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Topic Pair of Linear Equtions in two variables

Topic Pair of Linear Equtions in two variables. OBJECTIVES. To enable the learner to form linear equations in two variables from the verbal questions and to solve the problem using various methods. To develop the numerical,problem solving and comprehensive abilities.

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Topic Pair of Linear Equtions in two variables

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  1. Topic Pair of Linear Equtions in two variables

  2. OBJECTIVES • To enable the learner to form linear equations in two variables from the verbal questions and to solve the problem using various methods. • To develop the numerical,problem solving and comprehensive abilities. • To develop positive and critical thinking.

  3. AN OVERVIEW In earlier classes we have studied about the linear equations in one and twovariables,solving linear equations in one variable and to draw the graph of linear equations in one and two variables. In the following modules we will learn how to solve a pair of linear equations in two variables using different methods and apply the knowledge to solve practical and day to day problems

  4. MODULE 1SOLVING EQUATIONS GRAPHICALLY Steps to solve a problem graphically: • Take equation 1,give any value to one of the variables and find the value of the other variable and prepare a table of values. • Similarly take equation 2 and prepare a table of values. • Draw the graph of both the equations taking suitable scale. • Determine the coordinates of the point if they intersect,that gives the solution of the given equations • If the graph go parallel,there are no solutions. • If both the graph represent the same line,then there are infinite solutions.

  5. INTRODUCTION Naveen went to a fair in his town. He wanted to buy books and toys. The number of toys purchased is half the number of books. If a book costs Rs.3 and a toy costs Rs.4 and he had spent Rs20,then how can we find the number of books and toys bought? This problem can be solved graphically and algebraically .First let us solve graphically. To solve this let us apply the knowledge of equations learnt in class- 9.Now the situation can be represented by two equations. Let the number of books bought be = x And the number of toys bought be = y Now we have the equations y = 1/2x………….. (1) 3x + 4y = 20……………. (2)

  6. Let us the take the example given above. Naveen goes to a fair with Rs.20 and wants to buy books and toys. Represent this situation graphically. Solution: The pair of equations formed is : y = 1/2x x – 2y = 0 3x + 4y = 20 Let us represent these equations graphically. For this, we need at least two solutionsfor each equation. We give these solutions in table X 0 2 X 0 20/3 4 Y 5 0 2 Y = x/2 0 1 (i) (ii)

  7. Plot the points A(0,0),B(2,1) and P(0,5),Q(4,2),corresponding to the solutions in table.Now draw the lines AB and PQ representing the equations x – 2y = 0 and 3x + 4y = 20, as shown. In figure observe that the two lines representing the two equations are intersecting at the point(4,2) . We shall discuss what this means in the next section.

  8. P (0,5) 5 - 4 - 3 - 2 - 1 - o Y - , X- 2y = 0 P 1i Q(4,2) 3x + 4y = 20 . A(0,0) B(2,1) X’ i i i i i oo1 1 2 3 4 5 6 Solution X = 4 , y = 2 , (x = no.books)(y = no.toys) Y’

  9. Example ;1 Priya went to a stationery shop and purchased 2 pencils and 3 erasers for Rs 9. Her friend Preethi saw the new variety of pencils and erasers with Priya and she also bought 4 pencils and 6 erasers of the same kind for Rs 18. Represent this situation algebraically and graphically. Solution Let us denote the cost of 1 pencil by Rs x and 1 eraser by Rs y.Then the algebraic representation is given by the following equation 2x + 3y = 9 ……………….. [1] 4x + 6y = 18 ……………… [2] To obtain the equivalent geometric representation , we find 2 points on the line represening each equation . ie , we find 2 solutions of each equation.

  10. Graphicalrepresentation These solutions are given below, 2x + 3y = 9 4x + 6y = 18 X 0 3 X 0 4.5 Y 3 1 Y 3 0 [ii] [i] We plot these points in a graph paper and draw the lines.We find that both the lines coincide . This is so, because ,both the equations are equivalent ,i.e.,one can be derived from the other .

