- 1. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES CLASS X LEARNING OUTCOMES, EXPERIENTIAL CONTENT, EXPLANATION CONTENT, HOT SPOT, CURIOSITY QUESTIONS, MIND MAP, QUESTION BANK
- 2. LEARNING OUTCOMES • Understand to represent the algebraic situations algebraically and graphically. • Understanding of graphical method, substitution method, elimination method and cross multiplication method for solving linear equations. • Solving linear equations applicable in daily life. • Learners can describe system of pair of linear equations in two variables. • Learners can demonstrate proper graphical activity to solve the linear equations.
- 4. WHAT IS LINEAR EQUATION ? • A linear equation in two variables x and y is an equation that can be written in the form ax + by + c = 0, where a, b & c are real numbers and a and b are not both zero. • For an example, Linear equation in two variables i.e., x and y is -4x + 6y + 7 = 0. • We often denote the condition a and b are not both zero by (a2 + b2 ≠ 0). • We have also studied about that a solution of such an equation is a pair of values, one for x and one for y, which makes the two sides of the equation equal.
- 5. IMPORTANT NOTE • Note: A linear equation is an equation for a line. • Note: Each solution (x, y) of linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
- 6. HOW WE CAN SOLVE A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES? We have two methods for doing the same: 1. Graphical Method 2. Algebraic Method
- 7. GRAPHICAL METHOD Let us consider a system of two simultaneous linear equations in two variables. 2x + y = 5 -x + y = 2 For the equation 2x + y = 5 -------- (1) y = 5 – 2x x 0 1 2 y 5 3 1 And for the equation -x + y = 2 ------- (2) y = 2 + x x -1 0 1 y 1 2 3
- 9. TYPES OF SOLUTIONS OF SYSTEM OF EQUATIONS • One solution – the lines cross at one point (consistent pair of linear equations) • No solution – the lines do not cross anywhere (they are parallel) (inconsistent pair of linear equations) • Infinitely many solutions – the lines coincide (dependent pair of linear equations)
- 10. ALGEBRAIC METHOD To solve a pair of equations in two variables algebraically, we have three methods to do that: (i) Substitution Method (ii) Elimination Method (iii) Cross-multiplication Method
- 11. SUBSTITUTION METHOD We substitute the value of one variable by expressing it in terms of the other variable to solve the linear equations. Step 1: Pick any of the given equations, find the value of one variable, say x in terms of the other variable, i.e., y from the same equation, whichever is convenient. Step 2: Substitute the vale of x in other equation, and reduce it to an equation in one variable, i.e., in terms of y, which can be solved. Note: Sometimes we can get statements with no variable. If this statement is true, we can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent. Step 3: Substitute the values of y (or x) obtained in step 2 in the equation used in step 1, so that we can get the value of the other variable.
- 12. Example: - x + y = 2 -------- (i) 2x + y = 5 -------- (ii) Taking equation (i), - x + y = 2 Y = 2 + x -------- (iii) Substitute the value of y in equation (ii), we get 2x + (2 + x) = 5 3x + 2 = 5 3x = 5 – 2 3x = 3 x = 1 now, put the value of x in equation (iii), we get the value of y as well y = 2 + 1 y = 3 Hence, the solution of the pair of equation is (1, 3).
