substitution method,
15x – 5y = 10
15x – 5(3x – 2) = 10
15x – 15x + 10 = 10
10 = 10

This is true for any (x, y) pairs.
 Thus, there is an infinitely number of solutions.
MODELING WITH SYSTEMS OF EQUATIONS
Suppose a company produces and sells x iGizmos.
 Revenue function: R(x) = (price per unit sold)x
 Cost function:
C(x) = fixed cost
+ (cost per unit produced)x
 Break-even point: intersection of R(x) and C(x)

R(x) = fixed cost + (price per unit sold) x
Dollars
C(x) = fixed cost + (price per unit produced) x
Break-even point
x (iGizmo units)
FINDING A BREAK-EVEN POINT
A company plans to manufacture electronic
age wheelchairs. Fixed cost will be
$500,000, and the production cost for each
wheelchair is $400. The chairs will be sold
at $600 apiece.
 Write
the Cost function C(x).
 Write the Revenue function R(x).
 Graph the functions.
 Determine the beak-even point.
FINDING A BREAK-EVEN POINT
Cost function:
 C(x) = 500,000 + 400x
Revenue function:
 R(x) = 600x
 Graph
FINDING A BREAK-EVEN POINT
 Break-even
Point
C(x) = 500,000 + 400x
R(x) = 600x
or
y = 500,000 + 400x
y = 600x
600x = 500,000 + 400x
200x = 500,000
x = 2500
BREAK-EVEN POINT
x = 2500
y = 600x
y = 600(2500)
= 1,500,000
Thus,
Break-even point: (2500, 1,500,000)
I.e.,
$1,500,000 with 2500 units sold.
YOUR TURN
 The
profit function
P(x) = R(x) – C(x)
 For the preceding case,
P(x) = 600x – (500,000 + 400x)
P(x) = 200x – 500,000
1. Sketch the graph of the profit function.
2. What is the y-intercept of the function? How do
you interpret that value?
3. What is the slope of the function? How do you
interpret that value?
4. What is x-intercept of the function? How do you
interpret that value?
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