ICAI Notes- Equations | Quantitative Aptitude for CA Foundation PDF Download

EQUATIONS

LEARNING OBJECTIVES
After studying this chapter, you will be able to:

• Understand the concept of equations and its various degrees – linear, simultaneous, quadratic and cubic equations;
• Know how to solve the different equations using different methods of solution; and
• Know how to apply equations in co-ordinate geometry.

2.1 INTRODUCTION

Equation is defined to be a mathematical statement of equality. If the equality is true for certain value of the variable involved, the equation is often called a conditional equation and equality sign ‘=’ is used; while if the equality is true for all values of the variable involved, the equation is called an identity.

is an identity since it holds for all values of the variable x.
Determination of value of the variable which satisfies an equation is called solution of the equation or root of the equation. An equation in which highest power of the variable is 1 is called a Linear (or a simple) equation. This is also called the equation of degree 1. Two or more linear equations involving two or more variables are called Simultaneous Linear Equations. An equation of degree 2 (highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation.

For Example: 8x+17(x–3) = 4 (4x–9) + 12 is a Linear equation
3x² + 5x +6 = 0 is a quadratic equation.
4x³ + 3x² + x–7 = 1 is a Cubic equation.
x+2y = 12x+3y = 2 are jointly called simultaneous equations.

2.2 SIMPLE EQUATION

A simple equation in one unknown x is in the form ax + b = 0.
Where a, b are known constants and a¹0
Note: A simple equation has only one root.

Solution: By transposing the variables in one side and the constants in other side we have

Illustrations: 

1 . The denominator of a fraction exceeds the numerator by 5 and if 3 be added to both the
fraction becomes   Find the fraction

Let x be the numerator and the fraction be  . By the question   4x+12 = 3x+24 or x = 12

The required fraction is  

2. If thrice of A’s age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A’s present age.

Let x years be A’s present age. By the question
2x–3(x–6) = x
or 2x–3x+18 = x
or –x+18 = x
or 2x = 18
or x=9
∴ A’s present age is 9 years.

3. A number consists of two digits the digit in the ten’s place is twice the digit in the unit’s place. If 18 be subtracted from the number the digits are reversed. Find the number.
Let x be the digit in the unit’s place. So the digit in the ten’s place is 2x. Thus the number becomes 10(2x)+x. By the question

20x+x–18 = 10x + 2x
or 21x–18 = 12x
or 9x = 18
or x = 2
So the required number is 10 (2 × 2) + 2 = 42.

4. For a certain commodity the demand equation giving demand ‘d’ in kg, for a price ‘p’ in rupees per kg. is d = 100 (10 – p). The supply equation giving the supply s in kg. for a price p in rupees per kg. is s = 75(p - 3). The market price is such at which demand equals supply. Find the market price and quantity that will be bought and sold.
Given d = 100(10 – p) and s = 75(p – 3).
Since the market price is such that demand (d) = supply (s) we have
100 (10 – p) = 75 (p – 3) or 1000 – 100p = 75p – 225

So market price of the commodity is Rs. 7 per kg.
∴ the required quantity bought = 100 (10 – 7) = 300 kg.
and the quantity sold = 75 (7 – 3) = 300 kg.

2.3 SIMULTANEOUS LINEAR EQUATIONS IN TWO UNKNOWNS

The general form of a linear equations in two unknowns x and y is ax + by + c = 0 where a b are non-zero coefficients and c is a constant. Two such equations a₁x + b₁y + c₁ = 0 and a₂ x + b₂ x + c₂ = 0 form a pair of simultaneous equations in x and y. A value for each unknown which satisfies simultaneously both the equations will give the roots of the equations.

2.4 METHOD OF SOLUTION

1. Elimination Method: In this method two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknowns and then solving for the other unknown.
Example 1: Solve: 2x + 5y = 9 and 3x – y = 5.

Solution: 2x + 5y = 9 …….. (i)
3x – y = 5 ………(ii)
By making (i) x 1,  2x + 5y = 9
and by making (ii) x 5, 15x – 5y = 25
__________________________________
Adding 17x = 34 or x = 2. Substituting this values of x in (i) i.e. 5y = 9 – 2x 

we find,
5y = 9 – 4 = 5 

∴y = 1 

∴x = 2, y = 1.

