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Pair of Linear Equations in Two Variables : Exercise - 3.7 Optional (Mathematics NCERT Class 10th)


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Q.1      The ages of two friends Ani and Biju differ by 3 years. Ani's 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.
Sol.       Let the ages of Anil and Biju be x years and y years respectively. Then,
             x – y =  \pm 3 [Given]
             Dharam's age = 2x, and Cathy's age = {y \over 2}
             Clearly, Dharam is older than Cathy
             2x - {y \over 2} = 30
              \Rightarrow 4x – y = 60
             Thus, we have the following two systems of linear equations :
             x – y = 3 ...(1)
             and, 4x – y = 60 ...(2)
             x – y = – 3...(3)
             and, 4x – y = 80 ....(4)
             Subtracting (1) from (2), we get
             3x = 57
              \Rightarrow x = 19
             Putting x = 19 in (1), we get
             19 – y = 3
              \Rightarrow y = 19 – 3 = 16
             Subtracting (4) from (3),we get
              3x = 83
               \Rightarrow x = {{83} \over 3} = 27{2 \over 3}
              Putting {{83} \over 3} - y = -3
               \Rightarrow y = {{83} \over 3} + 3 = {{83 + 9} \over 3}
               = {{92} \over 3} = 30{2 \over 3}
              Hence, Ani's age = 19 years
              and Biju's age = 16 years
              or Ani's age = 27{2 \over 3} years
              and Biju's age = 30{2 \over 3} years

Q.2      One says, "Give me a hundred, friend ! I shall then become twice as rich as you." The other replies, " If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II]
Sol.        Let the friends be named as A and B. Let A has Rs. x and B has Rs. y.
              As per question, we have
              x + 100 = 2(y – 100)
               \Rightarrow x + 100 = 2y – 200
               \Rightarrow x – 2y + 300 = 0 ....(1)
              and, y + 10 = 6(x – 10)
               \Rightarrow y + 10 = 6x – 60
               \Rightarrow 6x – y – 70 = 0...(2)
              Multiplying (2) by 2, we get
              12x – 2y – 140 = 0 ...(3)
              Subtracting (3) from (1), we get
              – 11x + 440 = 0
              \Rightarrow – 11x = – 440
              \Rightarrow x = 40
             Putting x = 40 in (1), we get
             40 – 2y + 300 = 0
              \Rightarrow – 2y = – 340

              \Rightarrow y = 170

Q.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/m, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Sol.       Let the original speed of the train be x kmph and let the time taken to complete the journey be y hours.
             Then, the distance covered = xy km
              Case I : When speed = (x + 10) kmph
              and, time taken = (y – 2) hours
              In this case, distance = (x + 10) (y – 2)
               \Rightarrow xy = (x + 10) (y – 2)
               \Rightarrow xy = xy – 2x + 10y – 20
               \Rightarrow 2x – 10y + 20 = 0...(1)
              Case II : When speed = (x – 10) kmph
              and, time taken = (y + 3)hours
              In this case, distance = (x – 10) (y + 3)
               \Rightarrow xy = (x – 10) (y + 3)
               \Rightarrow xy = xy + 3x – 10y – 30
               \Rightarrow 3x – 10y – 30 = 0 ...(2)
              Subtracting (1) from (2), we get
              100 – 10y + 20 = 0
               \Rightarrow x = 50
              Putting x = 50 in (1), we get
              100 – 10y + 20 = 0
               \Rightarrow – 10y = – 120
              y = 12
              Therefore, The original speed of the train = 50kmph
              The time taken to complete the journey = 12 hours
              Therefore, The length of the journey = Speed × Time
              = (50 × 12) km = 600 km

Q.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.
Sol.       Let originally there x students in each row. Let there be y rows in total. Therefore, total number of students is xy.
              Clase I : When 3 students are taken extra in each row, then the number of rows in this case becomes (y –1).
              Therefore, xy = (x + 3) (y – 1)
               \Rightarrow xy = xy – x + 3y – 3
               \Rightarrow x – 3y + 3 = 0
              Case II : When 3 students are taken less in each row, then the number of rows becomes (y + 2).
               Therefore, xy = (x – 3) (y + 2)
                \Rightarrow xy = xy + 2x – 3y – 6
                \Rightarrow 2x – 3y – 6 = 0...(2)
                Solving (1) and (2), we get
                x = 9, y = 4
                Hence, the number of students = xy = 9 × 4 = 36.

Q.5     In a \Delta ABC, \angle C = 3\angle B = 2(\angle A + \angle B). Find the three angles.
Sol.       Let \angle A = x°,\angle B = y°.Then,
             \angle C = 3\angle B
                 \Rightarrow \angle C = 3y°
              \Rightarrow 3\angle B = 2(\angle A + \angle B)
              \Rightarrow 3y = 2(x + y)
             3y = 2x +2y
               \Rightarrow y = 2x
               \Rightarrow 2x – y = 0 ...(1)
              Since \angle A, \angle B, \angle C are angles of a triangle.
              Therefore, \angle A + \angle B + \angle C = 180°
               \Rightarrow x + y + 3y = 180° 
               \Rightarrow x + 4y = 180° ...(2)

              Putting y = 2x in (2), we get
              x + 8x = 180°
               \Rightarrow 9x = 180°
               \Rightarrow x = 20°
              Putting x = 20° in (1), we get
              y = 2 × 20° = 40°
              Hence,\angle A = 20°, \angle B = 40° and \angle C = 3y° =3 × 40° = 120°.

