Mathematical Equation

Equation is a mathematical statement of equality, containing one or more variables. An equation expresses the equality of things through the (=) equal sign

• Conditional Equation  : If the equality is true for certain value of the variable involved, the equation is often called a conditional equation

Ex : {(x+2) / 3} + {(x+3) / 2} = 3 is true only for value x=1. So, it is an conditional equation

• Identity :  If the equality is true for all values of the variable involved the equation is called an identity

Ex. {(x+1)/3} + {(x+3)/2} = (5x+11)/6 holds true for all values of the variable x. So, it is an Identity

Roots of Equation

Determination of value of the variable which satisfies an equation is called solution of the equation or root of the equation.

Types of Equation

• Linear 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.

Ex . 6x + 10(x−3) = 5 (4x − 9) + 12 is a Linear equation of one variable

An equation of the form ax + by + c =0, where a, b, c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y.

Ex. 2x + y – 15 =0 is a linear equation in the two variables (unknown) x and y.

• Simultaneous Linear Equations: Two or more linear equations involving two or more variables are called Simultaneous Linear Equations.

Ex. 5x+2y = 1 &   12x + 3y = 2 are jointly called simultaneous equations.

• Quadratic Equation : An equation of degree 2 (highest power of the variable is 2) is called Quadratic equation

Ex. 8x² + 15x +36 = 0 is a quadratic equation.

• Cubic Equation.: Equation of degree 3 is called Cubic Equation.

  Ex. 14x³ +3x² + 5x − 17 =1 is a Cubic equation.

Linear Equation of one variable : Problems

A linear equation of one variable takes the form ax+b=0, where ‘a’ and ‘b’ are known constants and ‘a≠0’

Method of solution : The variables are written in one side (normally in LHS) and the constants are written on another side (normally in LHS), by transposing to the respective sides. Cross multiplication is used to simplify fractions where necessary and the value of variable is obtained to get the solution (i.e root of the equation)

Ex Solve 4x + 5 = 9 

We have 4x + 5 = 9

So, 4x = 9 – 5 (Transposing 5 to RHS)

So, 4x = 4

So, x = 1 (Dividing both sides by 4).

So, x=1 Ans.

Ex. Solve x + 32 = 2x
We have x + 32 = 2x
So, x- 2x = -32 (Transposing 2x to LHS and 32 to RHS)
So, -x = -32 (Transposing LHS and RHS).
So, x=32

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Ex. Solve the following equation

1/ (x+8) = 2/ (6-x)

Step 1 : By cross multiplication we get

2 (x+ 8) = 1 (6 – x)

Step 2 : 2x + 16 = 6 – x

Step 3: 2x +  x = 6 – 16

Step 4  : 3x = -10. So x=-10/3

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Ex. Solve the following equation:

(x+1)/ (x-1) = (2x+8) / (2x-5)

Step 1 : By cross multiplication we get

(x – 1) (2x+8) = (x+1) (2x – 5)

Step 2 : 2x² + 8x  – 2x – 8 =2x² – 5x + 2x  –  5

Step 3  : 2x² + 6x – 8 = 2x² – 3x  – 5

Step 4 : 2x² + 6x – 2x² + 3x = – 5 + 8

Or 9x =  3, or x= 3/9, or x= 1/3

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Linear Equation of one variable : Problems

Linear Equation of one variable : Problems & Solutions

Ex. If thrice of Arun’s age 5 years ago be subtracted from twice his present age the result would be equal to his present age. Find Arun’s present age.

Let x years be Arun’s present age. So, Arun’s age five years ago is x-5. Thrice of Arun’s age 5 years ago is 3(x-5).  Twice of Arun’s present age is 2x.

So, as per the problems, the equation may be stated as

2x − 3(x−5) = x

2x − 3x+ 15 = x

−x + 15= x

2x = 15

x = 15/2 = 7.5  

Arun’s present age is 7.5 years.

Ex. 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 a be the digit in the unit’s place. So the digit in the ten’s place is 2a.

Step 1 : Thus the number becomes {10(2a) + a}.