2. Cross Multiplication Method: Let two equations be:

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

We write the coefficients of x, y and constant terms and two more columns by repeating the coefficients of x and y as follows:

Example 2: 3x + 2y + 17 = 0
5x – 6y – 9 = 0

Solution: 3x + 2y + 17 = 0 ....... (i)
5x – 6y – 9 = 0 ........(ii)
Method of elimination: By (i) x³ we get 9x + 6y + 51 = 0 ...... (iii)
Adding (ii) & (iii) we get 14x + 42 = 0

Putting x = –3 in (i) we get 3(–3) + 2y + 17 = 0
or, 2y + 8 = 0 or, y =  

So  x = –3  and  y = –4
Method of cross-multiplication: 3x + 2y + 17 = 0
5x – 6y – 9 = 0

2.5 METHOD OF SOLVING SIMULTANEOUS LINEAR EQUATION WITH THREE VARIABLES

Example 1: Solve for x, y and z:
2x–y + z = 3 x + 3y – 2z = 11 3x – 2y + 4z = 1

Solution: (a) Method of elimination
2x – y + z = 3 .......(i)
x + 3y – 2z = 11 .... (ii)
3x – 2y + 4z = 1 .... (iii)
By (i) × 2 we get
4x – 2y + 2z = 6 …. (iv)
By (ii) + (iv), 5x + y = 17 ….(v) [the variable z is thus eliminated]x
By (ii) × 2,  2x + 6y – 4z = 22 ….(vi)
By (iii) + (vi), 5x + 4y = 23 ....(vii)
By (v) – (vii), –3y = – 6 or y = 2
Putting y = 2 in (v) 5x + 2 = 17, or  5x = 15 or,  x = 3
Putting x = 3 and y = 2 in (i)
2 × 3 – 2 + z = 3
or 6–2+z = 3
or 4+z = 3
or z = –1
So x = 3, y = 2, z = –1 is the required solution.
(Any two of 3 equations can be chosen for elimination of one of the variables)

(b) Method of cross multiplication

We write the equations as follows:
2x – y + (z – 3) = 0
x + 3y + (–2z –11) = 0
By cross multiplication

Substituting above values for x and y in equation (iii) i.e. 3x - 2y + yz = 1, 

we have

or 60–3z–10z–38 + 28z = 7
or 15z = 7–22 or 15z = –15 or z = –1

Thus x = 3, y = 2, z = –1
Example 2: Solve for x, y and z :

u+v+w = 5 ........ (i)
2u–3v–4w = –11........ (ii)
3u+2v–w = –6 ........ (iii)
By (i) + (iii) 4u+3v = –1 ........ (iv)
By (iii) x 4 12u+8v–4w = –24 ......... (v)
By (ii) – (v) –10u–11v = 13
or 10u + 11v = –13 .......... (vi)
By (iv) × 11 44x+33v = –11 …..…(vii)
By (vi) × 3 30u + 33v = –39 ……..(viii)
By (vii) – (viii) 14u = 28 or u = 2
Putting u = 2 in (iv) 4 × 2 + 3v = –1
or 8 + 3v = –1
or 3v = –9 or v = –3
Putting u = 2, v = –3 in (i) or 2–3 + w = 5
or –1 + w = 5 or w = 5+1 or w = 6

Example 3:   Solve for x y and z:

Solution: We can write as

2.6 PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS

Illustrations :
1. If the numerator of a fraction is increased by 2 and the denominator by 1 it becomes 1.
Again if the numerator is decreased by 4 and the denominator by 2 it becomes 1/2 . Find the fraction

Solution: Let x/y be the required fraction.

Thus x + 2 = y + 1 or x – y = –1 ......... (i)
and 2x – 8 = y–2 or 2x – y = 6 ......... (ii)
By (i) – (ii) –x = –7 or x = 7
from (i) 7–y = –1 or y = 8
So the required fraction is 7/8.

