NCERT Solutions for Class 10 Math Chapter 3 - Pair of Linear Equations in Two Variables

Answer:

Let Aftab’s present age = x years and Aftab’s
daughter’s present age = y years
Algebraical situation
According to 1st condition,
x - 7 = 7 (y - 7)
or x - 7 = 7y - 49
or x - 7y + 42 = 0
According to 2nd condition,
x + 3 = 3 (y + 3)
or x + 3 = 3y + 9
or x - 3y - 6 = 0
∴ Pair of linear equation in two variables are
x - 7y + 42 = 0
and x - 3y - 6 = 0
Graphical situation
x - 7y + 42 = 0
x = 7y - 42 ...(1)
Putting y = 5 in (1), we get
x = 7 x 5 - 42
= 35 - 42 = - 7
Putting y = 6 in (1), we get:
x = 7 x 6 - 42
= 42 - 42 = 0
Putting y = 7 in (i), we get
x = 7 x 7 - 42
= 49 - 42 = 7

Plotting the points (-7, 5), (0, 6), (7, 7) and
drawing a line joining them, we get the graph
of the equation x - 7y + 42 = 0.
x - 3y - 6 = 0
x = 3y + 6 ...(2)
Putting x = 0 in (2), we get
0 = 3 x y + 6
y = -2
Putting x = 3 in (2), we get
3 = 3 x y + 6
y = -1
Putting x = 6 in (2), we get
6 = 3 x y + 6
y = 0

Plotting the points (0, -2), (3, -1), (6, 0) and
drawing a line joining them, we get the graph
of the equation x - 3y - 6 = 0
Hence, we get the graphical representation of
the situation.
We find that both the lines intersect at the
point (42, 12); therefore, x = 42, y = 12 is the
solution of the given system of equations.

Answer:

Let cost of one bat = ${}^{`}\mathrm{x}$
Cost of one ball= ${}^{`}\mathrm{y}$
Algebraical situation
According to 1st condition,
3x + 6y = 3900
According to 2nd condition,
1x + 3y = 1300
∴ Pair of linear equations in two variables are:
3x + 6y = 3900
and x + 3y = 1300
Graphical situation
3x + 6y = 3900
or x + 2y = 1300
or x = 1300 - 2y ....(1)
Putting y = 0 in (1), we get
x = 1300 - 2 × 0
x = 1300
Putting y = 500 in (1), we get
x = 1300 - 2 × 500
= 1300 - 1000 = 300
Putting y = 650 in (1), we get
x = 1300 - 2 × 650
= 1300 - 1300 = 0

Plotting the points A (1300, 0), B (300, 500)
and C (0, 650) drawing a line joining them we get the graph of the equation 3x + 6y =3900
x + 3y = 1300
x = 1300 - 3y ...(2)
Putting y = 0 in (2), we get
x = 1300 - 3 × 0 = 1300
Putting y = 500 in (2), we get:
x = 1300 - 3 × 500
= 1300 - 1500 = -200
Putting y = 300 in (2), we get:
x = 1300 - 3 × 300
= 1300 - 900 = 400

Plotting the points A (1300, 0), E (-200, 500),
F (400, 300) and drawing a line joining them
we get the graph of the equation.
x + 3y = 1300
From the graph it is clear that the two lines
intersect at A (1300, 0).
Hence x = 1300 and y = 0 is the solution of
given pair of linear equations.

Answer:

Let cost of 1 kg apples = ₹x
Cost of 1 kg grapes = ₹y
Algebraical situation
According to 1st condition,
2x + 1y = 160

Answer:

(i) Let the number of boys in the quiz = x
and the number of girls in the quiz = y
Total number of students took part in
quiz = 10
∴ x + y = 10
or x + y - 10 = 0
According to question,
y = x + 4
or x - y + 4 = 0
Now, draw the graph of linear equations
x + y = 10
and x = y - 4
x + y = 10
or x = 10 - y ...(1)
Putting y = 0 in (1), we get
x = 10 - 0 = 10
Putting y = 7 in (1), we get
x = 10 - 7 = 3
Putting y = 10 in (1), we get
x = 10 - 10 = 0

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For intersecting lines
Given linear equation is
2x + 3y - 8 = 0 ....(1)
There are many linear equation in two variables which satisfies the condition of intersecting lines i.e.,

Answer:

For parallel lines

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For coincident lines
Given linear equation is
2x + 3y - 8 = 0 ...(1)

Now, draw the graph of linear equations (1)
and (7).
Consider linear equation (7)
6x + 9y - 24 = 0
or 3 [2x + 3y - 8] = 0
or 2x + 3y - 8 = 0
∴ The points of both are same and line of both equations are same.

