Q1: For what value of p, the pair of linear equations px = 2y; 2x  y + 5 = 0 has unique solution?
Sol: We have:
px = 2y
⇒ p  2y = 0
2x = y + 5
⇒ 2x  y =  5
Here, a_{1} = p, b_{1} =  2,
c_{1} = 0
a_{2 }= 2, b_{2} =  1,
c_{2} =  5
For a unique solution,
⇒
⇒ p ≠ 2 × 2
⇒ p ≠ 4
Q2: In a cyclic quadrilateral PQRS, ∠P = (2x + 4)°, ∠Q = (y + 3)°, ∠R = (2y + 10)° and ∠S = (4x  5)°. Find its four angles.
Sol: In a cyclic quadrilateral, the opposite angles are supplementary.
∴∠P + ∠R = 180°
⇒ (2x + 4)° + (2y + 10)° = 180°
⇒ 2x + 2y + 14  180 = 0
⇒ 2x + 2y  166 = 0
⇒ x + y  83 = 0 ...(1)
Also ∠Q + ∠S = 180°
∴ (y + 3)° + (4x  5)° = 180°
⇒ y + 4x  2  180 = 0
⇒ y + 4x  182 = 0 ...(2)
From (1) and (2),
a_{1} = 1, b_{1} = 1, c_{1} =  83
a_{2} = 4, b_{2} = 1, c_{2} =  182
⇒
⇒
∴ ∠P = (2x + 4)° = [(2 × 33) + 4]° = 70°
∠Q = (y + 3)° = [50 + 3]° = 53°
∠R = (2y + 10)° = [2 × 50 + 10]° = 110°
∠S = (4x  5)° = [4 × 33  5°] = 127°
Q3: Solve:
23x + 35y = 209
35x + 23y = 197
Sol: We have:
23x + 35y = 209 ...(1)
35x + 23y = 197 ...(2)
58x + 58y = 406 [Adding (1) and (2)]
⇒ x + y = 7 ...(3) [Dividing by 58]
Subtracting (1) from (2),
35x + 23y = 197
23x + 35y = 209
() () ()
12x  12y =  12
⇒ x  y = 1 ...(4) [Dividing by 12]
Adding (3) and (4),
2x = 8 ⇒ x = 4
From (3) x + y = 7 ⇒ 4 + y = 7
⇒ y = 3
So, x = 4 and y = 3.
Q4: Solve:
3x + 5y = 70 ...(1)
7x  3y = 60 ...(2)
Sol: From (1) and (2), we have:
a_{1} = 3, b_{1} = 5, c_{1} =  70
a_{2} = 7, b_{2} =  3, c_{2} =  60
⇒
∴
Thus, the required solution is:
.
Q5: Without drawing the graphs, state whether the following pair of linear equations will represent intersecting lines, coinciding lines or parallel lines:
6x  3y + 10 = 0
2x  y + 9 = 0
Sol: Here, the given set of equations is:
6x  3y + 10 = 0 ...(1)
2x  y + 9 = 0 ...(2)
From (1) and (2), we have:
a_{1} = 6, b_{1} =  3, c_{1} = 10
a_{2} = 2, b_{2} =  1, c_{2} = 9
Now,
∴We have
=
This condition represents parallel lines. Hence, the given pair represents parallel lines.
Q6: Check graphically whether the pair of equations
3x  2y + 2 = 0
= 0
is consistent. Also find the coordinates of the points where the graphs of the equations meet the yaxis.
Sol: ∴ 3x  2y + 2 = 0 ⇒ y = ..(1)
Also = 0 ⇒ y = ...(2)
Plotting the points (0, 1), (2, 4), ( 2,  2) and (0, 3), (2, 6), (–2, 0) we get two straight lines l_{1} and l_{2} which are parallel.
∴ The given equations are inconsistent.
From the graph, we observe that line l_{1} meets the yaxis at (0, 1) and line l_{2} meets the yaxis at (0, 3).
Q7: A fraction becomes 1/3 if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes 2/5. Find the fraction.
Sol: Let the fraction be x/y.
From 1st condition,
⇒ 3x + 6 = y + 2
⇒ 3x − y + 4 = 0 ...(1)
From 2nd conditon,
⇒ 5x + 15 = 2y + 6
⇒ 5x − 2y + 9 = 0 ...(2)
From (1) and (2), we have:
Q8: Check graphically whether the pair of equations
3x + 5y = 15
x  y = 5
is consistent. Also, find the coordinates of the points where the graphs of equations meet the yaxis.
Sol: We have
3x + 5y = 15
⇒
∴
And from x  y = 5
⇒ y = x  5
Plotting the above two sets of points we get two straight lines l1 and l2 which intersect at the point (5, 0).
Thus, the given system is consistent.
Obviously, line l_{1} meets the yaxis at (0, 3) and line l_{2} meets the yaxis at (0,  5).
Q9: Places A and B are 160 km apart on the 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 8 hours, but if they travel towards each other, they meet in 2 hours. What are the speeds of the two cars?
