{"id":118967,"date":"2022-05-02T15:37:34","date_gmt":"2022-05-02T10:07:34","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118967"},"modified":"2022-05-02T15:37:34","modified_gmt":"2022-05-02T10:07:34","slug":"chapter-3-pair-of-linear-equations-in-two-variables-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-3-pair-of-linear-equations-in-two-variables-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 3 – Pair of Linear Equations in Two Variables Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 3.1<\/h2>\n

1. Aftab tells his daughter, \u201cSeven 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.\u201d (Isn\u2019t this interesting?) Represent this situation algebraically and graphically.<\/h2>\n

Solutions:Let the present age of Aftab be \u2018x\u2019.
\nAnd, the present age of his daughter be \u2018y\u2019.
\nNow, we can write, seven years ago,
\nAge of Aftab = x-7
\nAge of his daughter = y-7
\nAccording to the question,
\nx\u22127 = 7(y\u22127)
\n\u21d2x\u22127 = 7y\u221249
\n\u21d2x\u22127y = \u221242\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nAlso, three years from now or after three years,
\nAge of Aftab will become = x+3.
\nAge of his daughter will become = y+3
\nAccording to the situation given,
\nx+3 = 3(y+3)
\n\u21d2x+3 = 3y+9
\n\u21d2x\u22123y = 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026..\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nSubtracting equation (i) from equation (ii) we have
\n(x\u22123y)\u2212(x\u22127y) = 6\u2212(\u221242)
\n\u21d2\u22123y+7y = 6+42
\n\u21d24y = 48
\n\u21d2y = 12
\nThe algebraic equation is represented by
\nx\u22127y = \u221242
\nx\u22123y = 6
\nFor,\u00a0x\u22127y = \u221242 or x = \u221242+7y
\nThe solution table is<\/h3>\n

\"\"
\nFor, \u00a0x\u22123y = 6\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a0x = 6+3y
\nThe solution table is<\/h3>\n

\"\"
\nThe graphical representation is:<\/h3>\n

\"\"
\n2. The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.<\/h2>\n

Solutions:\u00a0Let us assume that the cost of a bat be \u2018Rs x\u2019
\nAnd,the cost of a ball be \u2018Rs y\u2019
\nAccording to the question, the algebraic representation is
\n3x+6y = 3900
\nAnd x+3y = 1300
\nFor,\u00a03x+6y = 3900
\nOr x = (3900-6y)\/3
\nThe solution table is<\/h3>\n

\"\"
\nFor, x+3y = 1300
\nOr x = 1300-3y
\nThe solution table is<\/h3>\n

\"\"
\nThe graphical representation is as follows.<\/h3>\n

\"\"
\n3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.<\/h2>\n

Solutions:Let the cost of 1 kg of apples be \u2018Rs. x\u2019
\nAnd, cost of 1 kg of grapes be \u2018Rs. y\u2019
\nAccording to the question, the algebraic representation is
\n2x+y = 160
\nAnd\u00a04x+2y = 300
\nFor, 2x+y = 160\u00a0or y = 160\u22122x, the solution table is;<\/h3>\n

\"\"
\nFor 4x+2y = 300 or y = (300-4x)\/2, the solution table is;<\/h3>\n

\"\"
\nThe graphical representation is as follows;<\/h3>\n

\"\"
\nExercise 3.2<\/h2>\n

1. Form the pair of linear equations in the following problems, and find their solutions graphically.
\n(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.
\n(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.<\/h2>\n

Solution:
\n(i)Let there are x number of girls and y number of boys. As per\u00a0the given question,\u00a0the algebraic expression can be represented as follows.
\nx +y = 10
\nx\u2013 y = 4
\nNow, for x+y = 10 or x = 10\u2212y, the solutions are;<\/h3>\n

\"\"
\nFor x \u2013 y = 4 or x = 4 + y, the solutions are;<\/h3>\n

\"\"
\nThe graphical representation is as follows;<\/h3>\n

\"\"
\nFrom the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.
\n(ii)\u00a0Let 1 pencil costs Rs.x and 1 pen costs Rs.y.
\nAccording to the question, the algebraic expression cab be represented as;
\n5x + 7y = 50
\n7x + 5y = 46
\nFor, 5x + 7y = 50 or \u00a0x = (50-7y)\/5, the solutions are;<\/h3>\n

\"\"
\nFor 7x + 5y = 46 or x = (46-5y)\/7, the solutions are;<\/h3>\n

\"\"
\nHence, the graphical representation is as follows;<\/h3>\n

\"\"
\nFrom the graph, it is can be seen that the given\u00a0lines cross each other at point (3, 5).
\nSo, the cost of a pencil is 3\/- and cost of a pen is 5\/-.<\/h3>\n

2. On comparing the ratios a1\/a2 , b1\/b2 , c1\/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
\n(i) 5x \u2013 4y + 8 = 0
\n7x + 6y \u2013 9 = 0
\n(ii) 9x + 3y + 12 = 0
\n18x + 6y + 24 = 0
\n(iii) 6x \u2013 3y + 10 = 0
\n2x \u2013 y + 9 = 0<\/h2>\n

Solutions:
\n(i) Given expressions;
\n5x\u22124y+8 = 0
\n7x+6y\u22129 = 0
\nComparing these equations with a1x+b1y+c1 = 0
\nAnd a2x+b2y+c2 = 0
\nWe get,
\na1 = 5, b1 = -4, c1 = 8
\na2 = 7, b2 = 6, c2 = -9
\n(a1\/a2) = 5\/7
\n(b1\/b2) = -4\/6 = -2\/3
\n(c1\/c2) = 8\/-9
\nSince, (a1\/a2) \u2260 (b1\/b2)
\nSo, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.<\/h3>\n

(ii) Given expressions;
\n9x + 3y + 12 = 0
\n18x + 6y + 24 = 0
\nComparing these equations with a1x+b1y+c1 = 0
\nAnd a2x+b2y+c2 = 0
\nWe get,
\na1 = 9, b1 = 3, c1 = 12
\na2 = 18, b2 = 6, c2 = 24
\n(a1\/a2) = 9\/18 = 1\/2
\n(b1\/b2) = 3\/6 = 1\/2
\n(c1\/c2) = 12\/24 = 1\/2
\nSince (a1\/a2) = (b1\/b2) = (c1\/c2)
\nSo, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.<\/h3>\n

(iii) Given Expressions;
\n6x \u2013 3y + 10 = 0
\n2x \u2013 y + 9 = 0
\nComparing these equations with a1x+b1y+c1 = 0
\nAnd a2x+b2y+c2 = 0
\nWe get,
\na1 = 6, b1 = -3, c1 = 10
\na2 = 2, b2 = -1, c2 = 9
\n(a1\/a2) = 6\/2 = 3\/1
\n(b1\/b2) = -3\/-1 = 3\/1
\n(c1\/c2) = 10\/9
\nSince (a1\/a2) = (b1\/b2) \u2260 (c1\/c2)
\nSo, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.<\/h3>\n

3. On comparing the ratio, (a1\/a2) , (b1\/b2) , (c1\/c2) find out whether the following pair of linear equations are consistent, or inconsistent.
\n(i) 3x + 2y = 5 ; 2x \u2013 3y = 7
\n(ii) 2x \u2013 3y = 8 ; 4x \u2013 6y = 9
\n(iii)(3\/2)x+(5\/3)y = 7; 9x \u2013 10y = 14
\n(iv) 5x \u2013 3y = 11 ; \u2013 10x + 6y = \u201322
\n(v)(4\/3)x+2y = 8 ; 2x + 3y = 12<\/h2>\n

Solutions:
\n(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0
\nand 2x \u2013 3y = 7 or 2x \u2013 3y -7 = 0
\nComparing these equations with\u00a0a1x+b1y+c1 = 0
\nAnd a2x+b2y+c2 = 0
\nWe get,
\na1 = 3, b1 = 2, c1 = -5
\na2 = 2, b2 = -3, c2 = -7
\n(a1\/a2) = 3\/2
\n(b1\/b2) = 2\/-3
\n(c1\/c2) = -5\/-7 = 5\/7
\nSince, (a1\/a2) \u2260 (b1\/b2)
\nSo, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.<\/h3>\n

(ii) Given 2x \u2013 3y = 8 and 4x \u2013 6y = 9<\/h3>\n

Therefore,
\na1 = 2, b1 = -3, c1 = -8
\na2 = 4, b2 = -6, c2 = -9
\n(a1\/a2) = 2\/4 = 1\/2
\n(b1\/b2) = -3\/-6 = 1\/2
\n(c1\/c2) = -8\/-9 = 8\/9
\nSince , (a1\/a2) = (b1\/b2) \u2260 (c1\/c2)
\nSo, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.<\/h3>\n