Step 2 : The number on reversing the digit is {10a + 2a}

On subtracting 18 from the number, it comes to {10(2a) + a} -18}, Or 20a +a-18 =21a – 18

Step 3 : So, as per the problem

21a−18 = 10a + 2a, Or 21a -18 = 12a, Or 9a = 18 or a=2

So, at the unit place of the number, the digit  is 2, and at the decimal lace, the digit is 2×2=4. So, the number is 42

You can verify the answer. The number is 42. Subtract 18 from the number. You get 42-18=24. The unit and ten place in the number is reversed. So, the answer is correct

Ex. The difference between the two numbers is 48. The ratio of the two numbers is 7:3. Find the two numbers. 

Solution: 

1. According to the question, Ratio between the numbers is 7:3. Assuming the common divisor as x, we may write the first number as 7x, and second number as 3x.
2. According to the question, the difference between the two numbers is 48
   so, we can write 7x – 3x = 48 
   or, 4x = 48 
   or x = 48/4 
   0r, x = 12
   Putting the value of x, we get the first number = 7x = 7 × 12 = 84
   Again, Putting the value of x, we get the second number 3x = 3 × 12 = 36 

Therefore, the two numbers are 84 and 36.

Linear Equation of two variables

The general form of simultaneous equation of 2 variable is ax+by+c =0, a,b, & c are any known numbers and a≠0, and b≠0.

Two simultaneous equation of 2 variable is a₁x+b₁y+c₁ =0 and a₂x+b₂y+c₂ =0, from a pair of simultaneous equations.

We need at least 2 simultaneous equations of 2 variables, to find the roots, satisfying both the paired equations (of x & y).

Method of solution : There are two methods of solution ofsimultaneous equation of 2 variables.

• Elimination Method : In Elimination Method, one of the variable is eliminated by multiplying both equation by a constant to equalize the value of one of the variables in the two equations, and then subtracting the values of one equation from another, to eliminate one of the variable to get a simple  equation of one variable.

The simple equation is then solved to get the value of one variable, as explained earlier. On obtaining value of one variable, the value is substituted in any of the equations, to get the value of the second variable.

This method is more commonly used, as it is easier to understand and solve.

• Cross Multiplication Method : The co-efficients of the equation pair are written in sets and then cross multiplied between the adjacent cells

a1	b1	c1
a2	b2	c2

To get the form [x / (b₁c₂-b₂c₁)] = [y / (c₁a₂-c₂a₁)] = [1 / (a₁b₂-a₂b₁)]

The solution would be x= [(b₁c₂-b₂c₁)] / [(a₁b₂-a₂b₁)], and y=[(c₁a₂-c₂a₁)] / [(a₁b₂-a₂b₁)]

This method used to quickly find out the answer. Mostly used to select right option in Multiple Choice Questions, without showing the detailed steps of solution

Linear Equation of two variables – Problems

Linear Equation of two variables : Problems & Solutions

Ex. Solve 4a + 3b =14, 9a – 5b =55

a. Solution by elimination method

Equation 1: 4a + 3b = 14

Equation 2 : 9a – 5b = 55

Step 1 : To make coefficients of  b equal (numerically), multiplying equation (1) by 5 and equation (2) by 3, we get

Step 2

Equation 3: 20a + 15b = 70

Equation 4: 27a – 15b = 165

Step 3

Adding both the equations (3) & (4), the variable  b is eliminated and we get another linear equation  (Equation 5), of single value  (a linear equation of single variable a)

Equation  5 : 47a = 235

Step 4

From equation 5, we get 47a=235. Or a=5

Step 5

Substituting the value of a in Equation 1, we get

4×5 + 3b = 14, or 20+3b = 14, or 3b=-6, or b=-2

Hence the solution is a = 5, b = – 2.

b. Solution by cross multiplication method

By cross multiplication method, you can get the values of the variables a and b, directly, by putting the values of the co-efficient

We can write the equations as :

4a + 3b -14 =0

9a – 5b – 55 =0

Put the respective co-efficient into the tbale

4	3	-14
9	-5	-55

a= [{3x(-55)} – {(-14)x(-5)} / [[{(4x (-5)} – (3 x 9)]

So, a=5

b=[{(-14)x9}- {4 x (-55)}] /  [[{(4x (-5)} – (3 x 9)] = [(-126) – (-220)] / [(-20) – 27)]= 94/-47 =-2. So, b=-2

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Ex. Solve:  2a + 5b = 26 and 3a − b = 5.

