# Algebraic Expressions and Inequalities Exercise Notes 2021 Download Study Materials BOOK PDF

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Applications of Linear Equations With One Variables

Example 2:

The sum of the digits of a two digit number is 16, If the number formed by reversing the digits is less than the original number by 18. Find the original number.

Solution:

Let unit digit be x.

Then tens digit = 16- x

∴Original number =10 × (16 – x)+x = 160 -9x.

On reversing the digits, we have x at the tens place and (16-x)at the unit place.

∴New number=10x+(16-x) = 9x+16

Original number-New number = 18

(160-9x) – (9x+16)= 18

160-18x-16=18

-18x+144 = 18

– 18x = 18 – 144           ⇒18x=126    ⇒ x = 7

∴In the original number, we have unit digit = 7

Tens digit = (16 – 7) = 9

Thus, original number=97

Linear equation in two variables: General equation of a linear equation in two variables is ax + by + c = 0, where a, b 0 and c is, a constant,’ and x and y are the two variables

The sets of values of x and y satisfying any equation are called its solution(s).

Consider the equation 2x+ y=4. Now, if we substitute x = – 2 in the equation, we obtain 2.(-2) +y = 4or-4 + y = 4 or y = 8. Hence (-2,8) is a solution. If we substitute x = 3 in the equation, we obtain 2.3 + y = 4 or 6 + y = 4 or y = -2

Hence (3, -2) is a solution. The following table lists six possible values for x and the corresponding values for y, i.e. six solutions of the equation.

If we plot the solutions of the equation 2x+y – 4which appear in the above table then we see that they all lie on the same line. We call .this line the graph of the equation since it corresponds precisely to the solution set of the equation.

Elimination Method:

4x + 3y = 25                            ….. (i)

x + 5y =19                               ….. (ii)

Multiply equation (ii) by 4 on both sides. We find 4x+20y =76

Subtracting equation (i) from equation (iii), we           have

4x+ 20y =76

4x + 3y=25

−   −  −

17 y = 51

⇒ y =  = 3

Substituting value of y in equation (i), we get 4x + 3 × 3 = 25

4x =16 ⇒ x =  = 4

∴ x=4 and y=3 is the solution.

Now consider two linear equations in two unknowns,

a1x + b1y = c1               ….. (i)

a2x + b2y = c2               ….. (ii)

The above equations arc nothing else but equations of 2 lines. Any pair (x, y) which satisfy both the equation is called a solution to the above system of equations.

Systems of Linear Equation

Consistent System: A system (of 2 or.3 or more equations taken together) of linear equations is said to be consistent, if it has at least one solution.

Inconsistent System: A system of simultaneous linear equations is said to be inconsistent, if it has no solutions at all

e.g. X+Y = 9;   3X+3Y = 8

Clearly there are no values of X & Y which simultaneously satisfy the given equations. So the system is inconsistent.

An equation of the degree two of one variable is called quadratic equation.

General form: ax2+bx+c =0….. (l) where a,b and c are all real number and a ≠ 0.      For Example:

2x2–5x+3=0; 2x2–5=0; x+ 3x = 0

A quadratic equation gives two and only two values of the unknown variable and both these values are called the roots of the equation.

Nature of Roots: The nature of roots of the equation depends upon the nature of its discriminant D.

1. If D < 0, then the roots are non-real complex, Such roots are always conjugate to one another. That is, if one root is p + iq then other is p-iq,q ≠ 0.
2. If D = 0, then the roots are real and equal. Each root of the equation becomes . Equal roots are referred as repeated roots or double roots also:
3. If D > 0 then, the roots are real and unequal.
4. In particular, if a, b, c are rational number, D> 0 and D is a perfect square, then the roots of the equation are rational number and unequal.
5. If a, b, c, are rational number, D>0 but D is not a perfect square, then the roots of the equation are irrational (surd). Surd roots are always conjugate to one another, that is if one root is then the other is ,q>0.
6. If a = 1, b and c are integers, D > 0 and perfect square, then the roots of the equation are integers.

Sign of Roots:

Let  are real roots of the quadratic equation ax2+bx + c = 0 that is D = b2-4ac≤0. Then

1. Both the roots are positive if a and c have the same sign and the sign of b is Opposite.
2. Both the roots are negative if a, b and c all have the same sign.
3. The Roots have opposite sign if sign of a and c are opposite.
4. The Roots are equal in magnitude and opposite in sign if b = 0 [that is its roots and ]
5. The roots are reciprocal if a = c.

