Welcome dear students! Today we are going to learn about Pair of Linear Equations in Two Variables from Class 10 Maths.
Let us begin with section 3.1 Introduction. You must have come across situations like the one given below. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla, which is a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs 3, and a game of Hoopla costs 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent 20. May be you will try it by considering different cases. If she has one ride, is it possible? Is it possible to have two rides? Or you may use the knowledge of Class 9, to represent such situations as linear equations in two variables. Let us try this approach. Denote the number of rides that Akhila had by x, and the number of times she played Hoopla by y. Now the situation can be represented by the two equations: y = 1/2 x, and 3x + 4y = 20. Can we find the solutions of this pair of equations? There are several ways of finding these, which we will study in this chapter.
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Moving to section 3.2 Graphical Method of Solution of a Pair of Linear Equations. A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent. We can now summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions as follows: first, the lines may intersect in a single point. In this case, the pair of equations has a unique solution, which is a consistent pair of equations. Second, the lines may be parallel. In this case, the equations have no solution, which is an inconsistent pair of equations. Third, the lines may be coincident. In this case, the equations have infinitely many solutions, which is a dependent consistent pair of equations.
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Consider the following three pairs of equations. First, x − 2y = 0 and 3x + 4y − 20 = 0. The lines intersect. Second, 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0. The lines coincide. Third, x + 2y − 4 = 0 and 2x + 4y − 12 = 0. The lines are parallel. Let us now write down, and compare, the values of a₁/a₂, b₁/b₂, and c₁/c₂ in all the three examples. Here, a₁, b₁, c₁ and a₂, b₂, c₂ denote the coefficients of equations given in the general form. For the first pair, x − 2y = 0 and 3x + 4y − 20 = 0, the ratios are 1/3, −2/4, and 0/−20. Here, a₁/a₂ ≠ b₁/b₂. The graphical representation shows intersecting lines, and the algebraic interpretation is exactly one unique solution. For the second pair, 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0, the ratios are 2/4, 3/6, and −9/−18. Here, a₁/a₂ = b₁/b₂ = c₁/c₂. The graphical representation shows coincident lines, and the algebraic interpretation is infinitely many solutions. For the third pair, x + 2y − 4 = 0 and 2x + 4y − 12 = 0, the ratios are 1/2, 2/4, and −4/−12. Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂. The graphical representation shows parallel lines, and the algebraic interpretation is no solution.
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From the table above, you can observe that if the lines represented by the equation a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are intersecting, then a₁/a₂ ≠ b₁/b₂. If they are coincident, then a₁/a₂ = b₁/b₂ = c₁/c₂. If they are parallel, then a₁/a₂ = b₁/b₂ ≠ c₁/c₂. In fact, the converse is also true for any pair of lines. You can verify them by considering some more examples by yourself.
Let us now consider Example 1. Check graphically whether the pair of equations x + 3y = 6 and 2x − 3y = 12 is consistent. If so, solve them graphically. Solution: Let us draw the graphs of the equations. For this, we find two solutions of each equation. For x + 3y = 6, when x is 0, y is 2. When x is 6, y is 0. For 2x − 3y = 12, when x is 0, y is −4. When x is 3, y is −2. Plot the points A(0, 2), B(6, 0), P(0, −4) and Q(3, −2) on graph paper, and join the points to form the lines AB and PQ. We observe that there is a point B(6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, which means the given pair of equations is consistent.
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Example 2: Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions: 5x − 8y + 1 = 0 and 3x − 24/5 y + 3/5 = 0. Solution: Multiplying the second equation by 5/3, we get 5x − 8y + 1 = 0. But, this is the same as the first equation. Hence the lines represented by the equations are coincident. Therefore, the equations have infinitely many solutions. Plot few points on the graph and verify it yourself.
Example 3: Champa went to a Sale to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased. Help her friends to find how many pants and skirts Champa bought. Solution: Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are: y = 2x − 2 and y = 4x − 4. Let us draw the graphs by finding two solutions for each equation. For y = 2x − 2, when x is 2, y is 2. When x is 0, y is −2. For y = 4x − 4, when x is 0, y is −4. When x is 1, y is 0. Plot the points and draw the lines passing through them. The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution, meaning the number of pants she purchased is 1 and she did not buy any skirt. Verify the answer by checking whether it satisfies the conditions of the given problem.
