Quantitative MethodsSession 11 – 08.08.12Chapter 6 Linear Equations, Equation &Slope of a straight line & Quadratic Equations- Formulation & Solving

Pranjoy Arup Das

Part 1 1) Linear Equations – formulation & solving2) Equation & Slope of a straight line.

• An Equation shows or is used to show how two different things are related to each other or may become equal to each other. • For Eg. Apples and Rice are two different things. But if the cost of P no. of apples is Rs. 100 and the cost of Q kg of rice is also Rs. 100, then We can say that, cost of P apples = cost of Q kg of Rice Which means that, cost of P apples - cost of Q kg of Rice = 0 We can also say that, Cost of P Apples + Cost of Q kg rice = 200• An equation always contains variables (such as x, y, z, p, q, r, a, b, c etc.) the values of which are unknown. These variables establish the relationship between two or more things. • Their values are the solution of the equations. • Equations are of four types: - Linear- Quadratic- Cubic- Bi-quadratic

• Equations such as 2x = 8, 2y+6 = 14, 2p + 5q = 23, 3a + 5b – c = 20 are all examples of linear equations. • In linear equations the power of the unknown variable is 1, i.e. x1, y1, p1 etc.• 2x = 8 is an example of a linear equation with one variable, x. The general form of such an equation is ax+b=0, where a which is called the co-efficient of x, is a constant & a≠0. - So, in this case, b = -8• 2x+5y = 23 is an example of linear equation with two variables, x&y. The general form of such an equation is ax+by+c=0, where a,b & c are constant. a is the co-efficient of x, b is the co-efficient of y and a≠0 & b≠0.- So, in this case, c = -23 • 2x+5y-z = 20 is a linear equation with three variables, x,y &z. • There can be multiple solutions, i.e., values of x , y, z etc., in a linear equation with more than one variable.

• Two or more linear equations containing the same variables are said to be simultaneous linear equations.• Eg. x+y=5 y-2=2xSolving simultaneous linear equations: There are three methods : 1) Elimination by substitution :Lets take the above example x + y = 5 …which will be denoted by (i) y - 2 = 2x …which will be denoted by (ii)

From (i) , we can get x = 5 – y ….which we will call (iii)So substituting the value of x from (iii) in (ii) we get, y – 2 = 2(5-y) => y – 2 = 10 – 2y=> y+2y = 10 +2=> 3y = 12 => y = 4Substituting the value of y in (iii) , we get x = 5 – 4 = 1. So our required solution is x = 1, y = 4.

THESE ARE SIMULTANEOUS LINEAR EQUATIONS OF TWO VARIABLES

2) Elimination by equating the co-efficient :Eg. Solve: x+3y = 5 & y-2 = 2xx + 3y = 5 …which will be denoted by (i) y - 2 = 2x …which will be denoted by (ii)We can re-arrange (ii) as 2x –y = -2 ……..(iii)So now we have, x + 3y = 5 …………(i) 2x – y = -2 …………..(iii)We need to equate the co-efficients of x or the co-efficients of y in any one of the above equations. Multiplying (iii) by 3 we get, 3(2x - y) = 3 * (-2)=> 6x - 3y = -6 …………(iv) Now Adding Eqn (i) & (iv) we get, (x+3y) + (6x – 3y) = 5 + (-6) (6x + x) + (3y-3y) = 5 – 6 => 7x = -1 => x = - 1/7Putting the value of x in (i) , we get -1/7 + 3y = 5 => 3y = 5 + 1/7 => y = 12/7

3) Method of comparison :Eg. Solve: x+y = 5 & y-2 = 2xx + y = 5 …which will be denoted by (i) y - 2 = 2x …which will be denoted by (ii)From equation (i) we get , x = 5 –yFrom equation (ii) we get, x = (y-2) / 2From the above , it is clear that , 5 – y = (y – 2) / 2 => 2 ( 5 – y ) = y – 2 => 10 – 2y = y – 2=> 10 + 2 = y + 2y => 12 = 3y => 3y = 12=> y = 12/3 => y = 4 Putting the value of y in (i), we get x = 5-4 = 1 The required values are x = 1, y = 4.