- 13. ELIMINATION METHOD • We substitute the value of one variable by expressing it in terms of the other variable to solve the linear equations. • Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal. • Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If we get an equation in one variable, go to Step 3. • If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. • If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent. • Step 3: Solve the equation in one variable (x oy y) so obtained to get its value. • Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
- 14. Example: - x + y = 2 -------- (i) 2x + y = 5 -------- (ii) Multiply by -2 in equation (i) so that it can get same coefficient of x as in equation (ii). -2 (-x + y) = -2 (2) 2x – 2y = -4 ------ (iii) Subtract equation (iii) from equation (ii) 2x + y = 5 2x – 2y = -4 - + + 3y = 9 => y = 3 putting the value of y in equation (i) we get, - x + 3 = 2 - x = - 1 x = 1 Hence, the solution of the pair of equations is (1,3)
- 15. CROSS-MULTIPLICATION METHOD To solve the pair of equations for x and y using cross-multiplication, we have to arrange the variables and their coefficients. 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 𝑎1 , 𝑎2 and 𝑏1, 𝑏2 and the coefficients 𝑐1 and 𝑐2 ⇒ x = 𝑏1 𝑐2−𝑏2 𝑐1 𝑎1 𝑏2−𝑎2 𝑏1 ⇒ y = 𝑐1 𝑎2−𝑐2 𝑎1 𝑎1 𝑏2−𝑎2 𝑏1 x y 1 𝑏1 𝑐1 𝑎1 𝑏1 𝑏2 𝑐2 𝑎2 𝑏2
- 16. Example: - x + y = 2 -------- (i) 2x + y = 5 -------- (ii) 𝑎1 = −1 , 𝑏1 = 1 , 𝑐1 = −2 , 𝑎2 = 2 , 𝑏2 = 1 , 𝑐2 = −5 ⇒ x = 𝑏1 𝑐2−𝑏2 𝑐1 𝑎1 𝑏2−𝑎2 𝑏1 ⇒ y = 𝑐1 𝑎2−𝑐2 𝑎1 𝑎1 𝑏2−𝑎2 𝑏1 (Using this) ⇒ x = 1 ×−5 − 1×−2 −1×1 − 2×1 ⇒ y = 2 ×−2 − −5×−1 −1×1 − 2×1 ⇒ x = −5+2 −1−2 = −3 −3 = 1 ⇒ y = −4−5 −1−2 = −9 −3 = −3 Hence, the solution of the pair of equations is (1, 3) x y 1 𝑏1 𝑐1 𝑎1 𝑏1 𝑏2 𝑐2 𝑎2 𝑏2 x y 1 1 −2 −1 1 1 −5 2 1
- 18. • Aim: To verify the conditions of consistency/ inconsistency for a pair of linear equations in two variables by Graphical method. • Requirements: Graph Paper, Stationary
- 19. Method of Construction: • Take a pair of linear equations in the form: 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 ------ (i) 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 ------ (ii) where 𝑎1, 𝑎2, 𝑏1, 𝑏2, 𝑐1, 𝑐2 are real and 𝑎1, 𝑎2, 𝑏1, 𝑏2 are not simultaneously zero. • There may be 2 cases. • Case I: 𝑎1 𝑎2 ≠ 𝑏1 𝑏2 • Case II: 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 • Case III: 𝑎1 𝑎2 = 𝑏1 𝑏2 ≠ 𝑐1 𝑐2 • Obtain the ordered pairs satisfying the pair of linear equations (i) and (ii) for each of the above cases. Plot the points obtained on a cartesian plane.
- 20. Demonstration 1: We obtain the graph as Fig. 1. The 2 lines are intersecting at one- point P. coordinates of the point (Case I) P (x, y) give the unique solution for the pair of linear equations (i) & (ii). Example: 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 2x – 5y + 4 = 0 ----- (i) 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 2x + y - 8 = 0 ------ (ii) 2x – 5y + 4 = 0 ----- (i) => 2x = 5y – 4 => x = 2.5y – 2 Y -4 -2 0 2 4 6 X -12 -7 -2 3 8 13
- 21. 2x + y - 8 = 0 ------ (ii) => y = -2x + 8 X -3 -2 -1 0 1 2 3 Y 14 12 10 8 6 4 2 𝑎1 = 2, 𝑎2 = 2, 𝑏1 = −5, 𝑏2 = 1 𝑎1 𝑎2 = 2 2 = 1 𝑏1 𝑏2 = − 5 1 = −5 ∵ 𝑎1 𝑎2 ≠ 𝑏1 𝑏2 ∴ The pair of linear equations is consistent and has one unique solution.
