2. Introduction
In mathematics, an equation is a statement that asserts the equality of
two expressions. The word equation and its cognates in other languages may have
subtly different meanings; for example, in French an équation is defined as
containing one or more variables, while in English any equality is an equation.
Solving an equation containing variables consists of determining which values of the
variables make the equality true. Variables are also called unknowns and the values
of the unknowns that satisfy the equality are called solutions of the equation. There
are two kinds of equations: identities and conditional equations. An identity is true for
all values of the variable. A conditional equation is only true for particular values of
the variables.
An equation is written as two expressions, connected by an equals sign ("="). The
expressions on the two sides of the equals sign are called the "left-hand side" and
"right-hand side" of the equation.
3. Objectives
After completing this unit you should able to:
Solve single linear equations with one unknown
variable,
Solve sets of simultaneous linear equations with
two or more variable using the substitution and
row operations methods; and
Solve quadratic equations with different
methods.
4. Equations
Types:
1. Linear Equations
2. Simultaneous Equations
3. Quadratic Equations
Linear Equations
Simultaneous Equations
Quadratic Equations
A statement of equality
between two quantities is
called an Equations.
Meaning:
5. 1. Linear Equations
Example: Solve the equations: 5x + 6 – 7(x -2) =5x
The given equation can also be written as,
5x + 6 – 7x + 14 =5x
or - 7x = - 20
or x = 20 = 2.857
7
x = 20 = 2.857
7
Equations in which maximum (or greatest) power of the unknowns is only one
are called Linear Equations
6. 2. Simultaneous Equations
If an equation contains two unknown quantities say x and y, then by giving
definite values to one of the unknown quantities a corresponding value for
another can be obtained called Simultaneous Equations
Example: Solve the equations: 3x -2y = 5 …(1)
5x –y =3 …(2)
X = 1 , Y = 16
7 7
Multiply (1) by 5 and (2) by 3
.
. . 5(3x -2y) =5x5 =15x -10y = 25
3(5x –y ) = 3x3 =15x -3y = 9
-7y = 16 on Subtracting
Or y = -16 , Substitute value of y in (1)
7
3x – 2(-16) = 5, or x = 1
7 7
7. We follow another method to solve such equations,
Let 1 = A and 1 =B. so that the equations become,
x y
5A -8B =1…..(3)
4A -15B =2….(4)
[Now, we follow the usual method,
Multiply (3) by 4 and (4) by 5]
4(5A -8B = 1) = 20A -32B =4
5(4A -15B =2) =20A -75B =10
43B = -6 on subtracting
B = - 6, but B = 1
43 y
Substituting values we get, x= -43
and y = -43
6
Example: Solve 5 - 8 = 1 ……(1)
x y
4 - 15 = 2 ……(2)
x y
8. A quadratic equation is one that can be written in the form ax² + bx + c = 0
where x is an unknown variable and a, b and c are constant quantities.
For example, 6x² + 2.5x + 7 = 0 A quadratic equation that includes terms in
both x and x² cannot be rearranged to get a single term in x, so we cannot use
the method used to solve linear equations. There are three possible methods
one might try to use to solve for the unknown in a quadratic equation:
(i) By factorization
(ii) By completing the square
(iii) Using the quadratic ‘formula’
3. Quadratic Equations
9. (i) By factorization
Two values x =12 and x =4,
both would satisfy the given
equations
x² -16x + 48 =0
Example: Solve x² -16x + 48 =0
We find the factors of the expressions on LHS of equality.
x² -16x + 48 =0 is of the form x² + px + q
.
. . x² -16x + 48 = x² - 12x- 4x + 48
= x(x -12) – 4(x -12)
= ( x- 4)(x -12) are the factors.
The given equations is ( x- 4)(x -12) = 0
Since the product of two factors is Zero,
.
. . Either (x -12) =0 or (x -4) = 0
(x -12) =0, then x =12
(x -4) = 0, then x =4
10. Example: Solve x² -16x + 48 =0
(ii) By completing the square
L.H.S = x² -16x + 48 =0
= {x² -2(8)(x) + (8)²} -64 +48
= (x² -16x + 64) -16
= (x -8)² -16
Given equations becomes:
(x- 8)² - 16 = 0
or (x- 8)²= 16
or (x-8) = ±√16 = ± 4
Either x – 8 = ± 4
x = 12, 4
Solution of equations is
x = 12 or 4
11. (iii) By use of quadratic formula
Using the quadratic ‘formula’
12. The part of the quadratic formula under the radical sign (b2-4ac) is called the
discriminant.
If b2-4ac is positive (>0), then the equation has two real solutions.
If b2-4ac is zero, then the equation has one solution.
If b2-4ac is negative (<0), then the equation has no real solutions (only
imaginary).
13. Example: x² -16x + 48 =0
Comparing the given equations with that of general
Form of quadratic equations : ax² + bx + c = 0
We find,
a =1, b = -16 and c = 48
x = 16±√(-16)² - 4(1)(48)
2 x1
= 16± √64
2
x = 2 or 4
(iii) By use of quadratic formula
14. Unit End Questions
1. Exercises Solve the following pairs of simultaneous equations:
a. 7x + y = 25, 5x − y = 11
b. 8x + 9y = 3, x + y = 0
c. 2x + 13y = 36, 13x + 2y = 69
d. 7x − y = 15, 3x − 2y = 19
2. Exercises Solve the following pairs of Quadratic equations:
a. x² + 2x - 11 =0
b. x² -6x + 5 =0
c. 3x² -15x + 11 =0
15. Required Readings
David Huang :Introduction to the use of Mathematics in Economic
Analysis
Mehta, B.C. and Madnani ;Arthashastra me Prarmbhik Ganit
https://en.wikipedia.org/wiki/Equation
https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.dream
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fe&ved=0CAIQjRxqFwoTCLCGp46HtOsCFQAAAAAdAAAAABA
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