(iii)Given (3\/2)x + (5\/3)y = 7 and 9x \u2013 10y = 14
\nTherefore,
\na1 = 3\/2, b1 = 5\/3, c1 = -7
\na2 = 9, b2 = -10, c2 = -14
\n(a1\/a2) = 3\/(2\u00d79) = 1\/6
\n(b1\/b2) = 5\/(3\u00d7 -10)= -1\/6
\n(c1\/c2) = -7\/-14 = 1\/2
\nSince, (a1\/a2) \u2260 (b1\/b2)
\nSo, the equations are intersecting\u00a0 each other at one point and they have only one possible solution. Hence, the equations are consistent.<\/h3>\n

(iv) Given, 5x \u2013 3y = 11 and \u2013 10x + 6y = \u201322
\nTherefore,
\na1 = 5, b1 = -3, c1 = -11
\na2 = -10, b2 = 6, c2 = 22
\n(a1\/a2) = 5\/(-10) = -5\/10 = -1\/2
\n(b1\/b2) = -3\/6 = -1\/2
\n(c1\/c2) = -11\/22 = -1\/2
\nSince (a1\/a2) = (b1\/b2) = (c1\/c2)
\nThese linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.<\/h3>\n

(v)Given, (4\/3)x +2y = 8 and 2x + 3y = 12
\na1 = 4\/3 , b1= 2 , c1 = -8
\na2 = 2, b2 = 3 , c2 = -12
\n(a1\/a2) = 4\/(3\u00d72)= 4\/6 = 2\/3
\n(b1\/b2) = 2\/3
\n(c1\/c2) = -8\/-12 = 2\/3
\nSince (a1\/a2) = (b1\/b2) = (c1\/c2)
\nThese linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.<\/h3>\n

4. Which of the following pairs of linear equations are consistent\/inconsistent? If consistent, obtain the solution graphically:
\n(i) x + y = 5, 2x + 2y = 10
\n(ii) x \u2013 y = 8, 3x \u2013 3y = 16
\n(iii) 2x + y \u2013 6 = 0, 4x \u2013 2y \u2013 4 = 0
\n(iv) 2x \u2013 2y \u2013 2 = 0, 4x \u2013 4y \u2013 5 = 0<\/h2>\n

Solutions:
\n(i)Given, x + y = 5 and 2x + 2y = 10
\n(a1\/a2) = 1\/2
\n(b1\/b2) = 1\/2
\n(c1\/c2) = 1\/2
\nSince (a1\/a2) = (b1\/b2) = (c1\/c2)
\n\u2234The equations are coincident\u00a0and they have infinite number of possible solutions.
\nSo, the equations are consistent.
\nFor, x + y = 5 or x = 5 \u2013 y<\/h3>\n

\"\"
\nFor 2x + 2y = 10 or x = (10-2y)\/2<\/h3>\n

\"\"
\nSo, the equations are represented in graphs as follows:<\/h3>\n

\"\"
\nFrom the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).
\n(iv) Given, 2x \u2013 2y \u2013 2 = 0 and 4x \u2013 4y \u2013 5 = 0
\n(a1\/a2) = 2\/4 = \u00bd
\n(b1\/b2) = -2\/-4 = 1\/2
\n(c1\/c2) = 2\/5
\nSince, a1\/a2 = b1\/b2 \u2260 c1\/c2
\nThus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.<\/h3>\n

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.<\/h2>\n

Solutions:Let us consider.
\nThe width of the garden is x and length is y.
\nNow, according to the question, we can express the given condition as;
\ny \u2013 x = 4
\nand
\ny + x = 36
\nNow, taking y \u2013 x = 4 or y = x + 4<\/h3>\n

\"\"
\nFor y + x = 36, y = 36 \u2013 x<\/h3>\n

\"\"
\nThe graphical representation of both the equation is as follows;<\/h3>\n

\"\"
\nFrom the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.<\/h3>\n

6. Given the linear equation 2x + 3y \u2013 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
\n(i) Intersecting lines
\n(ii) Parallel lines
\n(iii) Coincident lines<\/h2>\n

Solutions:
\n(i) Given the linear equation 2x + 3y \u2013 8 = 0.
\nTo find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;
\n(a1\/a2) \u2260 (b1\/b2)
\nThus, another equation could be 2x \u2013 7y + 9 = 0, such that;
\n(a1\/a2) = 2\/2 = 1 and (b1\/b2) = 3\/-7
\nClearly, you can see another equation satisfies the condition.<\/h3>\n

(ii) Given the linear equation 2x + 3y \u2013 8 = 0.
\nTo find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;
\n(a1\/a2) = (b1\/b2) \u2260 (c1\/c2)
\nThus, another equation could be 6x + 9y + 9 = 0, such that;
\n(a1\/a2) = 2\/6 = 1\/3
\n(b1\/b2) = 3\/9= 1\/3
\n(c1\/c2) = -8\/9
\nClearly, you can see another equation satisfies the condition.<\/h3>\n

(iii) Given the linear equation 2x + 3y \u2013 8 = 0.
\nTo find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;
\n(a1\/a2) = (b1\/b2) = (c1\/c2)
\nThus, another equation could be 4x + 6y \u2013 16 = 0, such that;
\n(a1\/a2) = 2\/4 = 1\/2 ,(b1\/b2) = 3\/6 = 1\/2, (c1\/c2) = -8\/-16 = 1\/2
\nClearly, you can see another equation satisfies the condition.<\/h3>\n

7. Draw the graphs of the equations x \u2013 y + 1 = 0 and 3x + 2y \u2013 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.<\/h2>\n

Solution: Given, the equations for graphs are x \u2013 y + 1 = 0 and 3x + 2y \u2013 12 = 0.
\nFor, x \u2013 y + 1 = 0 or x = 1+y<\/h3>\n

\"\"
\nFor, 3x + 2y \u2013 12 = 0 or x = (12-2y)\/3<\/h3>\n

\"\"
\nHence, the graphical representation of these equations is as follows;<\/h3>\n

\"\"
\nExercise 3.3<\/h2>\n

1. Solve the following pair of linear equations by the substitution method
\n(i) x + y = 14
\nx \u2013 y = 4
\n(ii) s \u2013 t = 3
\n(s\/3) + (t\/2) = 6
\n(iii) 3x \u2013 y = 3
\n9x \u2013 3y = 9
\n(iv) 0.2x + 0.3y = 1.3
\n0.4x + 0.5y = 2.3
\n(v) \u221a2 x+\u221a3 y = 0
\n\u221a3 x-\u221a8 y = 0
\n(vi) (3x\/2) \u2013 (5y\/3) = -2
\n(x\/3) + (y\/2) = (13\/6)<\/h2>\n

Solutions:
\n(i) Given,
\nx + y = 14 and x \u2013 y = 4 are the two equations.
\nFrom 1st\u00a0equation, we get,
\nx = 14 \u2013 y
\nNow, substitute the value of x in second equation to get,
\n(14 \u2013 y) \u2013 y = 4
\n14 \u2013 2y = 4
\n2y = 10
\nOr y = 5
\nBy the value of y, we can now find the exact value of x;
\n\u2235 x = 14 \u2013 y
\n\u2234 x = 14 \u2013 5
\nOr x = 9
\nHence, x = 9 and y = 5.<\/h3>\n

(ii) Given,
\ns \u2013 t = 3 and (s\/3) + (t\/2) = 6 are the two equations.
\nFrom 1st\u00a0equation, we get,
\ns = 3 + t ________________(1)
\nNow, substitute the value of s in second equation to get,
\n(3+t)\/3 + (t\/2) = 6
\n\u21d2 (2(3+t) + 3t )\/6 = 6
\n\u21d2 (6+2t+3t)\/6 = 6
\n\u21d2 (6+5t) = 36
\n\u21d25t = 30
\n\u21d2t = 6
\nNow, substitute the value of t in equation (1)
\ns = 3 + 6 = 9
\nTherefore, s = 9 and t = 6.<\/h3>\n

(iii) Given,
\n3x \u2013 y = 3 and 9x \u2013 3y = 9 are the two equations.
\nFrom 1st\u00a0equation, we get,
\nx = (3+y)\/3
\nNow, substitute the value of x in the given second equation to get,
\n9(3+y)\/3 \u2013 3y = 9
\n\u21d29 +3y -3y = 9
\n\u21d2 9 = 9
\nTherefore, y has infinite values and since, x = (3+y) \/3, so x also has infinite values.<\/h3>\n

