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Quantitative Methods Session 11 – Chapter 6 Linear Equations, Equation &Slope of a straight line & Quadratic Equations- Formulation & Solving Pranjoy Arup Das

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Part 1 1) Linear Equations – formulation & solving 2) Equation & Slope of a straight line.

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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

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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. x 1, y 1, p 1 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.

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Two or more linear equations containing the same variables are said to be simultaneous linear equations. Eg. x+y=5 y-2=2x Solving 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 = => 3y = 12 => y = 4 Substituting 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

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2) Elimination by equating the co-efficient : Eg. Solve: x+3y = 5 & y-2 = 2x x + 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 = - 1/7 Putting the value of x in (i), we get -1/7 + 3y = 5 => 3y = 5 + 1/7 => y = 12/7

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3) Method of comparison : Eg. Solve: x+y = 5 & y-2 = 2x x + y = 5 …which will be denoted by (i) y - 2 = 2x …which will be denoted by (ii) From equation (i) we get, x = 5 –y From equation (ii) we get, x = (y-2) / 2 From the above, it is clear that, 5 – y = (y – 2) / 2 => 2 ( 5 – y ) = y – 2 => 10 – 2y = y – 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.

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Solved Examples Eg 1) From the two simultaneous eqns given below, solve for x &y: 9x – 7y = 3 5x – 4y = 1 Soln 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 = 6 Putting 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

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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 = 6 The required values are x= 5 & y = 6

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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.

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

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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. 4x Income of two women = 2 * (x+500) = Rs. 2x It is given that income of four men and two women = Rs That means, 4x + (2x+1000) = Solve for x which will be the income of one man. Find income of one woman.

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Eg. 3) Soln.2: Let the income of each man be Rs. x Let the income of each woman be Rs. y Income of four men = 4 * x = Rs. 4x Income of two women = 2 * y = Rs. 2y It is given that income of four men and two women = Rs That 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.

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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. y For 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 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.

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EQUATION & SLOPE OF A STRAIGHT LINE: All linear equations with two variables of the form ax+by+c=0 can be represented on a graph as a straight line. The equation ax+by+c=0 is therefore called the general equation of a line. 0X Y X/X/ Y/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.

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Suppose we have an equation ax+by+c=0 and suppose putting the value of x as x 1 we get value of y as y 1, putting x as x 2 we get y as y 2 and so on: (x 1, y 1 ), ( x 2,y 2 ) 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 (x 1, y 1 ) & ( x 2, y 2 ), 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.

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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.

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SOLVING LINEAR SIM EQUATIONS WITH THREE UNKNOWN VARIABLES EG. 5) 2x + y = 15, 2y + z = 25, 2z + x = 26. Find the values of x, y & z Soln: 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.

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Eg. 6 If 2x + 3y +z = 55, x+z–y= 4 and y–x+z = 12 Find 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 z Adding (ii) & (iii), we get (x + z – y) + (y – x + z) = => x – x – y + y + z + z = 16 => 2z = 16 => z = 8 Rock on……..

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Part 2 Quadratic Equations – formulation & solving

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In quadratic equations the power of the unknown variable is 2, i.e. x 2, m 2, p 2 etc. 2x 2 +6x – 14 =0, 5p 2 – 3p = 23, etc. are examples of quadratic equations. The general form of a quadratic equation is ax 2 +bx+c = 0, where ‘a’ is the co-efficient of x 2, ‘b’ is the co-efficient of x & a≠0. So in our example 2x 2 +6x – 14, a = 2, b = 6 and c = – 14 2x =0 is an example of a pure quadratic equation. The general form of such an equation is ax 2 +c=0, where a, the co- efficient of x 2, and c are constant & a≠0. - So, in an equation such as 5x 2 – 12= 0, a = 5, c = – 12

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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 (Alpha) & (Beta) By using formula: For a quadratic equation ax 2 +bx+c = 0 NOTE: PLEASE DO NOT CONFUSE & (the roots) with a & b (the co-efficients) The term (b 2 - 4ac) is called the determinant or discriminant of the equation. If b 2 - 4ac = 0, then the value of = value of = If b 2 - 4ac > 0, then the value of and value of will be unequal. The sum of the roots, i.e., + = The product of the roots, i.e., * =

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Solved Examples: Eg. 1) Solve the equation 6x 2 +7x-10 =0 Note: Solving a quadratic equation means finding out the roots of the equation, i.e., the two possible values of x : & Solution : In our equation, 6x 2 +7x-10 =0, a = 6, b = 7 & c = (-10) First find out the value of b 2 - 4ac which is 7 2 – 4 * 6 * (-10) = ______________ Use the formulas: Find the values of &. The required solution to be written as: x = Value of or x = Value of

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Solved Examples: Eg. 2) If 2 and 3 are the roots of the equation 3x 2 – 2mx +2n = 0, find m & n. Solution 1): In our equation, 3x 2 – 2mx +2n = 0 a = 3, b = & c = Also given that, = 2 and = 3. We know that + = => = => m = __________ We also know that * = => 2*3 = => n = ____________ - 2m2n

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

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Solved Examples: Eg. 3) If one root of the equation x 2 + px +12 = 0 is 4. Find the other root and the value of p. Solution 1) : In our equation, x 2 + px +12 = 0 a =, b = & c = Also given that, = 4. Let the other root be We know that * = From this find. Then find p. Using + =

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

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Solved Examples: Eg. 3) If the equation x 2 -7x +q = 0 has equal roots, find the value of q. Solution 1) : Since our equation, x 2 -7x +q = 0 has equal roots b 2 - 4ac = 0 => (-7) 2 – 4 * 1 * q =0 => q = __________

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Some helpful formulae (a+b) 2 = (a 2 +b 2 +2ab) (a-b) 2 = (a 2 +b 2 - 2ab) (a 2 – b 2 ) = (a – b).(a+b) (a + b) 3 = a 3 +b 3 +3ab(a+b) (a – b) 3 = a 3 - b 3 -3ab(a – b)

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End of chapter 7

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LINEAR EQUATIONS: PRACTICE PROBLEMS 1.3(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?

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