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Now we will solve Exercise 3.1 completely. Question 1(i): 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Let number of boys be x and girls be y. Equations: x + y = 10 and y = x + 4. For x + y = 10, points are (0, 10) and (10, 0). For y = x + 4, points are (0, 4) and (1, 5). Plotting these, lines intersect at (3, 7). Solution: 3 boys, 7 girls. Question 1(ii): 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen. Let pencil cost be x, pen cost be y. Equations: 5x + 7y = 50 and 7x + 5y = 46. For first, points (0, 50/7) and (10, 0). For second, points (0, 46/5) and (46/7, 0). Solving graphically, intersection is at (3, 5). Cost of pencil is 3, pen is 5.
Question 2(i): 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0. Ratios: a₁/a₂ = 5/7, b₁/b₂ = −4/6 = −2/3. Since 5/7 ≠ −2/3, lines intersect at a point. Question 2(ii): 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0. Ratios: a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2. All equal, so lines are coincident. Question 2(iii): 6x − 3y + 10 = 0 and 2x − y + 9 = 0. Ratios: a₁/a₂ = 6/2 = 3, b₁/b₂ = −3/−1 = 3, c₁/c₂ = 10/9. a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so lines are parallel.
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Question 3(i): 3x + 2y = 5 and 2x − 3y = 7. Ratios: a₁/a₂ = 3/2, b₁/b₂ = 2/−3. Not equal, so consistent. Question 3(ii): 2x − 3y = 8 and 4x − 6y = 9. Ratios: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −3/−6 = 1/2, c₁/c₂ = 8/9. a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so inconsistent. Question 3(iii): (3/2)x + (5/3)y = 7 and 9x − 10y = 14. Rewrite first: 9x + 10y = 42. Ratios: a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/−10 = −1. Not equal, so consistent. Question 3(iv): 5x − 3y = 11 and −10x + 6y = −22. Ratios: a₁/a₂ = 5/−10 = −1/2, b₁/b₂ = −3/6 = −1/2, c₁/c₂ = 11/−22 = −1/2. All equal, so consistent (dependent). Question 3(v): (4/3)x + 2y = 8 and 2x + 3y = 12. Ratios: a₁/a₂ = (4/3)/2 = 2/3, b₁/b₂ = 2/3, c₁/c₂ = 8/12 = 2/3. Since a₁/a₂ = b₁/b₂ = c₁/c₂, the pair is consistent and dependent.
Question 4(i): x + y = 5 and 2x + 2y = 10. Ratios: 1/2 = 1/2 = 5/10. Consistent. Graphically, coincident lines, infinitely many solutions. Question 4(ii): x − y = 8 and 3x − 3y = 16. Ratios: 1/3 = −1/−3 = 1/3, but 8/16 = 1/2. 1/3 ≠ 1/2. Inconsistent. No solution. Question 4(iii): 2x + y − 6 = 0 and 4x − 2y − 4 = 0. Ratios: 2/4 = 1/2, 1/−2 = −1/2. Not equal. Consistent. Intersect at (2, 2). Question 4(iv): 2x − 2y − 2 = 0 and 4x − 4y − 5 = 0. Ratios: 2/4 = 1/2, −2/−4 = 1/2, −2/−5 = 2/5. 1/2 ≠ 2/5. Inconsistent. No solution.