Solved Examples Eg 1) From the two simultaneous eqns given below, solve for x &y: 9x – 7y = 3 5x – 4y = 1Soln 1) Using substitution: The given equations: 9x – 7y = 3………….(i) 5x – 4y = 1………….(ii)From (ii) we get 5x = 1+ 4y => x = (1+4y) / 5 ……………..(iii) Now substituting the value of x from (iii) in (i), we get:

=>=> 9 + y = 15 => y = 6Putting the value of y in (iii) we get, x = (1+6*4) / 5 = 25/5 = 5 So the required values are x = 5 & y = 6

Soln 2) By equating the co-efficients: The given equations: 9x – 7y = 3………….(i) 5x – 4y = 1………….(ii)Multiplying Eqn (i) by 4, we get 4( 9x-7y) = 4*3 => 36x – 28y = 12…..(iii)Multiplying Eqn (ii) by 7, we get 7(5x – 4y) = 7* 1 => 35x – 28y = 7 …….(iv) Now subtracting (iv) from (iii) we get, 36x – 35 x = 12-7=> x = 5

Putting the value of x in (ii) we get, 25 – 4y = 1 => 4y = 24 => y = 6The required values are x= 5 & y = 6

Soln 3) By comparison: The given equations: 9x – 7y = 3………….(i) 5x – 4y = 1………….(ii)From Eqn (i), we get From Eqn (ii), we get

So it is clear that Find y, then find x.

Solved Examples Eg 2) Two pens and three pencils cost Rs. 86. Four pens and a pencil cost Rs. 112. What is the cost of one pen and one pencil?Soln: Let the cost of one pen be Rs. x. Let the cost of one pencil be Rs. y. Cost of two pens = 2 * x = Rs. 2xCost of three pencils = 3 * y = Rs. 3y It is given that cost of two pens and three pencils = Rs. 86That means , 2x + 3y = 86 ………..(i) Similarly, 4x + y = 112 ……………(ii) Solve for x & y.

Solved Examples Eg 3) The total monthly income of four men and two women is Rs. 46,000. The women earn Rs. 500 more than the men. What is the monthly income of a man and a woman? Soln.1: Let the income of one man be Rs. x. So income of one woman = Rs. (x + 500)Income of four men = 4 * x = Rs. 4xIncome of two women = 2 * (x+500) = Rs. 2x + 1000 It is given that income of four men and two women = Rs. 46000That means , 4x + (2x+1000) = 46000Solve for x which will be the income of one man. Find income of one woman.

Eg. 3) Soln.2: Let the income of each man be Rs. x Let the income of each woman be Rs. yIncome of four men = 4 * x = Rs. 4xIncome of two women = 2 * y = Rs. 2yIt is given that income of four men and two women = Rs. 46000That means , 4x + 2y = 46000…………….(i)And Income of each woman is Rs. 500 more than that of each man.That means , y = x+500 ……………………(ii)Solve for x & y.

Solved Examples Eg 4) The charges of a Taxi comprises of a fixed charge plus a per km charge for the distance covered. For a journey of 10 km, Rs. 75 is charged (fixed + per km charge) and for a journey of 15 km, Rs. 110 is charged (fixed + per km charge). What will be the charge for a journey of 25 km? Soln.1: Let the fixed charge be Rs. x. Let the per km charge be Rs. yFor a journey of 10 kms, the total charge is Rs. 75.That means, The fixed charge + Per km charge for 10 kms = 75=> x + y * 10 = 75 => x + 10y = 75 …………..(i)For a journey of 15 kms, the total charge is Rs. 110.That means, The fixed charge + Per km charge for 15kms = 110=> x + y * 15 = 110 => x + 15y = 110 ………….(ii)From (i) & (ii), find x & y. Then find the value of x+25y.

EQUATION & SLOPE OF A STRAIGHT LINE: • All linear equations with two variables of the form ax+by+c=0 can berepresented on a graph as a straight line.

• The equation ax+by+c=0 is therefore called the general equation of a line.

0 X

Y

X/

Y/

Let us consider an equation 5x + 2y - 20 = 0 Putting different values of x we can get values of y

If the values of x & y are plotted on a graph we get a straight line.

