Pair of Linear Equations in Two Variables : Exercise 3.5

Q.1      Which of the following pairs of linear equations has unique solution, no solution, or finitely many solutions ? In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0            3x – 9y – 2 = 0
(ii) 2x + y = 5                 3x + 2y = 8
(iii) 3x – 5y = 20             6x – 10y = 40
(iv) x – 3y – 7 = 0           3x – 3y – 15 = 0
Sol.       (i) The given system of equations is
x – 3y – 3 = 0
and, 3x – 9y – 2 = 0
These are of the form ${a_1}x + {b_1}y + {c_1} = 0$
and ${a_2}x + {b_2}y + {c_2} = 0$
where ${a_1} = 1,{b_1} = - 3,{c_1} = - 3$ and ${a_2} = 3,{b_2} = - 9,{c_2} = - 2$
We have, ${{{a_1}} \over {{a_2}}} = {1 \over 3},{{{b_1}} \over {{b_2}}} = {{ - 3} \over { - 9}} = {1 \over 3}and{{{c_1}} \over {{c_2}}} = {{ - 3} \over { - 2}} = {3 \over 2}$
Clearly, ${{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} \ne {{{c_1}} \over {{c_2}}}$
So, the given system of equations has no solution i.e., it is inconsistent.

(ii) The given system of equations may be written as
2x + y – 5 = 0 and 3x + 2y – 8 = 0
These are of the form ${a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0$
where ${a_1} = 2,{b_1} = 1,{c_1} = - 5\,and\,{a_2} = 3,{b_2} = 2\,\,and\,\,{c_2} = - 8$
We have, ${{{a_1}} \over {{a_2}}} = {2 \over 3},{{{b_1}} \over {{b_2}}} = {1 \over 2}$
Clearly, ${{{a_1}} \over {{a_2}}} \ne {{{b_1}} \over {{b_2}}}$
So, the given system of equations has a unique solution.
To find the solution, we use the cross - multiplication method. By cross - multiplication, we have ${x \over {1 \times - 8 - 2 \times \left( { - 5} \right)}} = {y \over { - 5 \times 3 - \left( { - 8} \right) \times 2}} = {1 \over {2 \times 2 - 3 \times 1}}$
$\Rightarrow {x \over { - 8 + 10}} = {y \over { - 15 + 16}} = {1 \over {4 - 3}}$
$\Rightarrow {x \over 2} = {y \over 1} = {1 \over 1}$
$\Rightarrow$ x = 2, y = 1

Hence, the given system of equations has a unique solution given by x = 2, y = 1.

(iii) The system of equations may be written as
3x – 5y – 20 = 0 and 6x – 10y – 40 = 0
The given equations are of the form
${a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0$
where, ${a_1} = 3,{b_1} = - 5,{c_1} = - 20\,\,and\,{a_2} = 6,{b_2} = - 10,{c_2} = - 40$
We have, ${{{a_1}} \over {{a_2}}} = {3 \over 6} = {1 \over 2},{{{b_1}} \over {{b_2}}} = {{ - 5} \over { - 10}} = {1 \over 2}and{{{c_1}} \over {{c_2}}} = {{ - 20} \over { - 40}} = {1 \over 2}$
Clearly, ${{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} = {{{c_1}} \over {{c_2}}}$
So, the given system of equations has infinitely many solutions.

(iv) The given system of equations is
x – 3y – 7 = 0 and 3x – 3y – 15 = 0
The given equations are of the form
${a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0$
where ${a_1} = 1,{b_1} = - 3,{c_1} = - 7\,\,and\,\,{a_2} = 3,{b_2} = - 3,{c_2} = - 15$
we have, ${{{a_1}} \over {{a_2}}} = {1 \over 3},{{{b_1}} \over {{b_2}}} = {{ - 3} \over { - 3}} = 1$               Clearly, ${{{a_1}} \over {{a_2}}} \ne {{{b_1}} \over {{b_2}}}$
So, the given system of equations has a unique solution.To find the solution, we use cross-multiplication method.            By cross-multiplication, we have ${x \over {45 - 21}} = {y \over { - 21 + 15}} = {1 \over { - 3 + 9}}$
${x \over {24}} = {y \over { - 6}} = {1 \over 6}$
$\Rightarrow$ $x = {{24} \over 6} = 4,y = {{ - 6} \over 6} = - 1$
Hence, the given system of equations has a unique solution given by
x = 4, y = – 1