(iv) Given,
\n0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.
\nFrom 1st\u00a0equation, we get,
\nx = (1.3- 0.3y)\/0.2 _________________(1)
\nNow, substitute the value of x in the given second equation to get,
\n0.4(1.3-0.3y)\/0.2 + 0.5y = 2.3
\n\u21d22(1.3 \u2013 0.3y) + 0.5y = 2.3
\n\u21d2 2.6 \u2013 0.6y + 0.5y = 2.3
\n\u21d2 2.6 \u2013 0.1 y = 2.3
\n\u21d2 0.1 y = 0.3
\n\u21d2 y = 3
\nNow, substitute the value of y in equation (1), we get,
\nx = (1.3-0.3(3))\/0.2 = (1.3-0.9)\/0.2 = 0.4\/0.2 = 2
\nTherefore, x = 2 and y = 3.<\/h3>\n

(v) Given,
\n\u221a2 x + \u221a3 y = 0 and \u221a3 x \u2013 \u221a8 y = 0
\nare the two equations.
\nFrom 1st\u00a0equation, we get,
\nx = \u2013 (\u221a3\/\u221a2)y __________________(1)
\nPutting the value of x in the given second equation to get,
\n\u221a3(-\u221a3\/\u221a2)y \u2013 \u221a8y = 0 \u21d2 (-3\/\u221a2)y- \u221a8 y = 0
\n\u21d2 y = 0
\nNow, substitute the value of y in equation (1), we get,
\nx = 0
\nTherefore, x = 0 and y = 0.<\/h3>\n

(vi) Given,
\n(3x\/2)-(5y\/3) = -2 and (x\/3) + (y\/2) = 13\/6 are the two equations.
\nFrom 1st\u00a0equation, we get,
\n(3\/2)x = -2 + (5y\/3)
\n\u21d2 x = 2(-6+5y)\/9 = (-12+10y)\/9 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\nPutting the value of x in the given second equation to get,
\n((-12+10y)\/9)\/3 + y\/2 = 13\/6
\n\u21d2y\/2 = 13\/6 \u2013( (-12+10y)\/27 ) + y\/2 = 13\/6<\/h3>\n

\"\"
\nNow, substitute the value of y in equation (1), we get,
\n(3x\/2) \u2013 5(3)\/3 = -2
\n\u21d2 (3x\/2) \u2013 5 = -2
\n\u21d2 x = 2
\nTherefore, x = 2 and y = 3.<\/h3>\n

2. Solve 2x + 3y = 11 and 2x \u2013 4y = \u2013 24 and hence find the value of \u2018m\u2019 for which y = mx + 3.<\/h2>\n

Solution:
\n2x + 3y = 11\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(I)
\n2x \u2013 4y = -24\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (II)
\nFrom equation (II), we get
\nx = (11-3y)\/2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.(III)
\nSubstituting the value of x in equation (II), we get
\n2(11-3y)\/2 \u2013 4y = 24
\n11 \u2013 3y \u2013 4y = -24
\n-7y = -35
\ny = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(IV)
\nPutting the value of y in equation (III), we get
\nx = (11-3\u00d75)\/2 = -4\/2 = -2
\nHence, x = -2, y = 5
\nAlso,
\ny = mx + 3
\n5 = -2m +3
\n-2m = 2
\nm = -1
\nTherefore the value of m is -1.<\/h3>\n

3. Form the pair of linear equations for the following problems and find their solution by substitution method.
\n(i) The difference between two numbers is 26 and one number is three times the other. Find them.<\/h2>\n

Solution:
\nLet the two numbers be x and y respectively, such that y > x.
\nAccording to the question,
\ny = 3x \u2026\u2026\u2026\u2026\u2026\u2026 (1)
\ny \u2013 x = 26 \u2026\u2026\u2026\u2026..(2)
\nSubstituting the value of (1) into (2), we get
\n3x \u2013 x = 26
\nx = 13 \u2026\u2026\u2026\u2026\u2026. (3)
\nSubstituting (3) in (1), we get y = 39
\nHence, the numbers are 13 and 39.<\/h3>\n

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.<\/h2>\n

Solution:
\nLet the larger angle by xo\u00a0and smaller angle be yo.
\nWe know that the sum of two supplementary pair of angles is always 180o.
\nAccording to the question,
\nx + y = 180o\u2026\u2026\u2026\u2026\u2026. (1)
\nx \u2013 y = 18o\u00a0\u2026\u2026\u2026\u2026\u2026..(2)
\nFrom (1), we get x = 180o\u00a0\u2013 y \u2026\u2026\u2026\u2026. (3)
\nSubstituting (3) in (2), we get
\n180o\u00a0\u2013 y \u2013 y =18o
\n162o\u00a0= 2y
\ny = 81o\u00a0\u2026\u2026\u2026\u2026.. (4)
\nUsing the value of y in (3), we get
\nx = 180o\u00a0\u2013 81o
\n= 99o
\nHence, the angles are 99o\u00a0and 81o.<\/h3>\n

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.<\/h2>\n

Solution:
\nLet the cost a bat be x and cost of a ball be y.
\nAccording to the question,
\n7x + 6y = 3800 \u2026\u2026\u2026\u2026\u2026\u2026. (I)
\n3x + 5y = 1750 \u2026\u2026\u2026\u2026\u2026\u2026. (II)
\nFrom (I), we get
\ny = (3800-7x)\/6\u2026\u2026\u2026\u2026\u2026\u2026..(III)
\nSubstituting (III) in (II). we get,
\n3x+5(3800-7x)\/6 =1750
\n\u21d23x+ 9500\/3 \u2013 35x\/6 = 1750
\n\u21d23x- 35x\/6 = 1750 \u2013 9500\/3
\n\u21d2(18x-35x)\/6 = (5250 \u2013 9500)\/3
\n\u21d2-17x\/6 = -4250\/3
\n\u21d2-17x = -8500
\nx = 500 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (IV)
\nSubstituting the value of x in (III), we get
\ny = (3800-7 \u00d7500)\/6 = 300\/6 = 50
\nHence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.<\/h3>\n

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?<\/h2>\n

Solution:
\nLet the fixed charge be Rs x and per km charge be Rs y.
\nAccording to the question,
\nx + 10y = 105 \u2026\u2026\u2026\u2026\u2026.. (1)
\nx + 15y = 155 \u2026\u2026\u2026\u2026\u2026.. (2)
\nFrom (1), we get x = 105 \u2013 10y \u2026\u2026\u2026\u2026\u2026\u2026. (3)
\nSubstituting the value of x in (2), we get
\n105 \u2013 10y + 15y = 155
\n5y = 50
\ny = 10 \u2026\u2026\u2026\u2026\u2026.. (4)
\nPutting the value of y in (3), we get
\nx = 105 \u2013 10 \u00d7 10 = 5
\nHence, fixed charge is Rs 5 and per km charge = Rs 10
\nCharge for 25 km = x + 25y = 5 + 250 = Rs 255<\/h3>\n

(v) A fraction becomes 9\/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5\/6. Find the fraction.<\/h2>\n

Solution:
\nLet the fraction be x\/y.
\nAccording to the question,
\n(x+2) \/(y+2) = 9\/11
\n11x + 22 = 9y + 18
\n11x \u2013 9y = -4 \u2026\u2026\u2026\u2026\u2026.. (1)
\n(x+3) \/(y+3) = 5\/6
\n6x + 18 = 5y +15
\n6x \u2013 5y = -3 \u2026\u2026\u2026\u2026\u2026\u2026. (2)
\nFrom (1), we get x = (-4+9y)\/11\u00a0\u2026\u2026\u2026\u2026\u2026.. (3)
\nSubstituting the value of x in (2), we get
\n6(-4+9y)\/11 -5y = -3
\n-24 + 54y \u2013 55y = -33
\n-y = -9
\ny = 9 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (4)
\nSubstituting the value of y in (3), we get
\nx = (-4+9\u00d79 )\/11 = 7
\nHence the fraction is 7\/9.<\/h3>\n

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob\u2019s age was seven times that of his son. What are their present ages?<\/h2>\n

Solutions:
\nLet the age of Jacob and his son be x and y respectively.
\nAccording to the question,
\n(x + 5) = 3(y + 5)
\nx \u2013 3y = 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)
\n(x \u2013 5) = 7(y \u2013 5)
\nx \u2013 7y = -30 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (2)
\nFrom (1), we get x = 3y + 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (3)
\nSubstituting the value of x in (2), we get
\n3y + 10 \u2013 7y = -30
\n-4y = -40
\ny = 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (4)
\nSubstituting the value of y in (3), we get
\nx = 3 x 10 + 10 = 40
\nHence, the present age of Jacob\u2019s and his son is 40 years and 10 years respectively.
\nExercise 3.4<\/h3>\n