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Question 5: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find dimensions. Let length be x, width be y. Equations: x + y = 36 and x = y + 4. Substitute: (y + 4) + y = 36, 2y = 32, y = 16. x = 20. Dimensions: length 20 m, width 16 m. Question 6: Given 2x + 3y − 8 = 0. (i) Intersecting: need a₁/a₂ ≠ b₁/b₂. Choose 3x + 2y − 4 = 0. (ii) Parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Choose 4x + 6y − 10 = 0. (iii) Coincident: a₁/a₂ = b₁/b₂ = c₁/c₂. Choose 4x + 6y − 16 = 0. Question 7: Draw x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine vertices of triangle with x-axis. For first line, points (0, 1), (−1, 0). For second line, points (0, 6), (4, 0). Intersection of lines: solve x − y = −1 and 3x + 2y = 12. Multiply first by 2: 2x − 2y = −2. Add to second: 5x = 10, x = 2. y = 3. Intersection is (2, 3). x-axis intercepts are (−1, 0) and (4, 0). Vertices are (−1, 0), (4, 0), and (2, 3). Shade the triangular region bounded by these points.
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Now section 3.3 Algebraic Methods of Solving a Pair of Linear Equations. The graphical method is not convenient when coordinates are non-integral. Is there any alternative method? There are several algebraic methods. Subsection 3.3.1 Substitution Method. Example 4: Solve 7x − 15y = 2 and x + 2y = 3 by substitution. Step 1: From equation 2, x = 3 − 2y. Step 2: Substitute in equation 1: 7(3 − 2y) − 15y = 2. 21 − 14y − 15y = 2. −29y = −19. y = 19/29. Step 3: Substitute y in x = 3 − 2y: x = 3 − 2(19/29) = 49/29. Solution: x = 49/29, y = 19/29. Verification: Substituting these values satisfies both equations.
To understand the substitution method stepwise: Step 1: Find the value of one variable, say y in terms of x from either equation. Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable. Sometimes you get statements with no variable. If true, infinitely many solutions. If false, inconsistent. Step 3: Substitute the value obtained in Step 2 in the equation used in Step 1 to obtain the other variable.
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Example 5: Aftab tells his daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. Represent algebraically and graphically. Let ages be s and t. Equations: s − 7 = 7(t − 7), so s − 7t + 42 = 0. And s + 3 = 3(t + 3), so s − 3t = 6. Using second equation, s = 3t + 6. Substitute in first: (3t + 6) − 7t + 42 = 0. −4t = −48. t = 12. s = 3(12) + 6 = 42. Aftab is 42, daughter is 12.
Example 6: Cost of 2 pencils and 3 erasers is 9. Cost of 4 pencils and 6 erasers is 18. Find cost of each. Equations: 2x + 3y = 9 and 4x + 6y = 18. Express x: x = (9 − 3y)/2. Substitute in second: 4((9 − 3y)/2) + 6y = 18. 18 − 6y + 6y = 18. 18 = 18. True for all y. Infinitely many solutions. Cannot find unique cost.
Example 7: Two rails: x + 2y − 4 = 0 and 2x + 4y − 12 = 0. Will they cross? Express x: x = 4 − 2y. Substitute in second: 2(4 − 2y) + 4y − 12 = 0. 8 − 12 = 0. −4 = 0. False statement. No common solution. Rails will not cross.
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Exercise 3.2. Question 1(i): x + y = 14, x − y = 4. From second, x = y + 4. Substitute in first: y + 4 + y = 14. 2y = 10. y = 5. x = 9. Solution (9, 5). Question 1(ii): s − t = 3, s/3 + t/2 = 6. From first, s = t + 3. Substitute: (t + 3)/3 + t/2 = 6. Multiply by 6: 2t + 6 + 3t = 36. 5t = 30. t = 6. s = 9. Solution (9, 6). Question 1(iii): 3x − y = 3, 9x − 3y = 9. From first, y = 3x − 3. Substitute in second: 9x − 3(3x − 3) = 9. 9x − 9x + 9 = 9. 9 = 9. True. Infinitely many solutions. Question 1(iv): 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3. From first, 0.2x = 1.3 − 0.3y, x = 6.5 − 1.5y. Substitute in second: 0.4(6.5 − 1.5y) + 0.5y = 2.3. 2.6 − 0.6y + 0.5y = 2.3. −0.1y = −0.3. y = 3. x = 6.5 − 4.5 = 2. Solution (2, 3). Question 1(v): √2 x + √3 y = 0, √3 x − √8 y = 0. From first, x = −(√3/√2)y. Substitute in second: √3(−√3/√2 y) − √8 y = 0. −3/√2 y − 2√2 y = 0. Combine: −(3 + 4)/√2 y = 0. y = 0. x = 0. Solution (0, 0). Question 1(vi): (3/2)x − (5/3)y = −2, x/3 + y/2 = 13/6. From second, x = 13/2 − 3/2 y. Substitute in first: (3/2)(13/2 − 3/2 y) − (5/3)y = −2. 39/4 − 9/4 y − 5/3 y = −2. Multiply by 12: 117 − 27y − 20y = −24. 117 + 24 = 47y. 141 = 47y. y = 3. x = 13/2 − 9/2 = 2. Solution (2, 3).