  11. 5 - 4 - 3 - 2 - 1 - o Y - 4x + 6y = 18 2x + 3y = 9 (0,3) 1i . (3,1) (4.5,0) X’ i i i i i X oo11 2 3 4 5 6 Y’

  12. Example ;2 Two rails are represented by the equations x + 2y – 4 = 0 and 2x+ 4y - 12 = 0. Represent this situation graphically. X+2y – 4 = 0 [1] 2x+4y –12=0 [2] Two solutions for each equations are given below X 0 4 X 0 6 Y 2 0 Y 3 0

  13. To represent the equations graphically , we plot the points R(0,2) and S(4,0), to get the line RS and the points P(0,3) and Q(6,0) to get the line PQ. We observe in fig. ,that lines do not intersect any where , i.e ., they are parallel. So ,we have seen several situations which can be represented by a pair of linear equations . We have seen their algebraic and geometric representations .In the next few sections, we will discuss how these representations can be used to look for solutions of the linear equations .

  14. 5 - 4 - 3 - 2 - 1 - o Y - P(0,3) 1i 2x + 4y – 12 = 0 R(0,2) Q(6,0) X +2y - 4 = 0 X’ i i i i i X oo11 2 3 4 5 6 S(4,0) Y’

  15. GENERAL FORM OF LINEAR EQUATIONS IN TWO VARIABLES a1x + b1y+c1 =0 a2x+ b2y +c2 =0 Where a1,a2,b1,b2,c1 and c2 are reals and a1≠0,a2≠0,b1≠0,b2≠0.

  16. CONDITIIONS FOR CONSISTENCY AND INCONSISTENCY • If a1 /a2 ≠ b1 /b2 • The system has a unique solution and is said to be consistent and independent. • If a1 /a2 = b1 /b2 =c1/c2 • The system has infinitely many solutions and is said to be consistent and dependent.

  17. CONDITIONS (CONT) • If a1 /a2 =b1 / b2 ≠ c1/ c2 • The system has no solution and is said to be • Inconsistent.

  18. MODULE : 1 Graphical method: ( summary ) The graph of a pair of linear equations in two variables is represented by two lines i) If the lines intersect at a point ,then the that point gives the unique solutions of the two equations .In this case the pair of equations is consistent. ii) If the line coincides, then there are infinitely many solutions. Each point on the line being a solution, in this case the pair of equations is dependent. iii) If the lines are parallel,then the pair of equations has no solutions. In this case the pair of equations is inconsistent.

  19. ASSIGNMENT :MODULE 1 Solve the following pair of linear equations graphically. • 1) 2x – 3y – 17 = 0, 4x + y – 13 = 0 • 2) 3x + 2y = 0 , 2x + y = -1 Answers • 1) x = 4, y = -3 • 2) x = -2, y = 3

  20. 3.) Prove graphically the system of equations x – 2y = 6 & 3x – 6y = 0 has no solution. . 4) Show graphically that 2y = 4x – 6 , 2x = y + 3 has infinite solutions. 5) .Draw the graph of equations 2x – y = -8 8x + 3y = 24 Determine the vertices of the triangle formed by the lines representing these equations and x-axis.Shade the triangular region so formed. Answer- the vertices of the triangle are: (0,8) , (-4,0) , (3,0)

  21. Module : 2 Solving a pair of linear equations by algebraic methodElimination by substitution In the previous section , we learnt how to solve a pair of linear equations graphically. It is true that anything visual has a certain appeal. This method will not be suitable if the solutions are in decimal or rationals. Therefore we need other methods to solve the equations. Hence the algebraic methods were introduced. Step 1: Take any of the given linear equations and find the value of one of the variables in terms of other variable.ie if the variables are x and y.Find the value of y in terms of x. Step 2: Substitute the value of y ,obtained in step 1,in the other equation which will give an equation only in x. Stap 3: Solve the equation obtained in step 2. To get the value of x. Step 4: Substitute the value of x as obtained in step 3 in the expression for y in terms of x & solve the equation to get the value of y.