2. The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man?

Solution: Let x years be the present age of the man and sum of the present ages of the two sons be y years.
By the condition x = 3y .......... (i)
and x + 5 = 2 ( y+5+5) ..........(ii)
From (i) & (ii) 3y + 5 = 2 (y+10)
or 3y + 5 = 2y + 20
or 3y – 2y = 20 – 5
or y = 15
∴ x = 3 × y = 3 × 15 = 45
Hence the present age of the main is 45 years

3. A number consist of three digit of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297 find the number.

Solution: Let the number be 100x + y. we have x + y = 9……(i)
Also 100y + x = 100x + y + 297 …………………………….. (ii)
From (ii) 99(x – y) = –297
or x – y = –3 ….……………………………………………… (iii)

Adding (i) and (ii) 2x = 6 ∴ x = 3 ∴ from (i) y = 6
∴ Hence the number is 306.

2.7 QUADRATIC EQUATION

An equation of the form ax² + bx + c = 0 where x is a variable and a, b, c are constants with a ≠ 0 is called a quadratic equation or equation of the second degree.
When b=0 the equation is called a pure quadratic equation; when b ¹ 0 the equation is called an adfected quadratic.
Examples: i) 2x² + 3x + 5 = 0
ii) x² – x = 0
iii) 5x² – 6x –3 = 0
The value of the variable say x is called the root of the equation. A quadratic equation has got two roots.
How to find out the roots of a quadratic equation:
ax² + bx +c = 0 (a ≠ 0)

Let one root be and the other root be β

2.8 HOW TO CONSTRUCT A QUADRATIC EQUATION

For the equation ax² + bx + c = 0 we have

or x² – (Sum of the roots) x + Product of the roots = 0

2.9 NATURE OF THE ROOTS

i) If b²>4ac = 0 the roots are real and equal;
ii) If b²>4ac >0 then the roots are real and unequal (or distinct);
iii) If b²>4ac <0 then the roots are imaginary;
iv) If b²>4ac is a perfect square ( ≠ 0) the roots are real, rational and unequal (distinct);
v) If b²>4ac > 0 but not a perfect square the rots are real, irrational and unequal.
Since b² – 4ac discriminates the roots b² – 4ac is called the discriminant in the equation ax² + bx + c = 0 as it actually discriminates between the roots.

Note: (a) Irrational roots occur in pairs that is if   is a root then    is the other root of the same equation.

(b) If one root is reciprocal to the other root then their product is 1 and so   = 1 i.e. c = a

(c) If one root is equal to other root but opposite in sign then. their sum = 0 and so   = 0. i.e. b = 0. 

Example 1 : Solve x² – 5x + 6 = 0

Solution: 1st method : x² – 5x + 6 = 0
or x² –2x –3x +6 = 0
or x(x–2) – 3(x–2) = 0
or (x–2) (x–3) = 0
or x = 2 or 3
2nd method (By formula) x² – 5x + 6 = 0

Here a = 1 b = –5   c = 6 (comparing the equation with ax² + bx+c = 0)

Example 2: Examine the nature of the roots of the following equations.
i) x² – 8x² + 16 = 0
ii) 3x² – 8x + 4 = 0
iii) 5x² – 4x + 2 = 0
iv) 2x² – 6x – 3 = 0

Solution: (i) a = 1 b = –8 c = 16
b² – 4ac = (–8)² – 4.1.16 = 64 – 64 = 0
The roots are real and equal.
(ii) 3x² – 8x + 4 = 0
a = 3 b = –8 c = 4
b² – 4ac = (–8)² – 4.3.4 = 64 – 48 = 16 > 0 and a perfect square
The roots are real, rational and unequal

iii) 5x² – 4x + 2 = 0
b² – 4ac = (–4)² – 4.5.2 = 16–40 = –24 < 0
The roots are imaginary and unequal
(iv) 2x² – 6x – 3 = 0
b² – 4ac = (–6)2 – 4.2 (–3)
= 36 + 24 = 60 > 0
The root are real and unequal. Since b² – 4ac is not a perfect square the roots are real irrational and unequal.