• Suppose we have an equation ax+by+c=0 and suppose putting the value of x as x1 we get value of y as y1, putting x as x2 we get y as y2 and so on: • (x1, y1) , ( x2,y2) etc. are called co-ordinates of the line. • From the co-ordinates we can find the slope or gradient of a line. • The Slope or gradient of a line is the angle that a line makes with the x axis in the positive direction. It is denoted by m. • So from the two co-ordinate points (x1, y1) & ( x2, y2), we can find the slope of the line ax+by+c=0 with the following formula: • If m is > 0 , the slope is positive and the line is upward slopping.• If m is = 0 , the slope is nil and the line is parallel to x axis.• If m is < 0, the slope is negative and the line is downward slopping.

• Suppose we have an equation x+y-5=0 and suppose putting the value of x as 3 we get value of y as 2, putting x as 4 we get y as 1 and so on: • (3, 2) , ( 4,1) etc. are called co-ordinates of the line. • From the co-ordinates we can find the slope or gradient of a line.

=> m = -1 • m is < 0, so the slope of the line x+y-5=0 is negative and this line is downward slopping.

SOLVING LINEAR SIM EQUATIONS WITH THREE UNKNOWN VARIABLESEG. 5) 2x + y = 15, 2y + z = 25, 2z + x = 26. Find the values of x, y & zSoln: Given: 2x + y = 15 ……(i) 2y + z = 25…….(ii)2z + x = 26…….(iii)From (i) we get, y = (15 – 2x)Putting this value of y in (ii) we get, 2 ( 15 – 2x) + z = 25=> 30 – 4x + z = 25=> - 4x + z = 25 – 30 => - (4x – z) = - 5 => 4x – z = 5 …..(iv) From (iii) , we get , x = (26 – 2z)Putting this value of x in (iv) we get, 4(26 – 2z) – z = 5 Solve for z and then find x & y.

Eg. 6 If 2x + 3y +z = 55, x+z–y= 4 and y–x+z = 12Find the values of x, y & z.Soln: 2x+ 3y + z = 55 …..(i) x + z – y = 4 …... (ii) y – x + z = 12 …..(iii) Your aim should be to eliminate one variable either x or y or zAdding (ii) & (iii) , we get(x + z – y) + (y – x + z) = 4 + 12=> x – x – y + y + z + z = 16=> 2z = 16 => z = 8 Rock on……..

Part 2 Quadratic Equations – formulation & solving

• In quadratic equations the power of the unknown variable is 2, i.e. x2, m2, p2 etc.• 2x2+6x – 14 =0, 5p2 – 3p = 23, etc. are examples of quadratic equations. • The general form of a quadratic equation is ax2+bx+c = 0, where ‘a’ is the co-efficient of x2, ‘b’ is the co-efficient of x & a≠0.• So in our example 2x2+6x – 14, a = 2, b = 6 and c = – 14 • 2x2 + 8 =0 is an example of a pure quadratic equation. The general form of such an equation is ax2+c=0, where a, the co-efficient of x2, and c are constant & a≠0.

- So, in an equation such as 5x2 – 12= 0, a = 5, c = – 12

Solving Quadratic equations: Finding the roots of the equation, i.e. values of x. • Unlike linear equations which can have multiple solutions, a quadratic equation has only two solutions, i.e., two values of x which satisfy the equation.• The two possible solutions are called the roots of the quadratic equation. • The roots are generally denoted by 𝛂 (Alpha) & 𝛃 (Beta) • By using formula: For a quadratic equation ax2+bx+c = 0NOTE: PLEASE DO NOT CONFUSE 𝛂 & 𝛃 (the roots) with a & b (the co-efficients)• The term (b2 - 4ac) is called the determinant or discriminant of the equation. • If b2 - 4ac = 0, then the value of 𝛂 = value of 𝛃 =• If b2 - 4ac > 0, then the value of 𝛂 and value of 𝛃 will be unequal. • The sum of the roots, i.e., 𝛂 + 𝛃 =• The product of the roots, i.e., 𝛂 * 𝛃 =