  22. Example -1: Solve x + y = 7 & 3x – 2y = 11 using the method of substitution. Solution : Name the given equations as 1 & 2 X + y = 7 …………(1) 3x – 2y = 11 …………(2) From (1) we get y = 7 – x Substituting the value of y in the eqn (2) We get 3x – 2(7 – x) = 11 3x – 14 + 2x = 11 5x – 14 = 11 5x = 25, x = 5 Now y = 7 – 5 = 2 , solution is x = 5, y = 2

  23. Alternatively x + y = 7…(1) , 3x – 2y = 11…(2) , From equation (1) , find the value of x x = 7 – x ……………………… (3) substitute the value of x in eqn 3x – 2y = 11 ……………………… (2) 3(7 – y) – 2y = 11 21 – 3y – 2y = 11 21 –5y =11 -5y = 11-21 -5y = -10 y = 2 Put y= 2 in eqn (3),x=7 – 2 = 5 . Solution x = 5 ,y = 2 . Verification:put x = 5, y = 2 in any of the eqn (1) or (2) x + y = 7, 5 + 2 = 7.LHS = RHS . Thus the solution is correct.

  24. THANK YOU

  25. MODULE : 3Elimination by equating coefficients Step 1:Multiply both the given eqns by suitable non zero no.(constants) so thatcoefficients of x (or coefficients of y ) in both the eqns become equal. Step 2 :Add or subtract as required ,one eqn from the other ,so that the variable with equal coefficients gets eliminated. Step 3 : Solve the eqn obtained in step 2,to find the value of the variable which is left in step 2. Step 4 : Substitute the value of the variable obtained in step 3,in any one the given eqn and get the value of the other variable .

  26. MODULE-3 :EXAMPLE 1 Solve : 2x + 3y = 9 …… (1) 3x +4y = 5………(2) ,by using the method of elimination by equating the coefficients Solution Step 1 : In oder to make the coefficients of variable x equal multiply eqn (1) by 3 and multiply the eqn (2) by 2. Thus (2x + 3y) 3 = 9 x 3. 6x + 9y = 27 …………………………(3) (3x + 4y ) 2 = 5 x 2 6x + 8y = 10 ……………………..(4) On subtracting 4 from 3 ,we get y = 17 Subtituting the value of y in (1) we get 2x + 3y = 9 2x + 3 (17 ) = 9 2x = 9 – 51 = -42 x = - 21

  27. ALTERNATE METHOD We name the eqns 2x + 3y = 9 ……..(1) 3x + 4y = 5 ……(2) Multiply first eqn by 4 and by 3 . We get 8x + 12y = 36 ………….(3) 9x +12y = 15 ………….(4) On subtracting eqn 4 from 3, -x = 21 x = -21 Subtituing x = -21 in eqn (1) we get 2(-21) + 3y = 9 -42 + 3y = 9 y = 17 Verification : put x = -21 and y = 17 in eqn (2) 3(-21) + 4(17) = 5 ,we get 5 = 5, LHS = RHS . Thus the solution is correct.

  28. MODULE – 4 By cross multiplication Solution for the general form of pair of linear equation. Step 1 ; leta1x + b1y +c1 = 0 ……………………………… (1) a2x + b2y +c2 = 0 ………………………………(2) Be the 2 equations Step 2 ; 1) Write the coefficients of x , y & constant term as a1 b1 c1 a2 b2c2 2) Rewrite the first two columns after third column. As shown below. a1 b1 c1 a1 b1 a2 b2 c2 a2 b2 3) Now write x between 2nd and 3rd columns , y between 3rd and 4th & 1 between 4th and 5th.Also mark arrows 1 from left top to right bottom and other from left bottom to right top . x y +1 a1 b1 c1 a1 b1 a2 b2 c2 a2 b2

  29. Step 3 obtain three equal fractions with numerator x ,y,1 For their denominators multiply the numbers with down arrows and from their products subtract the product of the nos.with upward arrows. Thus we get x = y = 1 b1-c2-b2c1 c1 a2 –c2 a1a1 b2 –a1 b1 Step 4: Obtain the value of x by equating the first and third expressions. Obtain the value of y by equating the second and third expressions. Thus x = b1-c2-b2c1 y = c1 a2 –c2 a1 a1 b2 –a1 b1 a1 b2 –a1 b1 Note if the constant are on RHS of the eqns, then put –1 instead of +1in step 2.