- 23. Demonstration 2: 4x + 6y = 24 ----- (i) (Case II) => 2(2x + 3y) = 2 (12) => 2x = 12 – 3y => x = 6 – 1.5y Y -4 -2 0 2 4 X 12 9 6 3 0 2x + 3y = 6 ----- (ii) => 2x = 6 – 3y => x = 3 – 1.5y Y -4 -2 0 2 4 6 X -9 -6 3 0 -3 -6 We obtain the graph as shown in Fig.2. The two lines are parallel. The pair of equations has no solution. (It is inconsistent) 𝑎1 𝑎2 = 𝑏1 𝑏2 ≠ 𝑐1 𝑐2 => 4 2 = 6 3 ≠ 24 6 => 2 = 2 ≠ 4
- 25. Demonstration 3: x – 2y = 5 ----- (i) (Case III) => x = 5 + 2y Y -4 -3 -2 -1 0 1 2 3 X -3 -1 1 3 5 7 9 11 3x – 6y = 15 ----- (ii) => 3(x – 2y) = 3(5) => x = 5 + 2y Y -4 -3 -2 -1 0 1 2 3 X -3 -1 1 3 5 7 9 11 𝑎1 𝑎2 = 3 1 , 𝑏1 𝑏2 = 6 2 = 3 1 , 𝑐1 𝑐2 = 15 5 = 3 1 We obtain the graph as shown in Fig. 3. The two lines are coincident. Thus, the pair of linear have infinitely many solutions. Therefore, the pair of equations with 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 .
- 27. Observation: 1. 𝑎1 = 2 𝑎2 = 2 𝑏1 = −5 𝑏2 = 1 𝑐1 = 4 𝑐2 = −8 2. 𝑎1 = 4 𝑎2 = 2 𝑏1 = 6 𝑏2 = 3 𝑐1 = −24 𝑐2 = −6 3. 𝑎1 = 1 𝑎2 = 3 𝑏1 = −2 𝑏2 = −6 𝑐1 = −5 𝑐2 = −15 Applications: Consistency helps to check whether a pair of linear equations have solutions or not, and if yes, how many. S.no 𝑎1 𝑎2 𝑏1 𝑏2 𝑐1 𝑐2 Case I, II, III Types of lines No. of solutions Conclusions 1. 𝟐 𝟐 −𝟓 𝟏 𝟒 −𝟖 Case I Intersecting 1 Consistent 2. 𝟒 𝟐 𝟔 𝟑 𝟐𝟒 −𝟔 Case II Coincident Infinite Inconsistent 3. 𝟏 𝟑 −𝟐 −𝟔 −𝟓 −𝟏𝟓 Case III Parallel 0 Dependent
- 28. HOT SPOT Focus Spot: Representation of an equation on a graph. Why? • Learners are confused as to how the solutions of linear equations can be derived using graph or graphical method. • They are unable to understand the reason for two equations having one set of solution, infinitely many solutions and no solutions at all. Impact: • Learners will find it difficult to derive the solution of two linear equations.
- 29. MEASURING ATTAINMENT OF LEARNING OUTCOMES The first step to create positive learning outcomes is to devise a plan with specific goals and determinations for learning outcomes. A strong teaching plan is student focused and includes: • Detailed information that spells out the goals and expectations for the students • What each student should know and be able to do upon completion of the class • Measurable assessment tools that gauge learning outcomes • Class and home assignments that help students clearly understand the subject matter that is being taught. Good student development and learning outcomes should include student feedback. During the planning process and throughout the course, it is important to include the students in the learning process by asking what they expect to get out of the course and the best way to help them understand the material. By including the students, it fosters an atmosphere of trust and shows that you are there for them and you care about each individual student’s ability to learn and retain the subject matter. Assessment tools • Assessment tools should be put into place to determine if learning outcomes are being met. Assessment tools can be the test of assignments given. Based on grades and the ability to understand the assignments given, the teacher can determine if each student understands the course material.