1. Solve the following pair of linear equations by the elimination method and the substitution method:
\n(i) x + y = 5 and 2x \u2013 3y = 4
\n(ii) 3x + 4y = 10 and 2x \u2013 2y = 2
\n(iii) 3x \u2013 5y \u2013 4 = 0 and 9x = 2y + 7
\n(iv) x\/2+ 2y\/3 = -1 and x-y\/3 = 3<\/h2>\n

Solutions:
\n(i) x + y = 5 and 2x \u2013 3y = 4
\nBy the method of elimination.
\nx + y = 5\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)
\n2x \u2013 3y = 4 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nWhen the equation (i) is multiplied by 2, we get
\n2x + 2y = 10 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)
\nWhen the equation (ii) is subtracted from (iii) we get,
\n5y = 6
\ny = 6\/5\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iv)
\nSubstituting the value of y in eq. (i) we get,
\nx=5\u22126\/5 = 19\/5
\n\u2234x = 19\/5 , y = 6\/5
\nBy the method of substitution.
\nFrom the equation (i), we get:
\nx = 5 \u2013 y\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (v)
\nWhen the value is put in equation (ii) we get,
\n2(5 \u2013 y) \u2013 3y = 4
\n-5y = -6
\ny =\u00a06\/5
\nWhen the values are substituted in equation (v), we get:
\nx =5\u2212 6\/5 = 19\/5
\n\u2234x = 19\/5\u00a0,y = 6\/5<\/h3>\n

(ii) 3x + 4y = 10 and 2x \u2013 2y = 2
\nBy the method of elimination.
\n3x + 4y = 10\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\n2x \u2013 2y = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)
\nWhen the equation (i) and (ii) is multiplied by 2, we get:
\n4x \u2013 4y = 4 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iii)
\nWhen the Equation (i) and (iii) are added, we get:
\n7x = 14
\nx = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iv)
\nSubstituting equation (iv) in (i) we get,
\n6 + 4y = 10
\n4y = 4
\ny = 1
\nHence, x = 2 and y = 1
\nBy the method of Substitution
\nFrom equation (ii) we get,
\nx = 1 + y\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (v)
\nSubstituting equation (v) in equation (i) we get,
\n3(1 + y) + 4y = 10
\n7y = 7
\ny = 1
\nWhen y = 1 is substituted in equation (v) we get,
\nA = 1 + 1 = 2
\nTherefore, A = 2 and B = 1<\/h3>\n

(iii) 3x \u2013 5y \u2013 4 = 0 and 9x = 2y + 7
\nBy the method of elimination:
\n3x \u2013 5y \u2013 4 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i)
\n9x = 2y + 7
\n9x \u2013 2y \u2013 7 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nWhen the equation (i) and (iii) is multiplied we get,
\n9x \u2013 15y \u2013 12 = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)
\nWhen the equation (iii) is subtracted from equation (ii) we get,
\n13y = -5
\ny = -5\/13\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iv)
\nWhen equation (iv) is substituted in equation (i) we get,
\n3x +25\/13 \u22124=0
\n3x = 27\/13
\nx =9\/13
\n\u2234x = 9\/13 and y = -5\/13
\nBy the method of Substitution:
\nFrom the equation (i) we get,
\nx = (5y+4)\/3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (v)
\nPutting the value (v) in equation (ii) we get,
\n9(5y+4)\/3 \u22122y \u22127=0
\n13y = -5
\ny = -5\/13
\nSubstituting this value in equation (v) we get,
\nx = (5(-5\/13)+4)\/3
\nx = 9\/13
\n\u2234x =\u00a09\/13, y = -5\/13<\/h3>\n

(iv) x\/2 + 2y\/3 = -1 and x-y\/3 = 3
\nBy the method of Elimination.
\n3x + 4y = -6 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)
\nx-y\/3 = 3
\n3x \u2013 y = 9 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (ii)
\nWhen the equation (ii) is subtracted from equation (i) we get,
\n5y = -15
\ny = -3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iii)
\nWhen the equation (iii) is substituted in (i) we get,
\n3x \u2013 12 = -6
\n3x = 6
\nx = 2
\nHence, x = 2 , y = -3
\nBy the method of Substitution:
\nFrom the equation (ii) we get,
\nx = (y+9)\/3\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(v)
\nPutting the value obtained from equation (v) in equation (i) we get,
\n3(y+9)\/3 +4y =\u22126
\n5y = -15
\ny = -3
\nWhen y = -3 is substituted in equation (v) we get,
\nx = (-3+9)\/3 = 2
\nTherefore, x = 2 and y = -3<\/h3>\n

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
\n(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?<\/h2>\n

Solution:
\nLet the fraction be a\/b
\nAccording to the given information,
\n(a+1)\/(b-1) = 1
\n=> a \u2013 b = -2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\na\/(b+1) = 1\/2
\n=> 2a-b = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nWhen equation (i) is subtracted from equation (ii) we get,
\na = 3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iii)
\nWhen a = 3 is substituted in equation (i) we get,
\n3 \u2013 b = -2
\n-b = -5
\nb = 5
\nHence, the fraction is 3\/5.<\/h3>\n

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?<\/h2>\n

Solution:
\nLet us assume, present age of Nuri is x
\nAnd present age of Sonu is y.
\nAccording to the given condition, we can write as;
\nx \u2013 5 = 3(y \u2013 5)
\nx \u2013 3y = -10\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(1)
\nNow,
\nx + 10 = 2(y +10)
\nx \u2013 2y = 10\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(2)
\nSubtract eq. 1 from 2, to get,
\ny = 20 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(3)
\nSubstituting the value of y in eq.1, we get,
\nx \u2013 3.20 = -10
\nx \u2013 60 = -10
\nx = 50
\nTherefore,
\nAge of Nuri is 50 years
\nAge of Sonu is 20 years.<\/h3>\n

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.<\/h2>\n

Solution:
\nLet the unit digit and tens digit of a number be x and y respectively.
\nThen, Number (n) = 10B + A
\nN after reversing order of the digits = 10A + B
\nAccording to the given information, A + B = 9\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\n9(10B + A) = 2(10A + B)
\n88 B \u2013 11 A = 0
\n-A + 8B = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)
\nAdding the equations (i) and (ii) we get,
\n9B = 9
\nB = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(3)
\nSubstituting this value of B, in the equation (i) we get A= 8
\nHence the number (N) is 10B + A = 10 x 1 +8 = 18<\/h3>\n

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.<\/h2>\n

Solution:
\nLet the number of Rs.50 notes be A and the number of Rs.100 notes be B
\nAccording to the given information,
\nA + B = 25 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (i)
\n50A + 100B = 2000 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nWhen equation (i) is multiplied with (ii) we get,
\n50A + 50B = 1250 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iii)
\nSubtracting the equation (iii) from the equation (ii) we get,
\n50B = 750
\nB = 15
\nSubstituting in the equation (i) we get,
\nA = 10
\nHence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.<\/h3>\n

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.<\/h2>\n

Solution:
\nLet the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.
\nAccording to the information given,
\nA + 4B = 27 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)
\nA + 2B = 21 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)
\nWhen equation (ii) is subtracted from equation (i) we get,
\n2B = 6
\nB = 3 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026<\/h3>\n

(iii)
\nSubstituting B = 3 in equation (i) we get,
\nA + 12 = 27
\nA = 15
\nHence, the fixed charge is Rs.15
\nAnd the Charge per day is Rs.3<\/h3>\n

Exercise 3.5<\/h2>\n

1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
\n(i) x \u2013 3y \u2013 3 = 0 and 3x \u2013 9y \u2013 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8
\n(iii) 3x \u2013 5y = 20 and 6x \u2013 10y = 40 (iv) x \u2013 3y \u2013 7 = 0 and 3x \u2013 3y \u2013 15 = 0<\/h2>\n

Solutions:
\n(i) Given, x \u2013 3y \u2013 3 =0 and 3x \u2013 9y -2 =0
\na1\/a2=1\/3 , \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b1\/b2= -3\/-9 =1\/3, \u00a0\u00a0\u00a0 c1\/c2=-3\/-2 = 3\/2
\n(a1\/a2) = (b1\/b2) \u2260 (c1\/c2)
\nSince, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.<\/h3>\n