Q.6      Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinate of the vertices of the triangle formed by these lines and the y axis.
Sol.        The given system of equations is
               5x – y = 5 ...(1)
               and, 3x – y = 3 ...(2)
               Since both the equations are linear. so their graphs must be a straight line.
               Consider the equation (1),
               5x – y = 5
                \Rightarrow y = 5x – 5
               When x = 1, y = 5 – 5 = 0 ;
               When x = 2, y = 10 – 5 = 5
               Thus, we have the following table :

108
               Plot the points A(1, 0) and (2, 5) on the graph paper. Join AB and extend it on both sides to obtain the graph of 5x – y = 5.
               Now, consider the equation (2),
               3x – y = 3
                \Rightarrow y = 3x – 3
               When x = 0, y = 0 – 3 = 3;
               When x = 2, y = 3(2) – 3 = 3
                Thus, we have the following table :

48
                Plot the points C(0, – 3) and D(2, 3) on the same graph paper. Join CD and extend it on both sides to obtain the graph of 3x – y = 3.
49
                    We know that when a line meets y- axis, the value of x is zero. In the equation 5x – y = 5, if we put x = 0, we get y = – 5. Thus, the line 5x – y = 5 meets the y-axis at (0, – 5).
                     Similarly, the line 3x – y = 3 meets the y - axis at (0, –3)
                     The area of the \Delta ACE = {1 \over 2} × CE × OA
                                                             = \left( {{1 \over 2} \times 2 \times 1} \right) sq. units
                                                             = 1sq. units


Q.7      Solve the following pair of linear equations :
             (i) px + qy = p – q qx – py = p + q
             (ii) ax + by = c bx + ay = 1+ c
             (iii) {x \over a} - {y \over b} = 0 ax + by = {a^2} + {b^2}
             (iv) Solve for x and y :
             (a – b)x + (a + b)y = {a^2} - 2ab - {b^2}
             (a + b) (x + y) = {a^2} + {b^2}
             (v) 152x – 378y = – 74 – 378x + 152y = – 604
Sol.         (i) The given system of equations is
                px + qy = p – q
                 \Rightarrow px + qy – (p – q) = 0
                qx – py = p + q
                 \Rightarrow qx – py – (p + q) = 0
                By cross - multiplication, we get 

50
                 {x \over { - q(p + q) - p(p - q)}}  = {y \over { - q(p - q) + p(p + q)}}  = {1 \over { - {p^2} - {q^2}}}
                    \Rightarrow {x \over { - pq - {q^2} - {p^2} + pq}}  = {y \over { - pq + {q^2} + {p^2} + pq}}   = {1 \over { - \left( {{p^2} + {q^2}} \right)}}
                    \Rightarrow {x \over { - \left( {{p^2} + {q^2}} \right)}} = {y \over {{p^2} + {q^2}}} = {1 \over { - \left( {{p^2} + {q^2}} \right)}}
                    \Rightarrow x = {{ - \left( {{p^2} + {q^2}} \right)} \over { - \left( {{p^2} + {q^2}} \right)}} = 1
                   and y = {{{p^2} + {q^2}} \over { - \left( {{p^2} + {q^2}} \right)}} = - 1
                   Hence, the required solution is x = 1, y = – 1.
                  (ii) The given system of equations may be written as
                  ax + by – c = 0
                  bx + ay – (1 + c) = 0
                  By cross - multiplication, we have

51
                   \Rightarrow {x \over { - b\left( {1 + c} \right) + ac}} = {y \over { - bc + a\left( {1 + c} \right)}} = {1 \over {{a^2} - {b^2}}}
                    \Rightarrow {x \over { - b - bc + ac}} = {y \over { - bc + a + ac}} = {1 \over {{a^2} - {b^2}}}
                    \Rightarrow {x \over {c\left( {a - b} \right) - b}} = {y \over {c\left( {a - b} \right) + a}} = {1 \over {\left( {a - b} \right)\left( {a + b} \right)}}
                    \Rightarrow x = {{c\left( {a - b} \right) - b} \over {\left( {a - b} \right)\left( {a + b} \right)}}
                    = {c \over {a + b}} - {b \over {\left( {a - b} \right)\left( {a + b} \right)}}
                   and, y = {{c\left( {a - b} \right) + a} \over {\left( {a - b} \right)\left( {a + b} \right)}}
                    = {c \over {a + b}} + {b \over {\left( {a - b} \right)\left( {a + b} \right)}}
                   Hence, the required solution is
                    x = {c \over {a + b}} - {b \over {\left( {a - b} \right)\left( {a + b} \right)}}
                    y = {c \over {a + b}} + {b \over {\left( {a - b} \right)\left( {a + b} \right)}}