Equation 1: 2a+ 5b = 26

Equation 2: 3a − b = 5

Step 1 : To make coefficients of  b equal (numerically), multiplying equation (1) by 1 and equation (2) by 5, we get

Equation 3: 2a+ 5b = 26

Equation 4: 15a − 5b = 25

Step 2

Adding both the equations (3) & (4), the variable  b is eliminated and we get another linear equation  (Equation 5), of single value  (a linear equation of single variable a)

Equation  5 : 17a = 51

Step 3

From equation 5, we get 17a=51. Or a=3

Step 4

Substituting the value of a in Equation 1, we get

2×3+ 5b = 26, or 6+5b = 26, or 5b=20, or b=4

So, the solution is : a = 3, b = 4.

Example
4x +5y=30 … (Eq 1)
6x – 3y= 12… (Eq 2)

Solving by elimination method

To eliminate x, we multiply both sides of the equation 1 by 3 and equation 2 by 2 to get 12 x in both the equation. We can then eliminate x by subtracting equation 2 from equation to get the value of y.
Then by putting the value of y in any of the equation 1 or 2, we can get the value of x
4x +5y=30, or 12x+15y=90 .. (Eq 4) (multiply both sides by 3)
6x – 3y= 12, or 12x-6y=24 .. (Eq 5) (multiply both sides by 2)

12x+15y=90 .. (Eq 4)
12x-6y=24 .. (Eq 5)
subtracting equation 5 from equation 4, we get
15y- (-6y)=90-24, or 21y=66, or y=66/21 =22/7 , or y=22/7

Now, put the value of y in Eq 1
4x + 5 X (22/7) = 30, or, 4x+110/7= 30. Or 4x = 30-110/7 = (210-110)/7
X=100/(7×4) = 100/28=25/7.
So, x=25/7, y=22/7 (Ans)

Alternately, we may put the value of y in Eq.. 2 also, to get the value of x
Putting the value of y in Eq 2, we get
6x-3y=12, or, 6x- 3 X (22/7)=12, or 6x= 12 +3 X (22/7), or 6x= 12 + 66/7
0r 6x=(84+66/7) = 150/7, or x= 150/ (6X7) = 25/7

So, we find that by putting the value of one variable in any of the two equation, we get the same value of the other variable. This proves that the solution is correct

Linear Equation of two variables – Problems

Linear Equation of two variables : Problems & Solutions

Ex. The cost of 7 kg. of wheat and 5kg. of oil is Rs. 234,and the cost of 6 kg. of wheat and 7 kg. of rice is Rs.263.  Find the cost of wheat and oil per kg.

Let cost of wheat / kg =  a, Cost of Oil / kg =  b

Step 1: Cost of 7 kg of Wheat = 7a, Cost of 5 kg of Oil= 5b

Equation 1: 7a+5b=234

Cost of 6 kg of Wheat = 6a, Cost of 7 kg of Oil= 7b

Equation 2: 6a+7b=263

Step 2: To eliminate the variable b in equation 1 and 2, equalise the co-efficient  of b. Multiply equation 1 by 7 and  equation 2 by 5, to get following equations

Equation 3: 49a + 35b =1638

Equation 4 : 30a + 35b =1315

Step 3: By deducting both sides of  equation 4 from both sides of  equation 3, we get