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Question 2: Solve 2x + 3y = 11 and 2x − 4y = −24. Subtract second from first: 7y = 35. y = 5. Substitute: 2x + 15 = 11. 2x = −4. x = −2. Solution (−2, 5). Find m for y = mx + 3. 5 = m(−2) + 3. 2 = −2m. m = −1. Question 3(i): Difference 26, one is 3 times other. x − y = 26, x = 3y. 3y − y = 26. 2y = 26. y = 13. x = 39. Numbers 39 and 13. Question 3(ii): Supplementary angles. Larger exceeds smaller by 18. x + y = 180, x − y = 18. Add: 2x = 198. x = 99. y = 81. Angles 99 and 81 degrees. Question 3(iii): 7 bats + 6 balls = 3800. 3 bats + 5 balls = 1750. 7x + 6y = 3800, 3x + 5y = 1750. From second, 3x = 1750 − 5y. x = (1750 − 5y)/3. Substitute in first: 7(1750 − 5y)/3 + 6y = 3800. Multiply by 3: 12250 − 35y + 18y = 11400. −17y = −850. y = 50. x = (1750 − 250)/3 = 500. Bat cost 500, ball cost 50. Question 3(iv): Taxi fixed charge x, per km y. 10 km: x + 10y = 105. 15 km: x + 15y = 155. Subtract: 5y = 50. y = 10. x = 5. Fixed 5, per km 10. For 25 km: 5 + 25(10) = 255. Question 3(v): Fraction becomes 9/11 if add 2 to both. (x+2)/(y+2) = 9/11. 11x − 9y = −4. Add 3 to both: (x+3)/(y+3) = 5/6. 6x − 5y = −3. Solve: Multiply first by 6, second by 11: 66x − 54y = −24. 66x − 55y = −33. Subtract: y = 9. 11x − 81 = −4. 11x = 77. x = 7. Fraction is 7/9. Question 3(vi): Jacob and son. 5 years hence: x + 5 = 3(y + 5). x − 3y = 10. 5 years ago: x − 5 = 7(y − 5). x − 7y = −30. Subtract: 4y = 40. y = 10. x = 40. Ages 40 and 10.
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Subsection 3.3.2 Elimination Method. Example 8: Incomes ratio 9:7, expenditures 4:3. Each saves 2000. Find incomes. Let incomes 9x, 7x. Expenditures 4y, 3y. Equations: 9x − 4y = 2000 and 7x − 3y = 2000. Step 1: Multiply first by 3, second by 4. 27x − 12y = 6000. 28x − 12y = 8000. Step 2: Subtract first from second: x = 2000. Step 3: Substitute in first: 9(2000) − 4y = 2000. 18000 − 2000 = 4y. 16000 = 4y. y = 4000. Incomes: 18000 and 14000. Verification: 18000:14000 = 9:7. Expenditures 16000:12000 = 4:3. Steps for elimination: Step 1: Multiply equations to make coefficients of one variable numerically equal. Step 2: Add or subtract to eliminate one variable. If true statement with no variable, infinitely many solutions. If false, no solution. Step 3: Solve for one variable. Step 4: Substitute to find the other.
Example 9: 2x + 3y = 8 and 4x + 6y = 7. Multiply first by 2: 4x + 6y = 16. Subtract second: 0 = 9. False. No solution.
