System of Equations
Why should
you care
about System
of Equations?
Why should you care??
● Finance: Investment analysis, Portfolio analysis, Break-even analysis
● Project management: Allocation of resources to projects, Project scheduling
● Facility planning: Facility size & location, Number of facilities in logistics,
Distribution network
● Production / Manufacturing: Product mix, Production planning / scheduling,
Lot size, Inventory planning & control, Procurement mix
● Marketing: Sales promotional planning, Sales allocation to markets
● OB/ HR: Manpower planning, Scheduling of training programs
● R&D: Product introduction timing, Appropriate product design
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Sample Problem
A company produces inexpensive chairs and tables requiring labor hours for
carpentry department and painting department.
● Each table takes 3 hours of carpentry work and 2 hours of painting works.
● Each chair, on the other hand, requires 4 hours of carpentry work and 1
hours of painting works.
● Only 2400 hours of carpentry time and 1000 hours of painting time are
available during current month.
● Existing stock levels and demand pattern show that company should not
manufacture more than 450 new chairs and at least 100 new tables are to be
produced during this month.
● Expected profit for each table is $7 and that for each chair is $5.
What should be the best combination of production of chairs and tables for the
current month?
System of Equations
● The focus of the discussion has thus far concentrated on the analysis of
individual equations.
● Many mathematical models, however, involve more than one equation. The
equations of such models are said to form a system of equations.
● The solution to a system of equations is a set of values which concurrently
satisfy all the equations of the system.
● For a unique solution to a system of equations to exist, there must be at least
as many equations as variables.
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• A system with an equal number of equations and variables (n = υ) is an
exactly constrained system.
y = 5x + 7
y = 8x − 3
• An underconstrained system has fewer equations than variables (n <
υ).
y = 3x1 + 7x2 − 4x3
y = 9x1 − 5x2 + 2x3
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• An overconstrained system has more equations than variables (n > υ).
y = 6x − 12
y = 9x + 15
y = 7x − 11
• An underconstrained system will have an unlimited number of
solutions or no solution, never a unique solution. An exactly
constrained or overconstrained system may have a unique solution, an
infinite number of solutions, or no solution.
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Graphical method
• The graph of a (2 × 2) linear system of equations consists of two
straight lines.
• If the two lines intersect at some point (x1, y1), then the point (x1, y1)
satisfies both equations and the coordinates of the point represent a
unique solution to the system.
• For a (2 × 2) system of linear equations in the slope-intercept form
presented in equation:
y = m1x + b1
y = m2x + b2
If m1 ≠ m2, there is a unique solution to the system.
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Graphical method
• If the two lines do not intersect, there is no point in common that satisfies
both equations and hence no solution exists. The two equations are
inconsistent.
If m1 = m2, and b1 ≠ b2, the equations are inconsistent and there is no
solution to the system.
• Finally, if the two equations have identical graphs, they have an unlimited
number of points in common and the system has an infinite number of
solutions. Such equations are called equivalent or dependent equations.
If m1 = m2, and b1 = b2, the equations are equivalent and there is an
infinite number of solutions to the system.
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EXAMPLE
● The different systems of equations given below are solved by graphing as follows:
(a)
x − 2y = −2
6x + 3y = 33
(b)
10x + 2y = 30
15x + 3y = 75
(c)
21x − 7y = 14
−15x + 5y = −10
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Break-even analysis
In planning a new business, it is important to know the minimum
volume of sales needed to turn a profit.
Estimating the point at which a firm’s balance sheet moves from a loss
to a profit is called breakeven analysis.
The break-even point can be defined as the level of sales at which (1)
profit equals zero or (2) total revenue equals total cost.