(ii) Given, 2x + y = 5 and 3x +2y = 8
\na1\/a2\u00a0= 2\/3 , b1\/b2\u00a0= 1\/2 , c1\/c2\u00a0= -5\/-8
\n(a1\/a2) \u2260 (b1\/b2)
\nSince they intersect at a\u00a0unique point these equations will have a unique solution by cross multiplication method:
\nx\/(b1c2-c1b2) = y\/(c1a2\u00a0\u2013 c2a1) = 1\/(a1b2-a2b1)
\nx\/(-8-(-10)) = y\/(-15-(-16)) = 1\/(4-3)
\nx\/2 = y\/1 = 1
\n\u2234 x = 2 and y =1<\/h3>\n

(iii) Given, 3x \u2013 5y = 20 and 6x \u2013 10y = 40
\n(a1\/a2) = 3\/6 = 1\/2
\n(b1\/b2) = -5\/-10 = 1\/2
\n(c1\/c2) = 20\/40 = 1\/2
\na1\/a2\u00a0= b1\/b2\u00a0= c1\/c2
\nSince the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.<\/h3>\n

(iv) Given, x \u2013 3y \u2013 7 = 0 and 3x \u2013 3y \u2013 15 = 0
\n(a1\/a2) = 1\/3
\n(b1\/b2) = -3\/-3 = 1
\n(c1\/c2) = -7\/-15
\na1\/a2\u00a0\u2260 b1\/b2
\nSince this pair of lines are intersecting each other at a unique point, there will be\u00a0a unique solution.
\nBy cross multiplication,
\nx\/(45-21) = y\/(-21+15) = 1\/(-3+9)
\nx\/24 = y\/ -6 = 1\/6
\nx\/24 = 1\/6 and y\/-6 = 1\/6
\n\u2234\u00a0x = 4 and y = 1.<\/h3>\n

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
\n2x + 3y = 7
\n(a \u2013 b) x + (a + b) y = 3a + b \u2013 2
\n(ii) For which value of k will the following pair of linear equations have no solution?
\n3x + y = 1
\n(2k \u2013 1) x + (k \u2013 1) y = 2k + 1<\/h2>\n

Solution:
\n(i) 3y + 2x -7 =0
\n(a + b)y + (a-b)y \u2013 (3a + b -2) = 0
\na1\/a2\u00a0= 2\/(a-b) ,\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b1\/b2\u00a0= 3\/(a+b) ,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 c1\/c2\u00a0= -7\/-(3a + b -2)
\nFor infinitely many solutions,
\na1\/a2\u00a0= b1\/b2\u00a0= c1\/c2
\nThus 2\/(a-b) = 7\/(3a+b\u2013 2)
\n6a + 2b \u2013 4 = 7a \u2013 7b
\na \u2013 9b = -4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\n2\/(a-b) = 3\/(a+b)
\n2a + 2b = 3a \u2013 3b
\na \u2013 5b = 0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026.(ii)
\nSubtracting (i) from (ii), we get
\n4b = 4
\nb =1
\nSubstituting this eq. in (ii), we get
\na -5 x 1= 0
\na = 5
\nThus at a = 5 and b = 1 the given equations will have infinite solutions.<\/h3>\n

(ii) 3x + y -1 = 0
\n(2k -1)x\u00a0 +\u00a0 (k-1)y \u2013 2k -1 = 0
\na1\/a2\u00a0= 3\/(2k -1) ,\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b1\/b2\u00a0= 1\/(k-1), c1\/c2\u00a0= -1\/(-2k -1) = 1\/( 2k +1)
\nFor no solutions
\na1\/a2\u00a0= b1\/b2\u00a0\u2260 c1\/c2
\n3\/(2k-1) = 1\/(k -1)\u00a0\u00a0\u00a0\u2260 1\/(2k +1)
\n3\/(2k \u20131) = 1\/(k -1)
\n3k -3 = 2k -1
\nk =2
\nTherefore, for k = 2 the given pair of linear equations will have no solution.<\/h3>\n

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
\n8x + 5y = 9
\n3x + 2y = 4<\/h2>\n

Solution:
\n8x + 5y = 9 \u2026\u2026\u2026\u2026\u2026\u2026\u2026..(1)
\n3x\u00a0+ 2y = 4 \u2026\u2026\u2026\u2026\u2026\u2026.\u2026.(2)
\nFrom equation (2) we get
\nx = (4 \u2013 2y )\/ 3\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (3)
\nUsing this value in equation 1, we get
\n8(4-2y)\/3 + 5y = 9
\n32 \u2013 16y +15y = 27
\n-y = -5
\ny = 5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(4)
\nUsing this value in equation (2), we get
\n3x + 10 = 4
\nx = -2
\nThus, x = -2 and y = 5.
\nNow, Using Cross Multiplication method:
\n8x +5y \u2013 9 = 0
\n3x + 2y \u2013 4 = 0
\nx\/(-20+18) = y\/(-27 + 32 ) = 1\/(16-15)
\n-x\/2 = y\/5 =1\/1
\n\u2234 x = -2 and y =5.<\/h3>\n

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
\n(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.
\n(ii) A fraction becomes 1\/3 when 1 is subtracted from the numerator and it becomes 1\/4 when 8 is added to its denominator. Find the fraction.
\n(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
\n(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
\n(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.<\/h2>\n

Solutions:
\n(i)\u00a0Let x be the fixed charge and y be the charge of food per day.
\nAccording to the question,
\nx + 20y = 1000\u2026\u2026\u2026\u2026\u2026\u2026.. (i)
\nx +\u00a026y = 1180\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nSubtracting (i) from\u00a0 (ii) we get
\n6y = 180
\ny = Rs.30
\nUsing this value in equation (ii) we get
\nx = 1180 -26 x 30
\nx= Rs.400.
\nTherefore, fixed charges is Rs.400 and charge per day is Rs.30.<\/h3>\n

(ii)\u00a0Let\u00a0the fraction be x\/y.
\nSo, as per the question given,
\n(x -1)\/y = 1\/3 =>\u00a03x \u2013 y = 3\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\nx\/(y + 8) = 1\/4\u00a0 => 4x \u2013y =8 \u2026\u2026\u2026\u2026\u2026\u2026..(2)
\nSubtracting equation (1) from (2) , we get
\nx = 5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(3)
\nUsing this value in equation (2), we get,
\n(4\u00d75)\u2013 y = 8
\ny= 12
\nTherefore, the fraction is 5\/12.<\/h3>\n

(iii) Let the number of right answers is x and number of wrong answers be y
\nAccording to the given question;
\n3x\u2212y=40\u2026\u2026..(1)
\n4x\u22122y=50
\n\u21d22x\u2212y=25\u2026\u2026.(2)
\nSubtracting equation (2) from equation (1), we get;
\nx = 15 \u2026.\u2026.(3)
\nPutting this in equation (2), we obtain;
\n30 \u2013 y = 25
\nOr y = 5
\nTherefore, number of right answers = 15 and number of wrong answers = 5
\nHence, total number of questions = 20<\/h3>\n

(iv)\u00a0Let x km\/h be the speed of car from point A and y km\/h be the speed of car from point B.
\nIf the car travels in the same direction,
\n5x \u2013 5y = 100
\nx \u2013 y = 20 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nIf the car travels in the opposite direction,
\nx + y = 100\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nSolving equation (i) and (ii), we get
\nx = 60 km\/h\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)
\nUsing this in equation (i), we get,
\n60 \u2013 y = 20
\ny = 40 km\/h
\nTherefore, the speed of car from point A = 60 km\/h
\nSpeed of car from point B = 40 km\/h.<\/h3>\n

(v) Let,
\nThe length of rectangle = x unit
\nAnd breadth of the rectangle = y unit
\nNow, as per the question given,
\n(x \u2013 5) (y + 3) = xy -9
\n3x \u2013 5y \u2013 6 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\n(x + 3) (y + 2) = xy + 67
\n2x + 3y \u2013 61 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(2)
\nUsing cross multiplication method, we get,
\nx\/(305 +18) = y\/(-12+183) = 1\/(9+10)
\nx\/323 = y\/171 = 1\/19
\nTherefore, x = 17 and y = 9.
\nHence, the length of rectangle = 17 units
\nAnd breadth of the rectangle = 9 units
\nExercise 3.6<\/h3>\n

1. Solve the following pairs of equations by reducing them to a pair of linear equations:
\n(i) 1\/2x + 1\/3y = 2
\n1\/3x + 1\/2y = 13\/6<\/h2>\n