                   (iii) The given system of equations is
                   {x \over a} - {y \over b} = 0
                    \Rightarrow bx – ay = 0 ...(1)
                   and, ax + by – \left( {{a^2} + {b^2}} \right) = 0 ...(2)
                   By cross-multiplication, we have
52
                      {x \over {a\left( {{a^2} + {b^2}} \right) - 0}} = {y \over {0 + b\left( {{a^2} + {b^2}} \right)}} = {1 \over {{b^2} + {a^2}}}
                       \Rightarrow {x \over {a\left( {{a^2} + {b^2}} \right)}} = {x \over {b\left( {{a^2} + {b^2}} \right)}} = {1 \over {{a^2} + {b^2}}}
                     x = {{a\left( {{a^2} + {b^2}} \right)} \over {{a^2} + {b^2}}} = a
                     y = {{b\left( {{a^2} + {b^2}} \right)} \over {{a^2} + {b^2}}} = b
                     Hence, the required solution is x = a, y = b.

                      (iv) The given system of equations may be rewritten as
                      (a – b)x + (a + b)y – \left( {{a^2} - 2ab - {b^2}} \right) = 0
                      and, (a + b)x + (a + b)y – \left( {{a^2} + {b^2}} \right) = 0
                      By cross - multiplication, we have

53
                        {x \over { - \left( {a + b} \right)\left( {{a^2} + {b^2}} \right) + \left( {a + b} \right)\left( {{a^2} - 2ab - {b^2}} \right)}}
                        {y \over { - \left( {a + b} \right)\left( {{a^2} - 2ab - {b^2}} \right)\left( {a - b} \right)\left( {{a^2} + {b^2}} \right)}}
                          = {1 \over {\left( {a - b} \right)\left( {a + b} \right) - {{\left( {a + b} \right)}^2}}}
                          = {x \over {\left( {a + b} \right) - {a^2} - {b^2} + {a^2} - 2ab - {b^2}}}
                          = {y \over { - {a^3} + 2{a^2}b + a{b^2} - {a^2}b + 2a{b^2} - {b^3} + {a^3} + {a^3} + a{b^2} - {a^2}b - {b^3}}}
                          = {1 \over {{a^2} - {b^2} - {a^2} - {b^2} - 2ab}}
                          \Rightarrow {x \over {\left( {a + b} \right)\left( { - 2ab - 2{b^2}} \right)}}
                          \Rightarrow {x \over {\left( {a + b} \right)\left( { - 2ab - 2{b^2}} \right)}} = {y \over {4a{b^2}}} = {1 \over { - 2{a^2} - 2ab}}
                          \Rightarrow {x \over {\left( {a + b} \right)\left( { - 2b} \right)\left( {a + b} \right)}} = {y \over {4a{b^2}}}
                          = {1 \over { - 2b\left( {a + b} \right)}}
                          \Rightarrow x = a + b
                         and y = {{ - 2ab} \over {a + b}}
                         Hence, the required solution is x = a + b, y = {{ - 2ab} \over {a + b}}
                         Hence, the required solution is x = a + b, y = {{ - 2ab} \over {a + b}}

                         (v) We have, 152x – 378y = – 74 ...(1)
                         and, – 378x + 152y = – 604 ...(2)
                         Adding (1) and (2), we get
                         – 226x – 256y = – 678
                          \Rightarrow x + y = 3 ...(3)
                         Subtracting (1) from (2), we get
                         – 530x + 530y = – 530
                          \Rightarrow x – y = 1 ...(4)
                         Adding (3) and (4), we get 2x = 4
                          \Rightarrow x = 2
                         Putting x = 2 in (1), we get 2 + y = 3
                          \Rightarrow y = 1
                         Hence, the required solution is x = 1, y = 2.


Q.8      ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

54
Sol.       We know that the sum of the opposite angles of cyclic quadrilateral is 180°. In the cyclic quadrilateral ABCD, angles A and C, angles B and D form pais of opposite angles
              Therefore, \angle A + \angle C = 180^\circ
              and \angle B + \angle D = 180^\circ
               \Rightarrow (4y + 20) – 4x = 180°
              and (3y – 5) + (–7x + 5) = 180°
                \Rightarrow – 4x + 4y – 160 = 0
               and – 7x + 3y – 180 = 0
                \Rightarrow – x + y – 40 = 0
               and, – 7x + 3y – 180 = 0
               Multiplying (1) by 3, we get
               – 3x + 3y – 120 = 0
               Subtracting (3) from (2), we get
              – 4x – 60 = 0
               \Rightarrow x = – 15
              Putting x = – 15 in (1), we get
              15 + y – 40 = 0
               \Rightarrow y – 25 = 0
               \Rightarrow y = 25
              Hence, \angle A = 4y + 20 = 4 × 25 + 20
              = 100 + 20 = 120°
              \angle B = 3y-5 = 3 \times 25-5
               = 75-5 = 70^\circ
              \angle C = -4x = \left( {-4} \right) \times \left( {-15} \right) = 60^\circ
              and, \angle D = -7x + 5 = -7 \times \left( {-15} \right) + 5
               = 105 + 5 = 110^\circ



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