Illustrations:

1. If α and ß be the roots of x² + 7x + 12 = 0 find the equation whose roots are ( α + ß )² and (α - ß)².

Solution : Now sum of the roots of the required equation
= (α + b)² + (α + b)² = (-7)² + (α + b)² - 4 ∝ b
= 49 + (–7)² – 4x12 = 49 + 49 – 48 = 50

Product of the roots of the required equation = (α + b)²  (α - b)² = 49 (49–48) = 49
Hence the required equation is x² – (sum of the roots) x + product of the roots = 0
or x² – 50x + 49 = 0

2. If a ,b be the roots of 2x² – 4x – 1 = 0 find the value of  

or 4p² + 8p – 45 = 0
or 4p² + 18p – 10p – 45 = 0
or 2p(2p + 9) – 5(2p + 9) = 0
or (2p – 5) (2p + 9) = 0.

or t² + 32 = 12t
or t² – 12t + 32 = 0
or t² – 8t – 4t + 32 = 0
or t(t–8) – 4(t–8) = 0
or (t–4) (t–8) = 0
∴ t = 48
For t = 4   2^x = 4 = 2² i.e. x = 2
For t = 8   2^x = 8 = 2³ i.e. x = 3

∴ Required equation is : x² – (sum of roots)x + (product of roots) = 0 or x² – 4x + 1 = 0.

7. If α β are the two roots of the equation x² – px + q = 0 form the equation whose roots are     

Solution: As α, β are the roots of the equation x² – px + q = 0 α + β = – (– p) = p and α β = q.

or qx² – (p² – 2q) x + q = 0
8. If the roots of the equation p(q – r)x² + q(r – p)x + r(p – q) = 0
are equal show that  

Solution: Since the roots of the given equation are equal the discriminant must be zero ie. q²(r – p)² – 4. p(q – r) r(p – q) = 0

2.10 PROBLEMS ON QUADRATIC EQUATION

1. Difference between a number and its positive square root is 12; find the numbers?

Solution: Let the number be x.
Then  = 12 …………… (i)

or (y – 4) (y + 3) = 0 ∴ Either y = 4 or y = – 3 i.e. Either   or  

If  x = 9 if does not satisfy equation (i) so or x=16.

2. A piece of iron rod costs Rs. 60. If the rod was 2 metre shorter and each metre costs Re 1.00 more, the cost would remain unchanged. What is the length of the rod?
Solution: Let the length of the rod be x metres. The rate per meter is Rs.

New Length = (x – 2); as the cost remain the same the new rate per meter is  

or x² – 2x = 120 or x² – 2x – 120 = 0 or (x – 12) (x + 10) = 0.
Either x = 12 or x = –10 (not possible) ∴ Hence the required length = 12m.

3. Divide 25 into two parts so that sum of their reciprocals is 1/6.
Solution: let the parts be x and 25 – x

or x(x–15) – 10(x–15) = 0
or (x–15) (x–10) = 0

or x = 10, 15
So the parts of 25 are 10 and 15.

2.11 SOLUTION OF CUBIC EQUATION

On trial basis putting some value of x to check whether LHS is zero then to get a factor.
This is a trial and error method. With this factor to factorise the LHS and then to get values of x .

Illustrations :

1. Solve x³ – 7x + 6 = 0
Putting x = 1 L.H.S is Zero. So (x–1) is a factor of x³ – 7x + 6
We write x³–7x +6 = 0 in such a way that (x–1) becomes its factor. This can be achieved by writing the equation in the following form.

or x³–x²+x²–x–6x+6 = 0
or x²(x–1) + x(x–1) – 6(x–1) = 0
or (x–1)(x²+x–6) = 0
or (x–1)(x²+3x–2x–6) = 0
or (x–1){x(x+3) – 2(x+3) } = 0
or (x–1)(x–2)(x+3) = 0

2. Solve for real x: x³ + x + 2 = 0

Solution: By trial we find that x = –1 makes the LHS zero. So (x + 1) is a factor of x³ + x + 2
We write x³ + x + 2 = 0 as x³ + x² – x² – x + 2x + 2 = 0
or x²(x + 1) – x(x + 1) + 2(x + 1) = 0
or (x + 1) (x² – x + 2) = 0.
Either x + 1 = 0
or x² – x + 2 = 0 i.e. x = –1

As x =   is not real, x = –1 is the required solution.