Solution:
\nLet us assume 1\/x = m and 1\/y = n \u00a0, then the equation will change as follows.
\nm\/2 + n\/3 = 2
\n\u21d2 3m+2n-12 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(1)
\nm\/3 + n\/2 = 13\/6
\n\u21d2 2m+3n-13 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(2)
\nNow, using cross-multiplication method, we get,
\nm\/(-26-(-36) ) = n\/(-24-(-39)) = 1\/(9-4)
\nm\/10 = n\/15 = 1\/5
\nm\/10 = 1\/5 and n\/15 = 1\/5
\nSo, m = 2 and n = 3
\n1\/x = 2 and 1\/y = 3
\nx = 1\/2 and y = 1\/3<\/h3>\n

(ii) 2\/\u221ax + 3\/\u221ay = 2
\n4\/\u221ax + 9\/\u221ay = -1<\/h2>\n

Solution:
\nSubstituting 1\/\u221ax = m and 1\/\u221ay = n in the given equations, we get
\n2m + 3n = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\n4m \u2013 9n = -1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nMultiplying equation (i) by 3, we get
\n6m + 9n = 6 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026..(iii)
\nAdding equation (ii) and (iii), we get
\n10m = 5
\nm =\u00a01\/2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026(iv)
\nNow by putting the value of \u2018m\u2019 in equation (i), we get
\n2\u00d71\/2 + 3n = 2
\n3n = 1
\nn =\u00a01\/3
\nm =1\/\u221ax
\n\u00bd = 1\/\u221ax
\nx = 4
\nn = 1\/\u221ay
\n1\/3 = 1\/\u221ay
\ny = 9
\nHence, x = 4 and y = 9<\/h3>\n

(iii) 4\/x + 3y = 14
\n3\/x -4y = 23<\/h2>\n

Solution:
\nPutting\u00a0in the given equation we get,
\nSo, 4m + 3y = 14\u00a0\u00a0\u00a0\u00a0 => 4m + 3y \u2013 14 = 0\u00a0 \u2026\u2026\u2026\u2026\u2026..\u2026..(1)
\n3m \u2013 4y = 23\u00a0\u00a0\u00a0\u00a0 => 3m \u2013 4y \u2013 23 = 0\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(2)
\nBy cross-multiplication, we get,
\nm\/(-69-56) = y\/(-42-(-92)) = 1\/(-16-9)
\n-m\/125 = y\/50 = -1\/ 25
\n-m\/125 = -1\/25 and y\/50 = -1\/25
\nm = 5 and b = -2
\nm =\u00a01\/x\u00a0= 5
\nSo , x = 1\/5
\ny = -2<\/h3>\n

(iv) 5\/(x-1) + 1\/(y-2) = 2
\n6\/(x-1) \u2013 3\/(y-2) = 1<\/h2>\n

Solution:
\nSubstituting 1\/(x-1) = m and 1\/(y-2) = n \u00a0in the given equations, we get,
\n5m + n = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\n6m \u2013 3n = 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)
\nMultiplying equation (i) by 3, we get
\n15m + 3n = 6 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iii)
\nAdding (ii) and (iii) we get
\n21m = 7
\nm =\u00a01\/3
\nPutting this value in equation (i), we get
\n5\u00d71\/3 + n = 2
\nn = 2- 5\/3 = 1\/3
\nm = 1\/ (x-1)
\n\u21d2 1\/3 = 1\/(x-1)
\n\u21d2 x = 4
\nn = 1\/(y-2)
\n\u21d2 1\/3 = 1\/(y-2)
\n\u21d2 y = 5
\nHence, x = 4 and y = 5<\/h3>\n

(v) (7x-2y)\/ xy = 5
\n(8x + 7y)\/xy = 15<\/h2>\n

Solution:
\n(7x-2y)\/ xy = 5
\n7\/y \u2013 2\/x = 5\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\n(8x + 7y)\/xy = 15
\n8\/y + 7\/x = 15\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nSubstituting\u00a01\/x =m\u00a0in the given equation we get,
\n\u2013 2m + 7n = 5\u00a0\u00a0\u00a0\u00a0 => -2 + 7n \u2013 5 = 0\u00a0 \u2026\u2026..(iii)
\n7m + 8n = 15\u00a0\u00a0\u00a0\u00a0 => 7m + 8n \u2013 15 = 0 \u2026\u2026(iv)
\nBy cross-multiplication method, we get,
\nm\/(-105-(-40)) = n\/(-35-30) = 1\/(-16-49)
\nm\/(-65) = n\/(-65) = 1\/(-65)
\nm\/-65 = 1\/-65
\nm = 1
\nn\/(-65) = 1\/(-65)
\nn = 1
\nm = 1 and n = 1
\nm =\u00a01\/x\u00a0= 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n =\u00a01\/x\u00a0= 1
\nTherefore, x = 1 and y = 1<\/h3>\n

(vi) 6x + 3y = 6xy
\n2x + 4y = 5xy<\/h2>\n

Solution:
\n6x + 3y = 6xy
\n6\/y + 3\/x = 6
\nLet 1\/x = m and 1\/y = n
\n=> 6n +3m = 6
\n=>3m + 6n-6 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)<\/h3>\n

2x + 4y = 5xy
\n=>\u00a02\/y + 4\/x = 5
\n=> 2n +4m = 5
\n=> 4m+2n-5 = 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\n3m + 6n \u2013 6 = 0
\n4m + 2n \u2013 5 = 0
\nBy cross-multiplication method, we get
\nm\/(-30 \u2013(-12)) = n\/(-24-(-15)) = 1\/(6-24)
\nm\/-18 = n\/-9 = 1\/-18
\nm\/-18 = 1\/-18
\nm = 1
\nn\/-9 = 1\/-18
\nn = 1\/2
\nm = 1 and n = 1\/2
\nm = 1\/x = 1 and n = 1\/y = 1\/2
\nx = 1\u00a0and\u00a0y = 2
\nHence, x = 1 and y = 2<\/h3>\n

(vii) 10\/(x+y) + 2\/(x-y) = 4
\n15\/(x+y) \u2013 5\/(x-y) = -2<\/h2>\n

Solution:
\nSubstituting\u00a01\/x+y = m\u00a0and\u00a01\/x-y = n\u00a0in the given equations, we get,
\n10m + 2n = 4\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0 10m + 2n \u2013 4 = 0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026..\u2026..(i)
\n15m \u2013 5n = -2\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0 15m \u2013 5n + 2 = 0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nUsing cross-multiplication method, we get,
\nm\/(4-20) = n\/(-60-(20)) = 1\/(-50 -30)
\nm\/-16 = n\/-80 = 1\/-80
\nm\/-16 = 1\/-80 and n\/-80 = 1\/-80
\nm = 1\/5 and n = 1
\nm = 1\/(x+y) = 1\/5
\nx+y = 5 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)
\nn = 1\/(x-y) = 1
\nx-y = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iv)
\nAdding equation (iii) and (iv), we get
\n2x = 6\u00a0\u00a0 => x = 3 \u2026\u2026.(v)
\nPutting the value of x = 3 in equation (3), we get
\ny = 2
\nHence, x = 3 and y = 2<\/h3>\n

(viii) 1\/(3x+y) + 1\/(3x-y) = 3\/4
\n1\/2(3x+y) \u2013 1\/2(3x-y) = -1\/8<\/h2>\n

Solution:
\nSubstituting 1\/(3x+y) = m and 1\/(3x-y) = n in the given equations, we get,
\nm + n = 3\/4 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026\u2026 (1)
\nm\/2 \u2013 n\/2 = -1\/8
\nm \u2013 n = -1\/4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u2026(2)
\nAdding (1) and (2), we get
\n2m = 3\/4 \u2013 1\/4
\n2m = 1\/2
\nPutting in (2), we get
\n1\/4 \u2013 n = -1\/4
\nn = 1\/4 + 1\/4 = 1\/2
\nm = 1\/(3x+y) = 1\/4
\n3x + y = 4\u00a0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(3)
\nn = 1\/( 3x-y) = 1\/2
\n3x \u2013 y = 2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(4)
\nAdding equations (3) and (4), we get
\n6x = 6
\nx = 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(5)
\nPutting in (3), we get
\n3(1) + y = 4
\ny = 1
\nHence, x = 1 and y = 1<\/h3>\n

2. Formulate the following problems as a pair of equations, and hence find their solutions:
\n(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
\n(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
\n(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.<\/h2>\n