Q.2      (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?  2x + 3y = 7 (a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1 (2k – 1)x + (k – 1)y = 2k + 1
Sol.       (i) We know that the system of equations
${a_1}x + {b_1}y = {c_1}\,and\,{a_2}x + {b_2}y = {c_2}$
has infinite number of solutions if
${{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} = {{{c_1}} \over {{c_2}}}$
Therefore, The given system of equations will have infinite number of solutions if
${2 \over {a - b}} = {3 \over {a + b}} = {7 \over {3a + b - 2}}$
$\Rightarrow$ ${2 \over {a - b}} = {3 \over {a + b}}and{2 \over {a - b}} = {7 \over {3a + b - 2}}$
$\Rightarrow$ 2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b
$\Rightarrow$ a = 5b and a – 9b = – 4
Putting a = 5b in a – 9b = – 4, we get
5b – 9b = – 4
$\Rightarrow$ – 4b = – 4
$\Rightarrow$ b = 1
Putting b = 1 in a – 9b = – 4, we get
a = 5(1) = 5
Hence, the given system of equations will have infinitely many solutions if
a = 5 and b = 1

(ii) We know that the system of equations
${a_1}x + {b_1}y = {c_1}\,and\,{a_2}x + {b_2}y = {c_2}$
has no solution if ${{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} \ne {{{c_1}} \over {{c_2}}}$
So, the given system of equations will have no solution if
${3 \over {2k - 1}} = {1 \over {k - 1}} \ne {1 \over {2k + 1}}$
$\Rightarrow$ ${3 \over {2k - 1}} = {1 \over {k - 1}}and{1 \over {k - 1}} \ne {1 \over {2k + 1}}$
Now, ${3 \over {2k - 1}} = {1 \over {k - 1}}$
$\Rightarrow$ 3k – 3 = 2k – 1
$\Rightarrow$ k = 2
Clearly, for k = 2, we have
${1 \over {k - 1}} \ne {1 \over {2k + 1}}$
Hence, the given system of equations will have no solution if k = 2.

Q.3      Solve the following pair of linear equations by the substitution and cross - multiplication methods :
8x + 5y = 9
3x + 2y = 4
Sol.         The given system of equations is
8x + 5y = 9...(1)
and, 3x + 2y = 4...(2)
By graphical method :
For the graph of 8x + 5y = 9
We have, 8x + 5y = 9
$\Rightarrow$ 5y = 9 – 8x
$\Rightarrow y = {{9 - 8x} \over 5}$
$\Rightarrow$ When x = 3, y = ${{9 - 24} \over 5} = {{ - 15} \over 5} = - 3$ ; when x = – 2, y = ${{9 + 16} \over 5} = {{25} \over 5} = 5$
Thus, we have the following table : Plot the points A(3, – 3) and B(–2, 5) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 8x + 5y = 9 as shown. For the graph of 3x + 2y = 4
We have,
3x + 2y = 4
$\Rightarrow$ 2y = 4 – 3x
$\Rightarrow$ y = ${{4 - 3x} \over 2}$
When x = 0, y = ${{4 - 0} \over 2} = {4 \over 2} = 2$;
When x = 2, y = ${{4 - 6} \over 2} = {{ - 2} \over 2} = - 1$
Thus, we have the following table : Plot the points C(0, 2) and D(2, – 1) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x + 2y = 4.
Clearly, the two lines intersect at B(–2, 5).
Hence, x = – 2, y = 5 is the solution of the given system of equations.
By substitution method :
Substituting y = ${{9 - 8x} \over 5}$ in (2), we get
$3x + 2\left( {{{9 - 8x} \over 5}} \right) = 4$
$\Rightarrow$ 15x + 18 – 16x = 20
$\Rightarrow$ – x = 2
$\Rightarrow$ x = – 2
Putting x = – 2 in (1), we get
8(–2) +5y = 9
$\Rightarrow$ 5y = 9 + 16 = 25
$\Rightarrow$ $y = {{25} \over 5} = 5$
Hence, x = – 2, y = 5 is the solution of the given system of equations.
By cross - multiplication method : ${x \over { - 20 + 18}} = {y \over { - 27 + 32}} = {1 \over {16 - 15}}$
$\Rightarrow$ ${x \over { - 2}} = {y \over 5} = {1 \over 1}$
$\Rightarrow$ $x = {{ - 2} \over 1} = - 2\,\,and\,\,y = {5 \over 1} = 5$
Hence, x = – 2, y = 5 is the solution of the given system of equations.
The method of cross - multiplication is more efficient.

