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

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 : 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 : 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. 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 ${x \over { - q(p + q) - p(p - q)}}$ $= {y \over { - q(p - q) + p(p + q)}}$ $= {1 \over { - {p^2} - {q^2}}}$
$\Rightarrow$ ${x \over { - pq - {q^2} - {p^2} + pq}}$ $= {y \over { - pq + {q^2} + {p^2} + pq}}$  $= {1 \over { - \left( {{p^2} + {q^2}} \right)}}$
$\Rightarrow {x \over { - \left( {{p^2} + {q^2}} \right)}} = {y \over {{p^2} + {q^2}}} = {1 \over { - \left( {{p^2} + {q^2}} \right)}}$
$\Rightarrow x = {{ - \left( {{p^2} + {q^2}} \right)} \over { - \left( {{p^2} + {q^2}} \right)}} = 1$
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 $\Rightarrow {x \over { - b\left( {1 + c} \right) + ac}} = {y \over { - bc + a\left( {1 + c} \right)}} = {1 \over {{a^2} - {b^2}}}$
$\Rightarrow {x \over { - b - bc + ac}} = {y \over { - bc + a + ac}} = {1 \over {{a^2} - {b^2}}}$
$\Rightarrow {x \over {c\left( {a - b} \right) - b}} = {y \over {c\left( {a - b} \right) + a}} = {1 \over {\left( {a - b} \right)\left( {a + b} \right)}}$
$\Rightarrow x = {{c\left( {a - b} \right) - b} \over {\left( {a - b} \right)\left( {a + b} \right)}}$
$= {c \over {a + b}} - {b \over {\left( {a - b} \right)\left( {a + b} \right)}}$
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 ${x \over {a\left( {{a^2} + {b^2}} \right) - 0}} = {y \over {0 + b\left( {{a^2} + {b^2}} \right)}} = {1 \over {{b^2} + {a^2}}}$
$\Rightarrow {x \over {a\left( {{a^2} + {b^2}} \right)}} = {x \over {b\left( {{a^2} + {b^2}} \right)}} = {1 \over {{a^2} + {b^2}}}$
$x = {{a\left( {{a^2} + {b^2}} \right)} \over {{a^2} + {b^2}}} = a$
$y = {{b\left( {{a^2} + {b^2}} \right)} \over {{a^2} + {b^2}}} = b$
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 ${x \over { - \left( {a + b} \right)\left( {{a^2} + {b^2}} \right) + \left( {a + b} \right)\left( {{a^2} - 2ab - {b^2}} \right)}}$
${y \over { - \left( {a + b} \right)\left( {{a^2} - 2ab - {b^2}} \right)\left( {a - b} \right)\left( {{a^2} + {b^2}} \right)}}$
$= {1 \over {\left( {a - b} \right)\left( {a + b} \right) - {{\left( {a + b} \right)}^2}}}$
$= {x \over {\left( {a + b} \right) - {a^2} - {b^2} + {a^2} - 2ab - {b^2}}}$
$= {y \over { - {a^3} + 2{a^2}b + a{b^2} - {a^2}b + 2a{b^2} - {b^3} + {a^3} + {a^3} + a{b^2} - {a^2}b - {b^3}}}$
$= {1 \over {{a^2} - {b^2} - {a^2} - {b^2} - 2ab}}$
$\Rightarrow {x \over {\left( {a + b} \right)\left( { - 2ab - 2{b^2}} \right)}}$
$\Rightarrow {x \over {\left( {a + b} \right)\left( { - 2ab - 2{b^2}} \right)}} = {y \over {4a{b^2}}} = {1 \over { - 2{a^2} - 2ab}}$
$\Rightarrow {x \over {\left( {a + b} \right)\left( { - 2b} \right)\left( {a + b} \right)}} = {y \over {4a{b^2}}}$
$= {1 \over { - 2b\left( {a + b} \right)}}$
$\Rightarrow$ 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. 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, $\angle D = -7x + 5 = -7 \times \left( {-15} \right) + 5$
$= 105 + 5 = 110^\circ$