2.12 APPLICATION OF EQUATIONS IN CO–ORDINATE GEOMETRY

Introduction: Co-ordinate geometry is that branch of mathematics which explains the problems of geometry with the help of algebra

Distance of a point from the origin.

P (x, y) is a point.
By Pythagoras’s Theorem OP² =OL² + PL² or OP² = x² + y²
So Distance OP of a point from the origin O is  

Distance between two points

By Pythagoras’s Theorem PQ²=PT² +QT²

So distance between two points (x₁ y₁) and (x₂ y₂) is given by  

2.13 EQUATION OF A STRAIGHT LINE

(I) The equation to a straight line in simple form is generally written as y=mx+c …… (i)
where m is called the slope and c is a constant.

If P₁(x₁, y₁) and P₂(x₂, y₂) be any two points on the line the ratio   is known as the
slope of the line.

We observe that B is a point on the line y = mx+c and OB is the length of the y-axis that is intercepted by the line and that for the point B x=0.
Substituting x=0 in y=mx+c we find y=c the intercept on the y axis.
This form of the straight line is known as slope–intercept form.

Note : (i) If the line passes through the origin (0, 0) the equation of the line becomes y = mx (or x=my)
(ii) If the line is parallel to x–axis, m=0 and the equation of the line becomes y = c (or x = b b is the intercept on x–axis)

(iii) If the line coincides with x–axis, m=0, c=0 then the equation of the line becomes y=0 which is the equation of x–axis. Similarly x=0 is the equation of y–axis.
(II) Let y = mx + c ………….. (i) be the equation of the line p₁p₂.

Let the line pass through (x₁, y₁). So we get
y₁ = mx₁+c ...(ii) By (i) – (ii) y–y₁ = m(x–x₁) … (iii)
which is another from of the equation of a line to be used when the slope(m) and any point (x₁ y₁) on the line be given. This form is called point–slope form.
(III) If the line above line (iii) passes through another point (x₂, y₂). we write y₂–y₁ = m(x₂–x₁)

Which is the equation of the line passing through two points (x₁, y₁) and (x₂, y₂)
(IV) We now consider a straight line that makes x-intercept = a and y-intercept = b
Slope of the line

If (x, y) is any point on this line we may also write the slope as 

The form    is called intercept form of the equation of the line and the same is to be used when x–intercept and y–intercept be given.

Note: (i) The equation of a line can also be written as ax+by+c = 0
(ii) If we write ax+by+c = 0 in the form y = mx+c

(iii) Two lines having slopes m₁ and m₂ are parallel to each other if and only if m₁ = m₂ and perpendicular to each other if and only if m₁m₂ = –1
(iv) Let ax + by + c  =  0 be a line. The equation of a line parallel to
ax + by + c = 0  is  ax + by + k = 0 and the equation of the line perpendicular to
ax + by + c = 0  is  bx– ay +k = 0
Let lines ax + by + c = 0 and a¹x+b¹y+c¹ = 0 intersect each other at the point (x₁, y₁).
So ax₁ + by₁ + c = 0
a¹x1 + b¹y¹ + c¹ = 0

By cross multiplication  

Example : Let the lines 2x+3y+5 = 0 and 4x–5y+2 = 0 intersect at (x₁ y₁). To find the point of intersection we do cross multiplication as
2x₁ + 3y₁ + 5 = 0
4x₁ + 5y₁ + 2 = 0

Solving x₁ =19/2 y₁=–8

(V) The equation of a line passing through the point of intersection of the lines ax + by + c = 0 and a₁x + b₁y + c = 0 can be written as ax+by+c+K (a₁x+b₁y+c) = 0 when K is a constant.

(VI) The equation of a line joining the points (x₁ y₁) and (x₂ y₂) is given as

If any other point (x₃ y₃) lies on this line we get

which is the required condition of collinearity of three points.

Illustrations:

1. Show that the points A(2, 3) B(4, 1) and C(–2, 7) are collinear.
Solution : Using the rule derived in VI above we may conclude that the given points are collinear if 2(1–7)+4(7–3)–2(3–1)=0 i.e. if –12+16–4=0 which is true.
So the three given points are collinear

2. Find the equation of a line passing through the point (5, –4) and parallel to the line 4x+7y+5 = 0

Solution : Equation of the line parallel to 4x+7y+5 = 0 is 4x+7y+K = 0
Since it passes through the point (5, –4) we write
4(5) + 7(–4) + k = 0
or 20 – 28 + k = 0
or –8 + k = 0
or k = 8
The equation of the required line is therefore 4x+7y+8 = 0.