  30. MODULE -4 : EXAMPLE -1 Solve the follwing pair of equations using cross multiplication method . 2x +3y – 6 = 0 & 6x – 5y – 4 = 0 Solution : x y +1 2 3 - 62+3 6 -5 -4 6 -5 x = y = 1 3(-4 ) –(-5 )(-6 ) -6 (6) – (-4 ) 2 2(-5 ) – 6 x 3 x = y = 1 -42 -28 -28 x = 3/2 y = 1

  31. ASSIGNMENT: FOR MODULES 2,3 & 4 Solve the following pair of equations algebraically,using substitution method,elimination method and cross multiplication method and check your answer. • 2x + 3y = 7 6x + 5y = 11 • 7x – 2y = 3 11x – 3/2y = 8 • 4x + 7y = 10 10x – 35/2y = 25 4.3x – 5y =20 7x + 2y = 17 5. 6x + 5y = 11 9x + 10y= 21 6. 4x + 2/3y – 1 = 0 6x – y + 2 = 0 7. ax + by = c bx + ay = 1+c 8. x/a + y/b = a+b x/a + y/b = 2 2 2

  32. ANSWERS FOR THE PRACTICE PROBLEMS (MODULE – 2,3& 4) • x = 17/7 , y = 5/7 2. x = 1 , y = 2 3. x = 5/2 , y = 0 4. x = 125/41 , y=89/41 • x = 1/3 , y = 9/5 • x = -1/24 , y = 7/4 • x=bc – ac+b /b - a y= bc – ac-a / b - a • x = a , y = b 2 2 2 2 2 2

  33. MODULE :5 SOLVING WORD PROBLEMS Step 1 : Read and re-read the statement of the problem carefully and determine what quantities are to be found. Step 2 : Represent the unknown quantities by letter of english alphabets say x & y. Step 3 : According to the condition given in the problem ; form two eqns involving the required unknown quantities. Step 4 : Solve the equation to get the values of required unknown quantities. Example1: The sum of two numbers is 11. Find the nos. if two times one of these nos. exceeds three times the other number by 2. Soln: Let the nos. be x & y. therefore x + y = 11………………… (1) By 2nd condition, 2x – 3y = 2…………………..(2) Eqn (1) x 3 3x + 3y = 33…………………(3) Adding (1) + (3) 5x = 35 x = 7 Substituting in (1) x + y = 11 7 + y = 11 y = 4 Therefore two nos. are 7 & 4

  34. EXAMPLE 2 : The sum of the digit of two digit no. is 12.The no. obtained by reversing order of the digit of the given no. exceeds the given no. by 18. Find the 2 digit no. Soln: Let the digit at tens place = x & the digit at the units place = y Therefore required no. = 10x + y x + y = 12 …………………….(1) & (10y + x) – (10x + y) = 18 9y – 9x = 18 & y – x = 2 ………(2) (1) + (2) we get 2y = 14, y = 7, Substituting y = 7 in(1) x + y = 12, x +7 = 12, x = 5 Required no.is 10x + y = 10 x 5 + 7 = 57

  35. ASSIGNMENT-MODULE:5 1. Ram is three time as old as Rahim.Five years later, Ram will be two-and-a-half times as old as Rahim.How old are Ram and Rahim now? 2. Five years ago, Neeta was thrice as old as Geeta.Ten years later, Neeta will be twice as old as Geeta. How old are Neeta and Geeta now. 3. Two audio cassettes and three video cassettes cost Rs.340. But three audio cassettes and two video cassettes cost Rs.260.Find the price of audio an cassettes and that of a video cassettes . 4. The area of a rectangle gets reduced by 9 square, if its length is reduced by5 units and the breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, then the area increased by 67 square units.Find the length and breadth of the rectangle.