- 30. MIND MAP
- 32. QUESTION BANK
- 34. 1. AARON IS 5 YEARS YOUNGER THAN RON. FOUR YEARS LATER, RON WILL BE TWICE AS OLD AS AARON. FIND THEIR PRESENT AGES. Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
- 35. 2. IN A TWO-DIGIT NUMBER. THE UNITS DIGIT IS THRICE THE TENS DIGIT. IF 36 IS ADDED TO THE NUMBER, THE DIGITS INTERCHANGE THEIR PLACE. FIND THE NUMBER. Solution: Let the digit in the unit’s place is x And the digit in the tens place be y. Then x = 3y and the number = 10y + x The number obtained by reversing the digits is 10x + y. If 36 is added to the number, digits interchange their places, Therefore, we have 10y + x + 36 = 10x + y or, 10y – y + x + 36 = 10x + y - y or, 9y + x – 10x + 36 = 10x - 10x or, 9y - 9x + 36 = 0 or, 9x - 9y = 36 or, 9(x - y) = 36 or, 9(x - y)/9 = 36/9 or, x - y = 4 ………. (i)
- 36. Substituting the value of x = 3y in equation (i), we get 3y - y = 4 or, 2y = 4 or, y = 4/2 or, y = 2 Substituting the value of y = 2 in equation (i), we get x - 2 = 4 or, x = 4 + 2 or, x = 6 Therefore, the number becomes 26.
- 37. 3. IF TWICE THE AGE OF SON IS ADDED TO AGE OF FATHER, THE SUM IS 56. BUT IF TWICE THE AGE OF THE FATHER IS ADDED TO THE AGE OF SON, THE SUM IS 82. FIND THE AGES OF FATHER AND SON. Solution: Let father’s age be x years Son’s ages = y years Then 2y + x = 56 …………… (i) And 2x + y = 82 …………… (ii) Multiplying equation (i) by 2, (2y + x = 56)× 2 , we get
- 38. or, 3y/3 = 30/3 or, y = 30/3 or, y = 10 (solution (ii) and (iii) by subtraction) Substituting the value of y in equation (i), we get; 2 × 10 + x = 56 or, 20 + x = 56 or, 20 – 20 + x = 56 – 20 or, x = 56 – 20 x = 36
- 39. 4. IF 2 IS ADDED TO THE NUMERATOR AND DENOMINATOR IT BECOMES 9/10 AND IF 3 IS SUBTRACTED FROM THE NUMERATOR AND DENOMINATOR IT BECOME 4/5. FIND THE FRACTIONS. Solution: Let the fraction be x/y. If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have (x + 2)/(y + 2) = 9/10 or, 10(x + 2) = 9(y + 2) or, 10x + 20 = 9y + 18 or, 10x – 9y + 20 = 9y – 9y + 18 or, 10x – 9x + 20 – 20 = 18 – 20 or, 10x – 9y = -2 ………. (i)
- 40. If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have (x – 3)/(y – 3) = 4/5 or, 5(x – 3) = 4(y – 3) or, 5x – 15 = 4y – 12 or, 5x – 4y – 15 = 4y – 4y – 12 or, 5x – 4y – 15 + 15 = – 12 + 15 or, 5x – 4y = 3 ………. (ii) So, we have 10x – 9y = – 2 ………. (iii) and 5x – 4y = 3 ………. (iv) Multiplying both sided of equation (iv) by 2, we get 10x – 8y = 6 ………. (v) Now, solving equation (iii) and (v) , we get
- 41. • 10x – 9y = -2 • 10x – 8y = 6 - y = - 8 • y = 8 • Substituting the value of y in equation (iv) • 5x – 4 × (8) = 3 • 5x – 32 = 3 • 5x – 32 + 32 = 3 + 32 • 5x = 35 • x = 35/5 • x = 7 • Therefore, fraction becomes 7/8.