Solved Examples: Eg. 1) Solve the equation 6x2+7x-10 =0Note: Solving a quadratic equation means finding out the roots of the equation, i.e., the two possible values of x : 𝛂 & 𝛃 Solution : In our equation, 6x2+7x-10 =0, a = 6, b = 7 & c = (-10)First find out the value of b2 - 4ac which is 72 – 4 * 6 * (-10) = ______________Use the formulas: Find the values of 𝛂 & 𝛃. The required solution to be written as: x = Value of 𝛂 or x = Value of 𝛃

Solved Examples: Eg. 2) If 2 and 3 are the roots of the equation 3x2 – 2mx +2n = 0, find m & n. Solution 1): In our equation, 3x2 – 2mx +2n = 0 a = 3, b = & c = Also given that, 𝛂 = 2 and 𝛃 = 3.We know that 𝛂 + 𝛃 =

=> 2 + 3 ==> m = __________

We also know that 𝛂 * 𝛃 = => 2*3 ==> n = ____________

- 2m 2n

Solution 2): Since roots of the equation, 3x2 – 2mx +2n = 0 are 2 & 3So if x = 2 then : 3 * 22 – 2m * 2 +2n = 0 => 12 – 4m + 2n = 0 …….(i)And if x = 3 then : 3 * 32 – 2m * 3 + 2n = 0 => 27 – 6m+ 2n = 0 …….(ii)Solve (i) & (ii) and get value of m & n.

Solved Examples: Eg. 3) If one root of the equation x2 + px +12 = 0 is 4. Find the other root and the value of p. Solution 1) : In our equation, x2 + px +12 = 0 a = , b = & c = Also given that, 𝛂 = 4. Let the other root be 𝛃We know that 𝛂 * 𝛃 = From this find 𝛃. Then find p.Using 𝛂 + 𝛃 =

Solution 2): Since one root of the equation, x2 +px +12 = 0 is 4. So if x = 4 then : 42 +p*4 +12 = 0 => p = _________Let the other root be 𝛃.Using the value of p, find 𝛃.

Solved Examples: Eg. 3) If the equation x2 -7x +q = 0 has equal roots, find the value of q. Solution 1) : Since our equation, x2 -7x +q = 0 has equal roots

b2 - 4ac = 0 => (-7)2 – 4 * 1 * q =0=> q = __________

Some helpful formulae

• (a+b)2 = (a2+b2+2ab)

• (a-b)2 = (a2+b2- 2ab)

• (a2 – b2) = (a – b).(a+b)

• (a + b)3 = a3+b3+3ab(a+b)

• (a – b)3 = a3 - b3-3ab(a – b)

End of chapter 7

• LINEAR EQUATIONS: PRACTICE PROBLEMS1. 3(2x-1) = 5 – (3x-2). Find the value of x. 2. 3x – 5y = 20, 7x + 2y = 17. Find the value of x & y.3. If 2x +y = 35 and 3x + 4y = 65. Find the value of x/y.4. If a + b = 5 and 3a+2b = 20. Find the value of (3a+b). 5. If a fraction’s numerator is increased by 1 and the denominator is increased

by 2, then the fraction becomes 2/3. But when the numerator is increased by 5 and the denominator is increased by 1, the fraction becomes 5 /4. What is the value of the original fraction?

6. In a classroom, there were equal number of boys and girls. Eight girls left the room which resulted in the number of boys becoming double the remaining number of girls. How many boys and girls were there initially?

7. The price of 10 chairs is equal to that of 4 tables. If the price of 15 chairs and 2 tables together is Rs. 4000, find the total price of 12 chairs and 3 tables.

8. In a group of horses and people, the number of legs is 24 more than twice the number of heads. How many horses are there?

9. In an examination of 90 questions, 5 marks were allotted for every right answer and 2 marks for every wrong answer. A candidate attempted all 90 questions and scored a total of 387 marks. Find the number of right answers and number of wrong answers he gave.

10. A man has 85 currency notes out of which some are 100 Rupees notes and the rest are 50 Rupees notes. The total amount he has is Rs. 5000/-. How many 50 Rupees notes does he have?