Sol: Let the carI and carII start from A and B at x km/hr and y km/hr respectively.
CaseI: [Cars are moving in the same direction]
Let the two cars meet at C after 8 hours.
Distance covered:
by carI = AC = 8x km
by carII = BC = 8y km
∴ AB = AC  BC
⇒ 160 = 8x  8y
⇒ x  y = 20 ...(1)
CaseII: [Cars are moving in opposite directions]
Let, after 2 hours, the cars meet at D.
∴ Distance cover after 2 hours,
by carI = AD = 2x km
by carII = BD = 2y km
⇒ AB = AD + BD
⇒ 160 = 2x + 2y
⇒ 80 = x + y
⇒ x + y = 80 ...(2)
Adding (1) and (2), we get
x + y = 80
x  y = 20
2x = 100
⇒ x = 100/2 = 50
⇒ Substituting x = 50 in (1), we get
x  y = 20 ⇒ 50  y = 20
⇒ y = 50  20 = 30
⇒ Speed of carI = 50 km/hr
Speed of carII = 30 km/hr.
Q10: Solve for x and y:
Sol: We have:
= ...(1)
ax − by = 2ab ...(2)
Dividing (2) by a, we have:
⇒ ...(3)
From (1) and (3), we have
⇒
From (2),
ab  by = 2ab
⇒  by = 2ab  ab = ab
⇒
y =  a
Thus, x = b and y =  a.
Q11: The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is 8/15.
Sol: Let the two numbers be x and y.
According to the conditions:
x + y = 8 ...(1)
= ..(2)
From (1), x = (8  y)
Substituting x = (8  y) in (2),
⇒
⇒ 8 × 15 = 8 × y (8  y)
⇒ 64y  8y^{2}  120 = 0
⇒ y^{2} + 8y  15 = 0
⇒ y^{2}  8y + 15 = 0
⇒ y^{2}  5y  3y + 15 = 0
⇒ y (y  5)  3 (y  5) = 0
⇒ (y  5) (y  3) = 0
⇒ If y  5 = 0 then y = 5
or if y  3 = 0, then y = 3
Since x = 8  y
⇒ when y = 5,
then x = 8  5 = 3
when y = 3, then
x = 8  3 = 5
⇒ The required numbers are (3, 5) or
(5, 3).
Q12: Solve the following pair of equations:
Sol: Let
⇒The given system of equations becomes:
5p + q = 2 ...(1)
6p  3q = 1 ...(2)
Multiplying (1) by 3 and adding to (2),
and 5p + q = 2 ⇒
⇒
⇒
Since,
∴
= 3 ⇒ x = 4
Also,
⇒ y  2 = 3 ⇒ y = 5
Thus, x = 4 and y = 4
⇒ 3y  6 = 1
⇒ 3y = 1 + 6 = 7
⇒ y = 7/3
Q13: Solve the following pair of equations:
Sol: Let
⇒ The given pair of equation is expressed as
10p + 2q = 4 ⇒ 5p + q = 2 ...(1)
15p  5q =  2 ...(2)
Multiplying (1) by 5 and adding to (2)
From (1),
⇒1 + q = 2 ⇒ q = 2  1 = 1
Since,
⇒
⇒ x + y = 5 ...(3)
And
⇒ x − y = 1 ...(4)
Adding (3) and (4),
From (3), 3 + y = 5 ⇒ y = 2
Thus, x = 3 and y = 2.
Ques 14: Solve for x and y:
37x + 43y = 123
43x + 37y = 117
Sol: We have:
37x + 43y = 123 ...(1)
43x + 37y = 117 ...(2)
Adding (1) and (2)
Dividing both sides by 80, we get
x + y = 3 ...(3)
Subtracting (2) from (1),
 6x + 6y = 6 ⇒  x + y = 1 ...(4)
Adding:
⇒ y = 4/2 = 2
Putting y = 2 in x + y = 3, we get
x + 2 = 3 ⇒ x = 3  2 = 1
Thus, x = 1 and y = 2.
Ques 15: Solve for ‘x’ and ‘y’:
(a  b) x + (a + b) y = a^{2}  2ab  b^{2}
(a + b) (x + y) = a^{2} + b^{2}
Sol: We have:
(a  b) x + (a + b) y = a^{2}  2ab  b^{2} ...(1)
(a + b) x + (a + b) y = a^{2} + b^{2 }...(2)
− 2b x = − 2ab − b
⇒ (− 2b) x = − 2b (a + b)
⇒
From (2),
(a + b) (a + b) + (a + b) y = a^{2} + b^{2}
⇒(a + b)2 + (a + b) y = (a^{2} + b^{2})
⇒ (a + b) y = (a^{2} + b^{2})  (a + b)^{2}
⇒ (a + b) y = a^{2} + b^{2}  (a^{2} + b^{2} + 2ab)
⇒ (a + b) y = a^{2} + b^{2}  a^{2}  b^{2}  2ab
⇒ (a + b) y =  2ab
⇒
Thus, x = (a + b) and
Q16: Represent the following pair of equations graphically and write the coordinates of points where the lines intersect yaxis:
x + 3y = 6, 2x  3y = 12
Sol: We have:
x + 3y = 6
and 2x  3y = 12
Plotting the above points, we get two straight lines l_{1} and l_{2} such that they intersect at (6, 0) as shown below:
Obviously,
The line l_{1} meets the yaxis at (0, 2).