[that Is the roots are α and ¹⁄∝]

1. If c = 0. then one root is zero.
2. If b = c = 0. then both the roots are zero.
3. If a = 0, then one root is infinite.
4. If a = b = 0, then both the roots are infinite.
5. If a = b = c = 0, then the equation becomes an identity
6. If a + b + c =0 then one root is always unity and the other

root is , Hence the roots are rational provided a, b, c, are rational.

Symmetric Functions of Roots:

An expression in is called asymmetric function of α, β if the function is not affected by inter changing  and . If  are the roots of the quadratic equation ax2+bx + c = 0, a ≠ 0 t

Formation of Quadratic Equation With Given Roots

• An equation whoso roots are and  can be written as (x – )(x – ) = 0 or x2– x + = 0 or x2-(sum of the roots) x. + product of the roots = 0.
• Further If and  are the roots of a quadratic equation ax2 + bx + c = 0, then

ax2+bx+c = a(x- )(x- ) is an identity.

A number of relations between the roots can be derived using this identity by substituting suitable values of x real or imaginary.

Condition of a Common Root between two quadratic equations:

a1x2 + b1x + c1 = 0                    ….. (i)

and      a2x2 + b2x + c2 = 0                    ….. (ii)

Let  be a common root of the two equations

Then  and

On solving we get

Example 13:

If a, b are two roots of a quadratic equation    such that a + b = 24 and a – b = 8, then find a quadratic equation having a and b as its roots.

Solution:

a + b = 24 and a – b = 8

⇒ a = 16 and b = 8 ⇒ 16 × 8 = 128

A quadratic equation with roots a and b is

x2 – (a + b) x + ab = 0  or x2 – 24x + 128 = 0

In equations: A statement or equation which states that one thing is not equal to another, is called an in equation.

Symbols:

‘<’ means “is less than”

‘>’ means “is greater than”

‘≤’ means “is less than or equal to”

‘≥’ means “is greater than or equal to”

For example:

• x < 3 means x is less than 3.
• y ≥ 9 means y is greater than or equal to 9.

PROPERTIES

1. Adding the same number to each side of an equation does not effect the sign of inequality, it remains same, i.e. if x > y then, x + a > y + a.
2. Subtracting the same number to each side of an inequation does not effect the sign of inequality, i.e. if x < y then, x – a < y – a.
3. Multiplying each side of an inequality with same number does not effect the sign of inequality, i.e., if x ≤ 1 then ax ≤ ay (where, a> 0).
4. Multiplying each side of an inequality with a negative number effects the sign of inequality or sign of inequality reverses, i.e., if x < y then ax > ay (where a < 0)
5. Dividing each side of an inequation by a positive number does not effect   the sign of inequality, i.e., if x ≤ y then (where, a> 0).
6. Dividing each side of an inequation by a negative number reserves the sign of inequality, i.e., if x > y then (where, a < 0).

Modulus:

1. If a is positive real number, x and y be the fixed real numbers, then

(i) |x – y| < a ⇔ y –a < x < y + a

(ii) |x – y| ≤ a ⇔ y –a ≤ x ≤ y + a

(iii) |x – y| > a ⇔ x> y + a or x < y – a

(iv) |x – y| ≥ a

⇔ x ≥ y + a or x ≤ y – a

1. Triangle Inequality:

(i) |x + y| ≤ |x| + |y|, ∀ x, y ∈ R

(ii) |x – y| ≥ |x| – |y|, ∀ x, y ∈ R

IMPORTANT SERIES TYPE FORMULA

Value of √(P+√(P+√(P+⋯∞)) ) =(√(4P+1)+1)/2
Value of √(P-√(P-√(P-…∞)) ) =(√(4P+1)-1)/2
Value of √(P.√(P.√(P…∞)) ) =P
Value of √(P√(P√(P√(P√P) ) ) ) = P^((2^n-1)+2^n )
Where n ¬→ no. of times P repeated.

SOME SPECIAL SERIES

• Sum of first n natural numbers 1 + 2 + 3 +…n = ((n)(n+1))/2
• Sum of the squares of first n natural numbers 12 + 22 + 32 + …. b2 = ((n)(n+1)(2n+1))/6
• Sum of the cubes of first n natural numbers 13 + 23 + 33 + … n3 = (((n)(n+1))/2)^2