  36. 5. A two digit number is obtained by either multiplying the sum of the digits by 8 and addung 1, or by multiplying the difference of the digits by 13& adding 2.Find the number . How many such members are there? 6. Points A and B are 100 km apart on a high way. one car starts from Aand from B at the same time.If the cars travel in the same direction at a constant speed ,they meet in 5 hrs. If the cars travel towards each other, they meet in 1hour. what are the speeds of the 2 cars. 7. A person can row downstream 20 km in 2hrs. an upstream 4km in2hrs. Find man’s speed of rowing in srill water and the speed of the current. 8. A person can row 8 km upstream and 24 km down stream in 4hrs. He can row 12 km down stream and 12km upstream in 4hrs . Find the speed of the person in still water as also the speed f the current

  37. MODULE- 6EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES Eg: solve the folowing pair of equations by reducing them to a pair of L.E : 5 + 1 = 2 x – 1 y – 2 6 - 3 = 1 x-1 y-2 Solution : let us put 1/ x-1 =p & 1/y-2 =q . Then the given equations 5[1/ x-1] + 1/y-2 = 2 ………….(1) 6(1/ x-1 ) - 3(1/y-2 ) =1 ……..(2)

  38. The given equations can be written as 5p+ q = 2……………(3) 6p-3q = 1……………(4) equations (3) & (4) are in the general form . We can solve by using any method discussed eariler. We get p =1/3 &q=1/3. now substituting 1/ x-1 for p, we have 1/ x-1 =1/3 i.e. X-1 =3 , i.e x =4 similarly, substituting 1/y-2 for q , we get 1/y-2 =1/3 , i.e. 3= y-2 , i.e. y =5 Hence , x=4, y=5 is the requried solution of the given pair of equations

  39. Frequently Asked Questions • What is the general form of a linear equation in two variables? An eqn. of the form ax + by+ c=0 or ax + by =c where a,b,c are real numbers & x&y are two variables , a =0, b=0 is called a linear equation in two variables. for eg: 3x+5y-7 = 0 . • What is called solution of a linear eqn.? Any pair of values m&n is said to be the solution of eqn. ax + by+ c=0 , if am+bn+c =0

  40. Why does a linear eqn. in two variables has infinite solution ? • Graph of a linear equation in two variables is always a straight line .. As there are infinite points on a straight line ,there are infinite solutions. • What are called solution of a pair of linear equation ? • It is an ordered pair of real nos. which satisfy both the linear equation in two variables.

  41. How do we know the solution is correct? • We substitute the values of the variables found in both the eqns. If L.H.S. are equal to R.H.S. in both the eqns ,then the solution is correct . • 6. What are called consistant equations? • A pair of equation is consistant if it has at least one solution. • 7. What are called inconsistant equations? • A pair of equation is inconsistant if it has no solution.

  42. 8. Can we algebrically findout the number of solutions and without solving and relate it with graph? • Algebraic condtions for number of solutions are given below. • if a1/a2 = b1/ b2 then the pair of equations have a unique solutions . Graphically the two lines intersect at a point. • if a1/a2 = b1/ b2 = c1/c2 then the pair of equations have a infinite solutions. Graphically the two lines coincide or overlape each other. • if a1/a2 = b1/ b2 = c1/c2 then the pair of equations have no solutions. Graphically the two lines are parallel

  43. References • Mathematics – A text book for class- 10 by NCERT 2003 edition & new book. • Comprehensive- Mathematics for class 10 by R.K Bansal. • Mathematics for class 10 by R.L.Arora. • Question bank in maths by M.L.Aggarwal & R.G.Gupta. 5. www.google.com

  44. GLOSSARY Linear eqution-an algebraic equation of degree one Variable-the quantity which varies ,normally variables are denoted by english alphabats. Constant- the quantity which remains same Solution- the pair of values which satisfy the given equation Solve- finding the values of the variables Coefficient –the constant or the factors except the Variable Real –the rational and irrational taken together are rational Number Infinite solutions-uncountable solutions

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