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Example 10: Sum of two-digit number and reversed digits is 66. Digits differ by 2. Find number. Let digits be x, y. Number 10x + y. Reversed 10y + x. Sum: 11(x + y) = 66. x + y = 6. Digits differ by 2: x − y = 2 or y − x = 2. If x − y = 2, solve with x + y = 6. Add: 2x = 8. x = 4, y = 2. Number 42. If y − x = 2, solve with x + y = 6. Add: 2y = 8. y = 4, x = 2. Number 24. Two numbers: 42 and 24.
Exercise 3.3. Question 1(i): x + y = 5, 2x − 3y = 4. Multiply first by 3: 3x + 3y = 15. Add to second: 5x = 19. x = 19/5. y = 5 − 19/5 = 6/5. Substitution: y = 5 − x. 2x − 3(5 − x) = 4. 2x − 15 + 3x = 4. 5x = 19. x = 19/5, y = 6/5. Question 1(ii): 3x + 4y = 10, 2x − 2y = 2. Multiply second by 2: 4x − 4y = 4. Add to first: 7x = 14. x = 2. y = 1. Substitution: y = x − 1. 3x + 4(x − 1) = 10. 7x = 14. x = 2, y = 1. Question 1(iii): 3x − 5y − 4 = 0, 9x = 2y + 7. Rewrite: 3x − 5y = 4, 9x − 2y = 7. Multiply first by 3: 9x − 15y = 12. Subtract second: −13y = 5. y = −5/13. x = (4 + 5(−5/13))/3 = (52 − 25)/39 = 27/39 = 9/13. Substitution gives same. Question 1(iv): x/2 + 2y/3 = −1, x − y/3 = 3. Multiply first by 6: 3x + 4y = −6. Multiply second by 3: 3x − y = 9. Subtract: 5y = −15. y = −3. x = 3 + (−3)/3 = 2. Substitution gives same.
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Question 2(i): Fraction. (x+1)/(y-1) = 1. x − y = −2. (x)/(y+1) = 1/2. 2x − y = 1. Subtract first from second: x = 3. y = 5. Fraction 3/5. Question 2(ii): Nuri and Sonu. 5 years ago: x − 5 = 3(y − 5). x − 3y = −10. 10 years later: x + 10 = 2(y + 10). x − 2y = 10. Subtract: −y = −20. y = 20. x = 50. Nuri 50, Sonu 20. Question 2(iii): Two-digit number. x + y = 9. 9(10x + y) = 2(10y + x). 90x + 9y = 20y + 2x. 88x − 11y = 0. 8x − y = 0. Add to first: 9x = 9. x = 1. y = 8. Number 18. Question 2(iv): Meena withdraws 2000. 50 and 100 notes. x + y = 25. 50x + 100y = 2000. Divide second by 50: x + 2y = 40. Subtract first: y = 15. x = 10. 10 notes of 50, 15 notes of 100. Question 2(v): Library charges. Fixed x, extra per day y. 7 days: x + 4y = 27. 5 days: x + 2y = 21. Subtract: 2y = 6. y = 3. x = 15. Fixed 15, extra 3 per day.
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Section 3.4 Summary. In this chapter, you have studied the following points: 1. A pair of linear equations in two variables can be represented, and solved, by the graphical method and the algebraic method. 2. Graphical Method: The graph of a pair of linear equations in two variables is represented by two lines. If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent. If the lines coincide, then there are infinitely many solutions, each point on the line being a solution. In this case, the pair of equations is dependent consistent. If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent. 3. Algebraic Methods: We have discussed the Substitution Method and the Elimination Method for finding the solution(s) of a pair of linear equations. 4. If a pair of linear equations is given by a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, then the following situations can arise: a₁/a₂ ≠ b₁/b₂: In this case, the pair of linear equations is consistent. a₁/a₂ = b₁/b₂ ≠ c₁/c₂: In this case, the pair of linear equations is inconsistent. a₁/a₂ = b₁/b₂ = c₁/c₂: In this case, the pair of linear equations is dependent and consistent. 5. There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]