(49a-30a) + (35b-35b) = 1638-1315 = 323

Step 4 : From Step 3, we get

19a=323, or a=17

Step 5 : So, Price of Wheat =a= 17 / kg

Step 6 : Substituting value of a in equation 1, we get

7×17 + 5b = 234

Or  119+5b = 234, or 5b= 234-119= 115, or b=23

So, Cost of Wheat Rs.17 / kg and Cost of Oil Rs 23/kg

Linear Equation of two variables – Problems

Linear Equation of two variables : Problems & Solutions

Ex. The weekly incomes of M and N are in the ratio 3:4 and their weekly expenditures are in the ratio 1:2 .If each saves Rs. 2000 per week, find their weekly incomes

Let weekly Incomes of M and N be Rs. 3x and Rs. 4x respectively (as they are in ratio 3:4)

Let weekly Expenditure of M and N be Rs. y and Rs. 2y respectively (as they are in ratio 1:2)

Step 1 : Weekly savings of M = 3x-y, Weekly savings of N = 4x-2y

Step 2

Equation 1 : 3x-y = 2000

Equation 2 : 4x-2y = 2000

To eliminate the variable y in equation 1 and 2, equalise the co-efficient  of b. Multiply equation 1 by 2 and  equation 2 by 1, to get following equations

Equation 3 : 6x-2y=4000

Equation 4 : 4x-2y = 2000

Step 3: By deducting both sides of  equation 4 from both sides of  equation 3, we get

(6x – 4x) – {(-2y -(-2y)} = 4000-2000 = 2000

Or 2x- 0 = 2000, or 2x=2000, or x=1000

Step 4:

Income of M is = 3x=3×1000 = 3000

Income of N is = 4x=4×1000 = 4000

Quadratic Equation

Solving Quadratic Equation

An equation of the form ax² + bx + c = 0 where x is a variable and a b c are constants with a ¹ 0   is called a quadratic equation or equation of the second degree.

Ex. 4x² + 8x + 5 = 0 is a Quadratic Equation. x² −7x = 0 is a Quadratic Equation (here c=0). 4x² + 5 = 0 is also a quadratic equation (here b=0).

The value of the variable (e.g x) is called the root of the equation. A quadratic equation has

two roots.

Ex. Finding the roots of a quadratic equation

ax² + bx + c = 0           (a ¹ 0)   

Step 1 : (Dividing both sides by a, we get) 

Step 2 : (Multiply and divide the middle term by 2, we get)

Step 3 : (Take the constant values at the RHS, we get) x² –  {-(b/a)x} + (c/a) = 0

Step 4 : (Add in both LHS and RHS, to bring the LHS into square form),

we get,

Step  5: (Convert into square form, we get )

Step  6: (Convert LHS into square form, add up terms in RHS, we get)

Step  7: Take the square root of both sides, we get,

Step  8: Take the constant value to RHS, we get, x = [{(b²-4ac)^1/2 } / 2a)] – (b/2a)

Step  9: Add the 2 expressions in RHS, we get,

Properties of Quadratic Roots

Roots of Quadratic Equation

A quadratic equation has two roots, Let us denote them as a & b, then

a = [-b + (b²-4ac)^1/2]/ 2a, b = [-b – (b²-4ac)^1/2]/ 2a

Sum of the roots

a + b = [{-b + (b²-4ac)^1/2} /2a+  {-b – (b²-4ac)^1/2}/ 2a] = -2b/ 2a = -b/a

a + b = -b/a.

So, sum of the roots = – {(coefficient of x) / (coefficient of x²)}

Product of the roots

ab = [{-b + (b²-4ac)^1/2}/ 2a] x [{-b – (b²-4ac)^1/2}]/ 2a]

ab =  [{b² – (b² – 4ac)} ]/ 2a, ab =  4ac/ 4a² = a/c. So ab = ac

So, product of the roots = (constant term) / (coefficient of x²)

Irrational roots occur in pairs, that is if (m+√ n ) is a root then, (m+√ n ) is the other root of the same equation.