Q.4      Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(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.

(ii) A fraction becomes ${1 \over 3}$ when 1 is subtracted from the numerator and it becomes ${1 \over 4}$ when 8 is added to its denominator. find the fraction.

(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 ?

(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 ?

(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.
Sol.       (i) Let the fixed charges be Rs. x and charge per day be Rs. y.
Therefore, By the given conditions,
x + 20y = 1000 ...(1)
and, x + 26y = 1180 ...(2)
Subtracting (1) from (2), we get
6y = 180
$\Rightarrow$ y = 30
From (1), x + 20 × 30 = 1000
$\Rightarrow$ x = 1000 – 600 = 400
Therefore, Fixed charge = Rs. 400
and cost of food per day = Rs. 30

(ii) Let the fraction be ${x \over y}$.
According to the given conditions :
${{x - 1} \over y} = {1 \over 3}$
$\Rightarrow$ 3x – 3 = y
$\Rightarrow$ 3x – y – 3 = 0...(1)
and, ${x \over {y + 8}} = {1 \over 4}$
$\Rightarrow$ 4x = y + 8
$\Rightarrow$ 4x – y – 8 = 0
Solving (1) and (2) by cross - multiplication method. ${x \over {8 - 3}} = {y \over { - 12 + 24}} = {1 \over { - 3 + 4}}$
$\Rightarrow$ ${x \over 5} = {y \over {12}} = {1 \over 1}$
$\Rightarrow$ x = 5, y = 12
Hence, the required fraction is ${5 \over {12}}$.

(iii) Let Yash answer x questions correctly and y questions incorrectly. According to the given conditions :
3x – y = 40 ...(1)
and, 4x – 2y = 50 ...(2)
Multiplying (1) by 2 and subtracting (2) from the result so obtained, we get
2x = 30
$\Rightarrow$ x = 15
From (1), 3 × 15 – y = 40
$\Rightarrow$ – y = 40 – 45
$\Rightarrow$ – y = – 5
$\Rightarrow$ y = 5
Therefore, Total number of questions in the test are 15 + 5 = 20.

(iv) Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of Y be y km/hr.
Case I : When two cars move in the same direction. Let these cars meet at point Q. Then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 5 hours.
Therefore, Distance travelled by car A in 5 hours = 5x km
$\Rightarrow$ AQ = 5x
Distance travelled by' car Y in 5 hours = 5y km
$\Rightarrow$ BQ = 5y
Clearly, AQ – BQ = AB
$\Rightarrow$ 5x – 5y = 100 [Since, AB = 100 km]
$\Rightarrow$ x – y = 20
Case II : When two cars move in opposite directions.
Let these cars meet at point P. Then,
Distance traveled be car X = AP

Distance traveled be car Y = BP
In this case, two cars meet in 1 hour.
Therefore, Distance traveled by car X in 1 hour = x km
$\Rightarrow$ AP = x
Distance travelled by car Y in 1 hour = y km
$\Rightarrow$ BP = y
Clearly, AP + PB = AB
$\Rightarrow$ x + y = 100 ...(2) [Since, AB = 100 km]
Solving (1) and (2), we have
x = 60, y = 40
Hence, speed of car X is 60 km/hr and speed of car Y 40 km/hr.

(v) Let the length and breadth of the rectangle be x and y units respectively. Then,
Area = xy sq. units
If length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq. units.
Therefore, xy – 9 = (x – 5) (y + 3)
$\Rightarrow$ xy – 9 = xy + 3x – 5y – 15
$\Rightarrow$ 3x – 5y – 6 = 0 ...(1)
When length is increased by 3 units and breadth by units, the area is increased by 67 sq. units.
Therefore, xy + 67 = (x + 3) (y + 2)
$\Rightarrow$ xy + 67 = xy + 2x + 3y + 6
$\Rightarrow$ 2x + 3y – 61 = 0 ...(2)
Solving (1) and (2), we get
${x \over {305 + 18}} = {{ - y} \over { - 183 + 12}} = {1 \over {9 + 10}}$
$\Rightarrow$ $x = {{323} \over {19}} = 17$, $y = {{171} \over {19}} = 9$
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.