Solutions:
\n(i) Let us consider,
\nSpeed of Ritu in still water = x km\/hr
\nSpeed of Stream = y km\/hr
\nNow, speed of Ritu during,
\nDownstream = x + y km\/h
\nUpstream = x \u2013 y km\/h
\nAs per the question given,
\n2(x+y) = 20
\nOr x + y = 10\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(1)
\nAnd, 2(x-y) = 4
\nOr x \u2013 y = 2\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(2)
\nAdding both the eq.1 and 2, we get,
\n2x = 12
\nx = 6
\nPutting the value of x in eq.1, we get,
\ny = 4
\nTherefore,
\nSpeed of Ritu rowing in still water = 6 km\/hr
\nSpeed of Stream = 4 km\/hr<\/h3>\n

(ii) Let us consider,
\nNumber of days taken by women to finish the work = x
\nNumber of days taken by men to finish the work = y
\nWork done by women in one day = 1\/x
\nWork done by women in one day = 1\/y
\nAs per the question given,
\n4(2\/x + 5\/y) = 1
\n(2\/x + 5\/y) = 1\/4
\nAnd, 3(3\/x + 6\/y) = 1
\n(3\/x + 6\/y) = 1\/3
\nNow, put 1\/x=m and 1\/y=n, we get,
\n2m + 5n = 1\/4 => 8m + 20n = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\n3m + 6n =1\/3 => 9m + 18n = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(2)
\nNow, by cross multiplication method, we get here,
\nm\/(20-18) = n\/(9-8) = 1\/ (180-144)
\nm\/2 = n\/1 = 1\/36
\nm\/2 = 1\/36
\nm = 1\/18
\nm = 1\/x = 1\/18
\nor x = 18
\nn = 1\/y = 1\/36
\ny = 36
\nTherefore,
\nNumber of days taken by women to finish the work = 18
\nNumber of days taken by men to finish the work = 36.<\/h3>\n

(iii) Let us consider,
\nSpeed of the train = x km\/h
\nSpeed of the bus = y km\/h
\nAccording to the given question,
\n60\/x + 240\/y = 4 \u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\n100\/x + 200\/y = 25\/6 \u2026\u2026\u2026\u2026\u2026.(2)
\nPut 1\/x=m and 1\/y=n, in the above two equations;
\n60m + 240n = 4\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(3)
\n100m + 200n = 25\/6
\n600m + 1200n = 25 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.(4)
\nMultiply eq.3 by 10, to get,
\n600m + 2400n = 40 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(5)
\nNow, subtract eq.4 from 5, to get,
\n1200n = 15
\nn = 15\/1200 = 1\/80
\nSubstitute the value of n in eq. 3, to get,
\n60m + 3 = 4
\nm = 1\/60
\nm = 1\/x = 1\/60
\nx = 60
\nAnd y = 1\/n
\ny = 80
\nTherefore,
\nSpeed of the train = 60 km\/h
\nSpeed of the bus = 80 km\/h
\nExercise 3.7<\/h3>\n

1. The ages of two friends Ani and Biju differ by 3 years. Ani\u2019s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.<\/h2>\n

Solution:
\nThe age difference between Ani and Biju is 3 yrs.
\nEither Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani\u2019s father\u2019s age is 30 yrs more than that of Cathy\u2019s age.
\nLet the ages of Ani and Biju be A and B respectively.
\nTherefore, the age of Dharam = 2 x A = 2A yrs.
\nAnd the age of Biju sister Ann B\/2 yrs
\nBy using the information that is given,
\nCase (i)
\nWhen Ani is older than that of Biju by 3 yrs then A \u2013 B = 3 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (1)
\n2A\u2212B\/2 = 30
\n4A \u2013 B = 60 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (2)
\nBy subtracting the equations (1) and (2) we get,
\n3A = 60 \u2013 3 = 57
\nA = 57\/3 = 19
\nTherefore, the age of Ani = 19 yrs
\nAnd the age of Biju is 19 \u2013 3 = 16 yrs.
\nCase (ii)
\nWhen Biju is older than Ani,
\nB \u2013 A = 3 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (1)
\n2A \u2212 B\/2 = 30
\n4A \u2013 B = 60 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (2)
\nAdding the equation (1) and (2) we get,
\n3A = 63
\nA = 21
\nTherefore, the age of Ani is 21 yrs
\nAnd the age of Biju is 21 + 3 = 24 yrs.<\/h3>\n

2. One says, \u201cGive me a hundred, friend! I shall then become twice as rich as you\u201d. The other replies, \u201cIf you give me ten, I shall be six times as rich as you\u201d. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y \u2013 100), y + 10 = 6(x \u2013 10)].<\/h2>\n

Solution:
\nLet Sangam have Rs A with him and Reuben have Rs B with him.
\nUsing the information that is given we get,
\nA + 100 = 2(B \u2013 100) \u21d2 A + 100 = 2B \u2013 200
\nOr A \u2013 2B = -300 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (1)
\nAnd
\n6(A \u2013 10) = ( B + 10 )
\nOr 6A \u2013 60 = B + 10
\nOr 6A \u2013 B = 70 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (2)
\nWhen equation (2) is multiplied by 2 we get,
\n12A \u2013 2B = 140 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (3)
\nWhen equation (1) is subtracted from equation (3) we get,
\n11A = 140 + 300
\n11A = 440
\n\u21d2 A = 440\/11 = 40
\nUsing A =40 in equation (1) we get,
\n40 \u2013 2B = -300
\n40 + 300 = 2B
\n2B = 340
\nB = 170
\nTherefore, Sangam had Rs 40 and Reuben had Rs 170 with them.<\/h3>\n

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km\/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km\/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.<\/h2>\n

Solution:
\nLet the speed of the train be A km\/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.
\nSpeed of the train = Distance travelled by train \/ Time taken to travel that distance
\nA = N (distance)\/ X (time)
\nOr, N = AX \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (1)
\nUsing the information that is given, we get:
\n(A+10) = X\/(N-2)
\n(A + 10) (N \u2013 2) = X
\nAN + 10N \u2013 2A \u2013 20 = X
\nBy using the equation (1) we get,
\n\u2013 2A + 10N = 20 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (2)
\n(A-10) = X\/(N+3)
\n(A \u2013 10) (N + 3) = X
\nAN \u2013 10N\u00a0+ 3A \u2013 30 = X
\nBy using the equation (1) we get,
\n3A \u2013 10N = 30 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (3)
\nAdding equation (2) and equation (3) we get,
\nA = 50
\nUsing the equation (2) we get,
\n(-2) x (50) + 10N = 20
\n-100 +10N = 20
\n=> 10N = 120
\nN = 12hours
\nFrom the equation (1) we get,
\nDistance travelled by the train, X = AN
\n= 50 x 12
\n= 600 km
\nHence, the distance covered by the train is 600km.<\/h3>\n

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.<\/h2>\n

Solution:
\nLet the number of rows be A and the number of students in a row be B.
\nTotal number of students = Number of rows x Number of students in a row
\n=AB
\nUsing the information, that is given,
\nFirst Condition:
\nTotal number of students = (A \u2013 1) ( B + 3)
\nOr AB = ( A \u2013 1 )(B + 3) = AB \u2013 B + 3A \u2013 3
\nOr 3A \u2013 B \u2013 3 = 0
\nOr 3A \u2013 Y = 3 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (1)
\nSecond condition:
\nTotal Number of students = (A + 2 ) ( B \u2013 3 )
\nOr AB = AB + 2B \u2013 3A \u2013 6
\nOr 3A \u2013 2B = -6 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (2)
\nWhen equation (2) is subtracted from (1)
\n(3A \u2013 B) \u2013 (3A \u2013 2B) = 3 \u2013 (-6)
\n-B + 2B = 3 + 6B = 9
\nBy using the equation (1) we get,
\n3A \u2013 9 =3
\n3A = 9+3 = 12
\nA = 4
\nNumber of rows, A = 4
\nNumber of students in a row, B = 9
\nNumber of total students in a class = AB = 4 x 9 = 36<\/h3>\n

5. In a \u2206ABC, \u2220 C = 3 \u2220 B = 2 (\u2220A + \u2220 B). Find the three angles.<\/h2>\n

Solution:
\nGiven,
\n\u2220C = 3 \u2220B = 2(\u2220B + \u2220A)
\n3\u2220B = 2 \u2220A+2 \u2220B
\n\u2220B=2 \u2220A
\n2\u2220A \u2013 \u2220B= 0- \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013\u00a0 (i)
\nWe know, the sum of all the interior angles of a triangle is 180O.
\nThus, \u2220 A +\u2220B+ \u2220C = 180O
\n\u2220A + \u2220B +3 \u2220B = 180O
\n\u2220A + 4 \u2220B = 180O\u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 -(ii)
\nMultiplying 4 to\u00a0 equation (i) , we get
\n8 \u2220A \u2013 4 \u2220B = 0- \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 (iii)
\nAdding equations (iii) and (ii) we get
\n9 \u2220A = 180O
\n\u2220A = 20O
\nUsing this in equation (ii), we get
\n20O+ 4\u2220B = 180O
\n\u2220B = 40O
\n3\u2220B =\u2220C
\n\u2220C = 3 x 40 = 120O
\nTherefore, \u2220A = 20O
\n\u2220B=40O
\n\u2220C = 120O<\/h3>\n