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

– If one root is equal to other root but opposite in sign then, their sum =0 and so (b/a)=0, or b=0

Constructing a Quadratic Equation

Formation of Quadratic Equation

The general form of Quadratic Equation is : ax² + bx + c = 0

or,

or,

or,

or x² − (Sum of the roots) X + Product of the roots = 0

Nature of Roots of Quadratic Equation

From the above equation, we may deduce :

If b² − 4ac = 0 the roots are real and equal;

If b² − 4ac > 0 then the roots are real and unequal (or distinct);

If b² − 4ac < 0 then the roots are imaginary;

If b² − 4ac is a perfect square (¹ 0) the roots are real rational and unequal (distinct);

If b² − 4ac > 0 but not a perfect square the roots are real irrational and unequal.

Equation Discriminant

b² − 4ac is called the discriminant in the equation ax² + bx + c = 0, as it actually discriminates between the roots.

Quadratic Equation – Problems

Quadratic Equation- Problems & Solutions

Ex. Find the roots of the equation  y² − 5y + 6 = 0

First Method : Solving the equation without applying formula

Step 1 : y² − 2y − 3y + 6 = 0 (split the middle term to make two parts with common factors

Step 2 : y(y − 2) − 3(y− 2) = 0 (regroup with common factors)

Step 3 : (y − 2)  (y − 3) = 0 (express in factors)

Step 4 :  So, y-2=0, or y-3=0 (evaluate each factor)

Step 5 : y-2=0, i.e y=2, y-3=0, i.e y=3 (Find the values from each factor)

So, the roots are y=2, or 3

2nd Method : Computing the roots applying formula

Step1 : Express the equation y² − 5y + 6 = 0, in terms of ay² + by + c = 0

So, here a = 1, b = −5,  c = 6

So, the value of roots are :

Step 2: .
Substituting the values of a, b & c, we get

Step 3 :

Step 4 : ,

Step 5 :

Step 6: y=3 or y=2

Quadratic Equation : Problems

Quadratic Equation- Problems & Solutions

Ex. Examine the nature of the roots of the following equations.  

(i) z² − 8z + 16 = 0,

Here a = 1   b = − 8   c = 16

b² − 4ac = (−8)² − 4.1.16 = 64 − 64 = 0

The roots are real and equal.

(ii) 3z² − 8z + 4 = 0

Here, a = 3   b = −8    c = 4

b² − 4ac = (−8)² − 4.3.4 = 64 − 48 = 16

The value is greater than zero and a perfect square. The roots are real rational and unequal

(iii) 5x² − 4x + 1 = 0

Here, a = 5   b = −4    c = 1

b² − 4ac = (−4)² − 4.5.1 = 16 − 20 = − 4 < 0

The value is negative. The roots are imaginary and unequal.

(iv) 2x² − 6x − 4 = 0

Here, a = 2   b = −6    c = 4

b² − 4ac = (−6)² − 4.2(−4) = 36 + 32 = 68 >0

The value is positive. Since the value is not a perfect square, the roots are real irrational and unequal.

Quadratic Equation : Problems

Quadratic Equation- Problems & Solutions

Ex 1. If a ,b    be the roots of 4x² − 4x − 1 = 0 find the value of (a² / b) + (b² / a)

Step 1: (a +b) = – (-4) / 4 = 1,  ab = -1 /4

Step 2: The value of expression  (a² / b) + (b² / a)

Step 3:  =  (a³ + b³) / ab (adding the 2 parts)

Step  4 : [(a +b)³ – 3ab(a+b)] /ab (applying the formula)

Step  5 : [ 1³ – 3´ (-1/4) ´ 1] / (-1/4)

Step  6 : [1+(3/4)] / -(1/4)

Step 7 : (7/4) ´ (-4) = -7

Ex 2: If a and b are the two roots of the equation x² − px +4q = 0,

Form the equation whose roots are (a/b) & (b /a)

Step 1 : As a and b are the roots of the equation x² − px +4q = 0

a+b = − (− p) = p and ab = 4q.

Step 2 : The roots of the equation to be evaluated are : (a/b) & (b /a)

Sum of the roots = (a/b) + (b /a) = [(a²+b²)] / ab] = [(a+b)² − 2ab]/ ab = = p²-8q

Products  of the roots = (a/b) ´ (b /a) =1

Step 3 : Required equation is x²  – [ ( p²  –  8q) / 4q ] x +1 =0

Or, 4q x² − (p² − 8q) x + 4q = 0  (multiplying both sides by 4q)     

Quadratic Equation – Problem

Quadratic Equation- Problems & Solutions

Ex. 1: Solve 9a² + 6a + 1 = 0.