EXAMPLE. Given total revenue R(x) = 80x, total cost C(x) = 30x + 2000,
then profit π(x) = R(x) − C(x) = 50x – 2000, and the breakeven point can be
found in any of the three following ways:
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(a) Setting π = 0 and solving for x,
50x − 2000 = 0
x = 40 Break-even level of output
(b) Setting R(x) = C(x) and solving for x,
80x = 30x + 2000
50x = 2000
x = 40 Break-even level of output
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(c) Graphing R(x) and C(x) and finding the point of intersection, as
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EXAMPLE
● Find the break-even point for the firms with the following linear total
revenue R(x) and total cost C(x) functions by
● R(x) = 50x
● C(x) = 35x + 90
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Graphing R(x) and C(x) and finding the point of intersection, as
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Using Graphical Method
A company produces inexpensive chairs and tables requiring labor hours for
carpentry department and painting department.
● Each table takes 3 hours of carpentry work and 2 hours of painting works.
● Each chair, on the other hand, requires 4 hours of carpentry work and 1
hours of painting works.
● Only 2400 hours of carpentry time and 1000 hours of painting time are
available during current month.
● Existing stock levels and demand pattern show that company should not
manufacture more than 450 new chairs and at least 100 new tables are to be
produced during this month.
● Expected profit for each table is $7 and that for each chair is $5.
What should be the best combination of production of chairs and tables for the
current month?
Formulate
An advertising agency wishes to reach two types of audiences – people with
annual income greater than one lakh (type A) and those with less than one
lakh (type B). The total advertising budget is Rs 2,00,000. One programme of
TV advertising costs Rs 50,000 and that of radio advertising costs Rs 20,000.
As per contract at least three number of programmes ought to be on TV and the
radio programmes must be limited to 5. Surveys indicate that a single TV
programme is supposed to reach 4,50,000 type A audience and 50,000 type B
audience. Whereas a single radio programme is expected to reach 20,000 type A
audience and 80,000 type B audience.
Determine the media mix to maximize the total reach.
Quadratic Equation
In mathematics, a quadratic equation is a polynomial equation
of the second degree.
The general form is:
ax2 + bx + c = 0
where x represents a variable or an unknown.
and a, b, and c are constants with a ≠ 0.
(If a = 0, the equation is a linear equation).
The constants a, b, and c are called,
a=the quadratic coefficient,
b=the linear coefficient and
Roots of quadratic equation
A real number α is called a root of the quadratic
equation ax2 + bx + c = 0, a≠0 if aα2 + bα + c = 0.
If α is a root of ax2 + bx + c = 0, then we say that:
(i) x= α satisfies the equation ax2+bx+c =0
or
(ii) x= α is a solution of the equation ax2+bx+c =0
The root of a quadratic equation ax2+bx+c=0 are
called zeros of the polynomial ax2+bx+c
Examples for Quadratic Equations
Other Examples of Quadratic Equation
How to solve quadratic equations?
1. Factorization
Factorization is typically one of the easiest and
quickest ways to solve the quadratic equations as
long as the binomial or trinomial is easily factorable.
However, not all quadratic polynomials can be
factored.
This means that factorization will not work to solve
many quadratic equations.
1. Factorization
Example 1 Example 2
x2 – 2x – 24 = 0 x2 – 8x + 11 = 0
(x + 4)(x – 6) = 0 11 is prime; therefore,
x+4=0 x–6=0 another method must be
used to solve this
x = –4 x=6 equation.
2. Square Root Property
We previously have used factoring to solve
quadratic equations.
Here we will introduce additional methods for
solving quadratic equations.
Square Root Property
If b is a real number and a2 = b, then
a b
Square Root Property Examples
Solve x2 = 49
x 49 7
Solve 2x2 = 4
x2 = 2
x 2
Solve (y – 3)2 = 4
y 3 4 2
y=32
y = 1 or 5
Square Root Property Examples
Solve (x + 2)2 = 25
x 2 25 5
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
Square Root Property Examples
Solve (3x – 17)2 = 28
3x – 17 = 28 2 7
3x 17 2 7
17 2 7
x
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3. Completing the Square
What constant term should be added to the
following expressions to create a perfect square
trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
2
7 49
add
2 4
Completing the Square
We now look at a method for solving
quadratics that involves a technique called
completing the square.