6. Draw the graphs of the equations 5x \u2013 y = 5 and 3x \u2013 y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.<\/h2>\n

Solutions:
\nGiven,
\n5x \u2013 y = 5
\n=> y = 5x \u2013 5
\nIts solution table will be.<\/h3>\n

\"\"
\nAlso given,3x \u2013 y = 3
\ny = 3x \u2013 3<\/h3>\n

\"\"
\nThe graphical representation o f these lines will be as follows:<\/h3>\n

\"\"
\nFrom the above graph we\u00a0can see that the triangle formed is \u2206ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) ,\u00a0 C(0,-5) and B(0,-3).<\/h3>\n

7. Solve the following pair of linear equations:
\n(i) px + qy = p \u2013 q
\nqx \u2013 py = p + q
\n(ii) ax + by = c
\nbx + ay = 1 + c
\n(iii) x\/a \u2013 y\/b = 0
\nax + by = a2\u00a0+ b2
\n(iv) (a \u2013 b)x + (a + b) y = a2\u00a0\u2013 2ab \u2013 b2
\n(a + b)(x + y) = a2\u00a0+ b2
\n(v) 152x \u2013 378y = \u2013 74
\n\u2013378x + 152y = \u2013 604<\/h2>\n

Solutions:
\n(i) px + qy = p \u2013 q\u2026\u2026\u2026\u2026\u2026(i)
\nqx \u2013 py = p + q\u2026\u2026\u2026\u2026\u2026\u2026.(ii)
\nMultiplying p to equation (1)\u00a0 and q to equation (2), we get
\np2x + pqy = p2\u00a0\u2212 pq \u2026\u2026\u2026\u2026 (iii)
\nq2x \u2212 pqy = pq + q2\u00a0\u2026\u2026\u2026\u2026 (iv)
\nAdding equation (iii) and equation (iv),we get
\np2x + q2\u00a0x = p2\u00a0 + q2
\n(p2\u00a0+ q2\u00a0) x = p2\u00a0+ q2
\nx = (p2\u00a0+ q2)\/ p2\u00a0+ q2\u00a0= 1
\nFrom equation (i), we get
\np(1) + qy = p \u2013 q
\nqy = p-q-p
\nqy = -q
\ny = -1<\/h3>\n

(ii) ax + by= c\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nbx + ay = 1+ c\u2026\u2026\u2026\u2026 ..(ii)
\nMultiplying a to equation (i) and\u00a0 b to equation (ii), we obtain
\na2x + aby = ac \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)
\nb2x + aby = b + bc\u2026\u2026\u2026\u2026\u2026 (iv)
\nSubtracting equation (iv) from equation (iii),
\n(a2\u00a0\u2013 b2) x = ac \u2212 bc\u2013 b
\nx = (ac \u2212 bc\u2013 b)\/ (a2\u00a0\u2013 b2)
\nx = c(a-b) \u2013b \/ (a2+b2)
\nFrom equation (i), we obtain
\nax +by = c
\na{c(a\u2212b)\u2212b)\/ (a2\u00a0\u2013 b2)} +by=c
\nac(a\u2212b)\u2212ab\/ (a2\u00a0\u2013 b2)+by=c
\nby=c\u2013ac(a\u2212b)\u2212ab\/(a2\u00a0\u2013 b2)
\nby=abc \u2013 b2\u00a0c+ab\/a2-b2
\ny = c(a-b)+a\/a2-b2<\/h3>\n

(iii) x\/a \u2013 y\/b = 0
\nax + by = a2\u00a0+ b2
\nx\/a \u2013 y\/b = 0
\n=>\u00a0bx \u2212 ay = 0 \u2026\u2026. (i)
\nax + by = a2\u00a0+ b2\u00a0\u2026\u2026.. (ii)
\nMultiplying a and b to equation (i) and (ii) respectively, we get
\nb2x \u2212 aby = 0 \u2026\u2026\u2026\u2026\u2026 (iii)
\na2x + aby = a\u00a03\u00a0+ ab3\u00a0\u2026\u2026 (iv)
\nAdding equations (iii) and (iv), we get
\nb2x + a2x = a\u00a03\u00a0+ ab2
\nx (b2\u00a0+ a2) = a (a2\u00a0+ b2) x = a
\nUsing equation (i), we get
\nb(a) \u2212 ay = 0
\nab \u2212 ay = 0
\nay = ab,
\ny = b<\/h3>\n

(iv) (a \u2013 b)x + (a + b) y = a2\u00a0\u2013 2ab \u2013 b2
\n(a + b)(x + y) = a2\u00a0+ b2
\n(a + b) y + (a \u2013 b) x = a2\u2212 2ab \u2212 b2\u00a0\u2026\u2026\u2026\u2026\u2026 (i)
\n(x + y)(a + b)\u00a0 = a\u00a02\u00a0+ b2
\n(a + b) y + (a + b) x\u00a0 = a2\u00a0+ b2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (ii)
\nSubtracting equation (ii) from equation (i), we get
\n(a \u2212 b) x \u2212 (a + b) x = (a\u00a02\u00a0\u2212 2ab \u2212 b\u00a02) \u2212 (a2\u00a0+ b2)
\nx(a \u2212 b \u2212 a \u2212 b) = \u2212 2ab \u2212 2b2
\n\u2212 2bx = \u2212 2b (b + a)
\nx =\u00a0b + a
\nSubstituting this value in equation (i), we get
\n(a + b)(a \u2212 b)\u00a0 +y (a + b)\u00a0 = a2\u2212 2ab \u2013 b2
\na2\u00a0\u2212 b2\u00a0+ y(a + b)\u00a0 = a2\u2212 2ab \u2013 b2
\n(a + b) y = \u2212 2ab
\ny = -2ab\/(a+b)
\n(v) 152x \u2212 378y = \u2212 74
\n76x \u2212 189y = \u2212 37
\nx =(189y-37)\/76\u2026\u2026\u2026\u2026\u2026..\u2026(i)
\n\u2212 378x + 152y = \u2212 604
\n\u2212 189x + 76y = \u2212 302 \u2026\u2026\u2026\u2026.. (ii)
\nUsing the value of x in equation (ii), we get
\n\u2212189(189y\u221237)\/76+76y=\u2212302
\n\u2212 (189)2y + 189 \u00d7 37 + (76)2\u00a0y = \u2212 302 \u00d7 76
\n189 \u00d7 37 + 302 \u00d7 76 = (189)2\u00a0y \u2212 (76)2y
\n6993 + 22952 = (189 \u2212 76) (189 + 76) y
\n29945 = (113) (265) y
\ny = 1
\nUsing equation (i), we get
\nx = (189-37)\/76
\nx = 152\/76 = 2<\/h3>\n

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.<\/h2>\n

\"\"
\nSolution:
\nIt is known that\u00a0the sum of the opposite angles of a cyclic quadrilateral is 180o
\nThus, we have
\n\u2220C +\u2220A = 180
\n4y + 20\u2212 4x = 180
\n\u2212 4x + 4y = 160
\nx \u2212 y = \u2212 40 \u2026\u2026\u2026\u2026\u2026(1)
\nAnd, \u2220B + \u2220D = 180
\n3y \u2212 5 \u2212 7x + 5 = 180
\n\u2212 7x + 3y = 180 \u2026\u2026\u2026..(2)
\nMultiplying 3 to equation (1), we get
\n3x \u2212 3y = \u2212 120 \u2026\u2026\u2026(3)
\nAdding equation (2) to equation (3), we get
\n\u2212 7x + 3x = 180 \u2013 120
\n\u2212 4x = 60
\nx = \u221215
\nSubstituting this value in equation (i), we get
\nx \u2212 y = \u2212 40
\n-y\u221215 = \u2212 40
\ny = 40-15
\n= 25
\n\u2220A = 4y + 20 = 20+4(25) = 120\u00b0
\n\u2220B = 3y \u2212 5 = \u2212 5+3(25)\u00a0= 70\u00b0
\n\u2220C = \u2212 4x = \u2212 4(\u2212 15) = 60\u00b0
\n\u2220D =\u00a05-7x
\n\u2220D=\u00a05\u2212 7(\u221215) = 110\u00b0
\nHence, all the angles are measured.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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