Step 1 : 9a² + 3a +3a + 1 = 0 (spit the middle part to form a factor)

Step  2 : 3a(3a + 1)  + 1 (3a + 1) = 0 (take common factor)

Step  3 : (3a+1) (3a+1) = 0 (factorise)

Step 4 :  3a + 1 = 0 or 3a+1 = 0 (evaluate each factor part)

Step 5 : 3a = -1, or

Bothe the roots have equal value,

Ex 2: Solve 2x² – 5x + 3 = 0.

Step 1 : Express 2x² – 5x + 3 =0 in the form of ax² + bx + c=0.

So, here a=2, b=-5, c=3

Step 2 : The value of the roots are

Step 3

Step 4

So, the value of roots are

Quadratic Equation – Problem

Quadratic Equation- Problems & Solutions

Ex. Divide 54 into two parts such that their product is 704.

Step 1 : Let one part be x, then the other part is 54 – x. So, According to the problem,

Step 2 : x(54 – x )= 704

Step 3  : 54x – x² – 704 = 0

Step 4  : x² – 54x + 704 = 0 (multiplying both side by -1 to make the coefficient of x² positive)

Step 5  : x² – 32x – 22x + 704 = 0 (split middle part for factorization)

Step 6  : (x – 32) (x – 22) = 0 (Factorise)

Step 7  :  (x – 32) =0 or (x – 22) =0

Step 8  :   x = 32, or x=22

So, one part is 32, The other part is 54-32=22

Quadratic Equation – Problem

Quadratic Equation- Problems & Solutions

then find value of b

Equation 1:(1/a) + (1/b) + (1/c) = 36

Equation 2: (1/a) + (3/b) – (1/c) = 28

Equation 3: (1/a) + (1/3b) + (1/2c) = 20

Step 1: Adding Equation 1 & 2 (to eliminate c), we get

(1/a) + (1/b) + (1/a) + (3/b) = 36+28

Equation  4 : (2/a) + (4/b) = 64

Step 2: Dividing Equation (2) by 2, we get

Equation 5 : (1/2a) + (3/2b) – (1/2c) = 14

Step 3 : Adding Eq (3) and Eq (5), we get

Eq 6 : (3/2a) + (11/6b) = 34

Step 4 : Multiplying Eq (3) by 3, we get

Equation 7 : (6/a) + (12/b) = 192

Step 5 : Multiplying Eq (6) by 4, we get

Equation 8 : (6/a) + (22/3b) = 136

Step 6 : Subtracting Eq  (8) from Eq (7), we get

14/3b= 56, or 1/b= 56 ´ (3/14) = 12. So, b=1/12

Ex2 . If a, b are the roots of 2x² + 7x + 8 = 0,

find the values of (i) a² + b², (ii) (1/ a) + (1/ b), (iii) a³ + b³

Here,  a + b = (-7/2), ab = 8/2 = 4

(i) a² + b²  = (a + b)² – 2a b =  (-7/2)² – 2×4 = 49/4 – 8 = 17/4

(ii) (1/ a) + (1/ b) = (a + b) / ab = [(-7/2)] / 4 =  (-7/8)

(iii) a³ + b³ =  (a + b ) (a² + b²  – ab) = (a + b) {(a + b ) ²  – 3ab} = (-7/2) [(-7/2)² – 3´4]

= (-7/2)´ (49/4 -12) = (-7/2) ´ (1/4) = – 7/8

Cubic Equation

Quadratic Equation- Problems & Solutions

An equation of the form ax³ +bx² + cx + d = 0 where a, b, c and d are real numbers and a ¹ 0, is called a cubic equation in variable x.

Ex. 5x³ +x² – x + 7 = 0, x³ – 16x +27 =0  are cubic equations.

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