3. Find the equation of the straight line which passes through the point of intersection of the straight lines 2x+3y = 5 and 3x+5y = 7 and makes equal positive intercepts on the coordinate axes.

Solution: 2x+3y–5 = 0
3x+5y–7 = 0
By cross multiplication

So the point of intersection of the given lines is (4, –1)
Let the required equation of line be

  (*for equal positive intercepts a=b)
∴ x + y = a
Since it passes through (4, –1) we get 4 – 1 = a or a = 3
The equation of the required line is therefore x + y = 3

4. Prove that (3, 1) (5, –5) and (–1, 13) are collinear and find the equation of the line through these three points.

Solution: If A (3, 1) B (5, – 5) and C (–1, 13) are collinear we may write
3(–5–13) +5(13–1) –1(1+5) = 0
or 3(–18) +5(12) – 6 = 0 which is true.
Hence the given three points are collinear.
As the points A, B, C are collinear, the required line will be the line through any of these two points. Let us find the equation of the line through B(5, – 5) and A (3, 1)
Using the rule derived in III earlier we find

or 3x + y = 10 is the required line.

5. Find the equation of the line parallel to the line joining points (7, 5) and (2, 9) and passing through the point (3, –4).

Solution : Equation of the line through the points (7, 5) and (2, 9) is given by

or –5y + 25 = 4x–28
or 4x+5y–53 = 0
Equation of the line parallel to 4x+5y–53=0 is 4x+5y+k = 0
If it passes through (3, –4) we have 12–20+k = 0 i.e. k=8
Thus the required line is 4x+5y+8 = 0

6. Prove that the lines 3x – 4y + 5 = 0, 7x – 8y + 5 = 0 and 4x + 5y = 45 are concurrent.

Solution: Let (x₁ y₁) be the point of intersection of the lines
3x – 4y +5 = 0 …………… (i)
7x – 8y + 5 = 0 …………... (ii)
Then we have 3x₁ – 4y₁ + 5 = 0
7x₁ – 8y₁ + 5 = 0

Hence (5, 5) is the point of intersection. Now for the line 4x + 5y = 45
we find 4.(5) + 5.5 = 45; hence (5, 5) satisfies the equation 4x+5y=45.
Thus the given three lines are concurrent.

7. A manufacturer produces 80 T.V. sets at a cost Rs. 220000 and 125 T.V. sets at a cost of Rs. 287500. Assuming the cost curve to be linear find the equation of the line and then use it to estimate the cost of 95 sets.

Solution: Since the cost curve is linear we consider cost curve as y = Ax + B where y is total cost. Now for x = 80 y = 220000.
 ∴ 220000 = 80A +B …..(i) and for x = 125 y=287500
∴ 287500 = 125A +B ……(ii)
Subtracting (i) from (ii) 45A = 67500 or A = 1500
From (i) 220000 – 1500 ´ 80 = B or B = 220000 – 120000 = 100000
Thus equation of cost line is y = 1500x + 100000.
For x = 95 y = 142500 + 100000 = Rs. 242500.
∴ Cost of 95 T.V. set will be Rs. 242500.

2.14 GRAPHICAL SOLUTION TO LINEAR EQUATIONS

1 . Drawing graphs of straight lines From the given equation we tabulate values of (x, y) at least 2 pairs of values and then plot them in the graph taking two perpendicular axis (x, y axis). Then joining the points we get the straight line representing the given equation.

Example 1 : Find the graph of the straight line having equation 3y = 9 – 2x

Solution: We have 2x + 3y = 9. We tabulate y =  

Here AB is the required straight line shown in the graph.

Example 2 : Draw graph of the straight lines 3x +4y = 10 and 2x – y = 0 and find the point of intersection of these lines.

Solution: For 3x + 4y = 10 we have y =  

For 2x – y = 0 we tabulate

From the graph, the point of intersection is (1, 2)