- 42. 5. TWO PENS AND ONE ERASER COST RS. 35 AND 3 PENCIL AND FOUR ERASERS COST RS. 65. FIND THE COST OF PENCIL AND ERASER SEPARATELY. Solution: Let the cost of pen = x and the cost of eraser = y Then 2x + y = 35 ……………(i) And 3x + 4y = 65 ……………(ii) Multiplying equation (i) by 4,
- 43. Subtracting (iii) and (ii), we get; 5x = 75 or, 5x/5 = 75/5 or, x = 75/5 or, x = 15 Substituting the value of x = 15 in equation (i) 2x + y = 35 we get; or, 2 × 15 + y = 35 or, 30 + y = 35 or, y = 35 – 30 or, y = 5 Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.
- 44. 6. A BOAT RUNNING DOWNSTREAM COVERS A DISTANCE OF 20 KM IN 2 HOURS WHILE FOR COVERING THE SAME DISTANCE UPSTREAM, IT TAKES 5 HOURS. WHAT IS THE SPEED OF THE BOAT IN STILL WATER? Solution: These types of questions are the real-time example of linear equations in two variables. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream. Let us consider the speed of a boat is u km/h and the speed of the stream is v km/h, then: Speed Downstream = (u + v) km/h Speed Upstream = (u – v) km/h We know that Speed = Distance/Time So, the speed of boat when running downstream = (20⁄2) km/h = 10 km/h The speed of boat when running upstream = (20⁄5) km/h = 4 km/h From above, u + v = 10 ……. (1) u – v = 4 ………. (2) Adding equation 1 and 2, we get: 2u = 1 u = 7 km/h Also, v = 3 km/h Therefore, the speed of the boat in still water = u = 7 km/h
- 46. 1. Using substitution method solve the below equation a. x−2y+300=0, 6x−y−70=0 b. 5x−y=5, 3x−y=3 Solution: a. From 1st equation x=2y−300 Substituting this in second equation 6(2y−300)−y−70=0 => y=170 Putting this 1st equation x=40 b. Similarly, this can be solved. Answer is (1,0)
- 47. 2. Using elimination method, solve the following a. x+y−40=0, 7x+3y=180 b. x+10y=68, x+15y=98 Solution: a. x+y−40=0 ---(1) 7x+3y=180 ---(2) Multiplying equation (1) by 7 7x+7y−280=0 ---(3) Subtracting equation 2 from equation 7 7x+7y−280=0 7x+3y=180 We get 4y=100 => y=25 Substituting this in (1) , we get x=15 b. ( 8,6)
- 48. Very Short Answer Questions
- 49. 1. Sum of two numbers is 35 and their difference is 13. Find the numbers. Solution: let the larger number be x and smaller be y. Given: x + y = 35 x – y = 13 adding both the equations we get 2x = 48 or x = 24 and y = 11 2. for what value of k, the system of equations 4x + 6y - 11 = 0 and 2x + ky -7 = 0 will be inconsistent? Solution: here 𝑎1 = 4, 𝑏1 = 6 𝑎𝑛𝑑 𝑐1 = −11 And 𝑎2 = 2, 𝑏2 = 𝑘 𝑎𝑛𝑑 𝑐2 = −7 𝑎1 𝑎2 = 4 2 𝑎𝑛𝑑 𝑏1 𝑏2 = 6 𝑘 For the system to be inconsistent Since 𝑎1 𝑎2 = 𝑏1 𝑏2 = 𝑐1 𝑐2 So 4 2 = 6 𝑘 => 𝑘 = 3
- 50. MULTIPLE CHOICE QUESTIONS The pair of linear equations 2x + 3y -5 = 0 and 3x – 2y + 5 has: (i) unique solution (ii) two solution (iii) no solution (iv) infinite solution
- 52. 1. Represent the following situation: Few friends of an apartment are made to stand in rows. If 3 friends are extra in a row, there would be one row less. If 3 friends are less in a row, there would be two rows more. 2. You are travelling 600 km apart by train and partly by car. It takes 8 hours and 40 minutes if you travel 320 km by train and rest by car. It would take 30 minutes more if you travel 200 km by train and the rest by the car/. Find the speed of the train and by car separately.
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