The line l_{2} meets the yaxis at (0,  4).
Q17: Solve for x and y:
Sol: Let
∴ We have:
5p + q = 2 ...(1)
6p  3q = 1 ...(2)
From (1) and (2), we have:
∴
⇒
∴
And
Now,
⇒ x − 1 = 3 ⇒ x = 4
And
⇒ y − 2 = 3 ⇒ y = 5
Thus x = 4 and y = 5
Q18: For what values of ‘a’ and ‘b’ does the following pair of equations have an infinite number of solutions?
2x + 3y = 7
a (x + y)  b (x  y) = 3a + b  2
Sol: We have:
2x + 3y = 7 ...(1)
a (x + y)  b (x  y) = 3a + b  2 ...(2)
From (2), we have:
a (x + y)  b (x  y) = 3a + b  2
⇒ ax + ay  bx + by = 3a + b  2
⇒ ax  bx + ay + by = 3a + b  2
⇒ (a  b) x + (a + b) y = 3a + b  2
Now, A_{1} = 2, B_{1}= 3,
C_{1} =  7
A_{2} = (a  b), B_{2}
= (a + b),
C_{2} =  [3a + b  2]
For infinite number of solutions,
i.e.,
∴
⇒ 2 (a + b) = 3 (a − b)
⇒ 2a + 2b − 3a + 3b = 0
⇒ − a + 5b = 0
⇒ a = 5b ...(3)
Also =
⇒ 3 (3a + b  2) = 7 (a + b)
⇒ 9a + 3b  6 = 7a + 7b
⇒ 9a  7a + 3b  7b = 6
⇒ 2a  4b = 6
⇒ a  2b = 3 ...(4)
From (3) and (4),
5b  2b = 3
⇒ 3b = 3 ⇒ b = 1
Thus, a = 5 × b
⇒ a = 5 × 1 = 5
i.e., a = 5 and b = 1.
Q19: Solve the following pairs of equations for x and y:
Sol: Let
∴ We have:
15p + 22q = 5 ...(1)
40p + 55q = 13 ...(2)
From (1) and (2), we get
Adding (3) and (4), we have
2x = 16 ⇒ x = 16/2 =8
From (4), 8 + y = 11 ⇒ y = 11  8 = 3
Thus, x = 8 and y = 3.
Q20: Draw the graph of the pair of linear equations x – y + 2 = 0 and 4x – y – 4 = 0. Calculate the area of triangle formed by the lines so drawn and the xaxis.
Sol: To draw the graph of the given pair of equations, we have the table of ordered pairs
Plot the points A(0, 2), B(–2, 0); C(0, –4) and D(1, 0) on the graph paper and join the points to form the lines AB and CD :
From the graph, we find that the points P(2, 4) is common to both the lines AB and CD.
These lines meet the xaxis at B(–2, 0) and D(1, 0).
Thus, the triangle BDP is formed by the lines and the xaxis.
The vertices of this Δ are
B(–2, 0), D(1, 0) and P(2, 4)
Now, the base of ΔBDP = BD
= (BO + OD)
= (2 + 1) units
= 3 units
Altitude of the ΔBDP = PQ
= 4 units
∴ Area of ΔBDP =
= 6 sq. units
Q21: It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?
Sol: Let the time taken by the pipe of larger diameter to fill the pool separately = x hours.
The time taken by the pipe of smaller diameter to fill the pool separately = y hours.
∴ Part of the pool fill by a pipe of larger diameter in 1 hour = 1/x.
Part of fool filled by the pipe of larger diameter in 4 hours = 4/x.
Similarly, Part of the pool filled by the pipe of smaller diameter in 9 hours = 9/y.
∴We have = ...(1)
Since the pool is filled by both the pipes together in 12 hours.
∴ ...(2)
To solve (1) and (2), multiplying (1) by 3 and subtracting (2) from it, we have
Substituting, y = 30 in (2), we have
⇒
∴
⇒ x = 20
⇒ Required time taken by pipe of larger diameter = 20 hours
Required time taken by pipe of smaller diameter = 30 hours
124 videos457 docs77 tests

1. What are the different methods to solve a pair of linear equations in two variables ? 
2. How can we represent a pair of linear equations graphically ? 
3. What is the significance of the coefficients in a pair of linear equations ? 
4. Can a pair of linear equations have no solution ? If so, how ? 
5. How do you verify the solution of a pair of linear equations ? 

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