It involves creating a trinomial that is a
perfect square, setting the factored
trinomial equal to a constant, then using the
square root property from the previous
section.
Completing the Square
If the coefficient of x2 is NOT 1, divide both
sides of the equation by the coefficient.
Isolate all variable terms on one side of the
equation.
Complete the square (half the coefficient of the
x term squared, added to both sides of the
equation).
Factor the resulting trinomial.
Use the square root property.
Solving Equations
Solve:
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y+3=± = ± 11
y = 3 ± 1
y = 4 or 2
Solving Equations
Solve:
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
29
(y + ½) =
2
4
1 29 29
y
2 4 2
1 29 1 29
y
2 2 2
4. The Quadratic Formula
Another technique for solving quadratic
equations is to use the quadratic formula.
The formula is derived from completing the
square of a general quadratic equation.
The Quadratic Formula
A quadratic equation written in
standard form, ax2 + bx + c = 0, has the
solutions
2
b b 4ac
x
2a
Discriminant
The Quadratic Formula
Solve 11n2 – 9n = 1 by the quadratic
formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
9 (9) 2 4(11)(1) 9 81 44 9 125
n
2(11) 22 22
95 5
22
The Quadratic Formula
1 2 5
Solve 8 x + x – 2 = 0 by the quadratic formula.
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
8 (8) 2 4(1)(20) 8 64 80 8 144
x
2(1) 2 2
8 12 20 4
or , 10 or 2
2 2 2
The Discriminant
The expression under the radical sign in
the formula (b2 – 4ac) is called the
discriminant.
The discriminant will take on a value that
is positive, 0, or negative.
The value of the discriminant indicates two
distinct real solutions, one real solution, or
no real solutions, respectively.
NATURE OF THE ROOTS OF QUADRATIC EQUATION
b2 – 4ac < 0
The Discriminant
Use the discriminant to determine the number and
type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.
Solving Equations
Solve 12x = 4x2 + 4.
0 = 4x2 – 12x + 4
0 = 4(x2 – 3x + 1)
Let a = 1, b = -3, c = 1
3 (3) 2 4(1)(1) 3 9 4 3 5
x
2(1) 2 2
Factorization method
6x2 − 48x − 54 = 0
x2 - 2x − 24 = 0
x2 + 6x − 10 = 30
Completing the square method
x2 + 2x − 24 = 0
p2 + 12p − 54 = 0
x2 − 8x + 15 = 0
3m2 − 16m − 2 = −7
m2 + 16m − 32 = −7
Sum and Product of the roots
1. Sum of the roots: α = +
∴ α + β = − b/a = - coefficient of x / coefficient of x2
2. Product of the roots: α =( ) ( )
=
= = = constant term
/ coefficient of x2
Sum and Product of the roots
Find the sum and product of roots of the quadratic equation given below.
x2 - 5x + 6 = 0
Solution :
Comparing
x2 - 5x + 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = -5 and c = 6
Therefore,
Sum of the roots = -b/a = -(-5)/1 = 5
Product of the roots = c/a = 6/1 = 6
Sum and Product of the roots
Find the sum and product of roots of the quadratic equation given below.
x2 - 6 = 0
Solution :
Comparing
x2 - 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = 0 and c = -6
Therefore,
Sum of the roots = -b/a = 0/1 = 0
Product of the roots = c/a = -6/1 = -6
Using roots of the equation to form quadratic equation
When two roots of a quadratic equation are given ,
the formula to form the quadratic equation is given by
x² - (sum of the roots)x + product of the roots = 0
If ∝ and ᵦ be the two roots of a quadratic equation are
given , then the formula to form the quadratic
equation is given by
x² - (α + β) x + αβ = 0
Using roots of the equation to form quadratic equation
Construct a quadratic equation whose two roots are -10 and -3
α + β = -10 + (-3) α β = -10(-3)
= 30
= -10 - 3
= -13
Solution :
Roots are α = - 10 and β = -3
x² - (-13) x + 30 = 0
x² + 13 x + 30 = 0
Solution
Problem
Solution
Solution
Problem
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