Answer:

Consider the pair of linear equation
x - y + 1 = 0
and 3x + 2y - 12 = 0
x - y + 1 = 0
or x = y -1 ...(1)
Putting y = 0 in (1), we get
x = 0 - 1 = -1
Putting y = 3 in (1), we get
x = 3 - 1 = 2
Putting y = 1 in (1), we get
x = 1 - 1 = 0

Answer:

Given pair of linear equations
x + y = 14 ...(1)
and x - y = 4 ...(2)
From (2), x = 4 + y ...(3)
Substitute this value of x in equation (1), we
get
4 + y + y = 14
or 2y = 14 - 4
or 2y = 10

Substitute this value of y in equation (3), we get
x = 4 + 5 = 9
Hence, x = 9 and y = 5

Answer:

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Given pair of linear equations is
3x - y = 3 ...(1)
and 9x - 3y = 9 ...(2)
From (1),
3x - 3 = y
or y = 3x - 3 .....(3)
Substitute this value of y in equation (2), we
get
9x - 3 (3x - 3) = 9
or 9x - 9x + 9 = 9
or 9 = 9
This statement is true for all values of x.
However, we do not get a specific value of x as a solution. Therefore we cannot obtain a specific value of y. This situation arises because both the given equations are same.
Therefore, equations (1) and (2) have infinitely many solutions.

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Given pair of linear equations is:
2x + 3y = 11 ...(1)
and 2x - 4y = -24 ...(2)
From (2),
2x = 4y - 24
or 2x = 2[2y - 12]
or x = 2y - 12 ...(3)
Substitute this value of x in (1), we get
2 (2y - 12) + 3y = 11
or 4y - 24 + 3y = 11
or 7y = 11 + 24
or 7y = 35
or y = 35

Substitute this value of y in (3), we get
x = 2(5) - 12
= 10 - 12 = -2
Now, consider y = mx + 3
Substitute the value of x = -2, y = 5, we get
5 = m (-2) + 3
or 5 - 3 = -2m
or 2 = -2m
or -2m = 2
or m = -1
Hence, x = -2, y = 5 and m = -1

Answer:

Let two numbers be x and y.
According to 1st condition,
x - y = 26 ...(1)
According to 2nd condition,
x = 3y ...(2)
Substitute this value of x in (1), we get
3y - y = 26
or 2y = 26

Substitute this value of y in (2), we get
x = 3 x 13 = 39
Hence, two numbers are 39, 13.

Answer:

Let, required two supplementary angles are
x, y and x > y.
According to 1st condition,
x + y = 180 ...(1)
According to 2nd condition, x = y + 18 ...(2)
Substitute this value of x in (1), we get
y + 18 + y = 180
or 2y = 180 - 18
or 2y = 162

Substitute this value of y in (2), we get
x = 81 + 18 = 99
Hence, required angles are 99$°$, 81$°$.

Answer:

Let cost of one bat = ₹x
and cost of one ball = ₹y
According to 1st condition,
7x + 6y = ₹3800 ...(1)
According to 2nd condition,
3x + 5y = ₹1750 ...(2)
From (1), 7x = 3800 - 6y

Answer:

Let the fixed charges for the taxi = ₹x
and charges for travelling one km = ₹y
According to 1st condition,
x + 10y = 105 ...(1)
According to 2nd condition,
x + 15y = 155 ...(2)
From (1), x = 105 - 10y ...(3)
Substitute the value of x in (2), we get
105 - 10y + 15y = 155
or 5y = 155 - 105
or 5y = 50

Substitute the value of y in (3), we get
x = 105 - 10 × 10
= 105 - 100 = 5
Hence, fixed charges for the taxi = ₹5
and charges for travelling one km = ₹10
Also, charges for travelling 25 km
= ₹(10 × 25) + ₹5
= ₹[250 + 5] = ₹255

Answer:

Answer:

Let Jacob’s present age = x years
and Jacob son’s present age = y years
Five years hence
Jacob’s age = (x + 5) years
His son’s age = (y + 5) years
According to 1st condition,
x + 5 = 3 (y + 5)
or x + 5 = 3y + 15
or x = 3y + 15 - 5
or x = 3y + 10 ...(1)
Five years ago
Jacob’s age = (x - 5) years
His son’s age = (y - 5) years
According to 2nd condition,
x - 5 = 7 (y - 5)
or x - 5 = 7y - 35
or x - 7y = -35 + 5
or x - 7y = -30
Substitute the value of x from (1), we get
3y + 10 - 7y = -30
or -4y = -30 - 10
or -4y = -40
or y = 10
Substitute this value of y in (1), we get
x = 3 (10) + 10

Answer:

Given pair of linear equations is:
x + y = 5 ..(1)
and 2x - 3y = 4 ...(2)

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Given pair of linear equations is
3x - 5y - 4 = 0 ...(1)
and 9x = 2y + 7
or 9x - 2y - 7 = 0 ...(2)
Elimination method
Multiplying (1) by 3, we get
9x - 15y - 12 = 0 ...(3)
Now, (3) - (2) gives
9x - 15y - 12 = 0
9x - 2y - 7 = 0

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Let numerator of fraction = x

Answer:

Let Nuri’s present age = x years
Sonu’s present age = y years
Five years ago
Nuri’s age = (x - 5) years
Sonu’s age = (y - 5) years
According to 1st condition,
x - 5 = 3 (y - 5)
or x - 5 = 3y - 15
or x - 3y + 10 = 0 ...(1)
Ten years later
Nuri’s age = (x + 10) years
Sonu’s age = (y + 10) years
According to 2nd condition,
x + 10 = 2(y + 10)
or x + 10 = 2y + 20
or x - 2y - 10 = 0 ...(2)

or -y = -20
or y = 20
Substitute this value of y in (2), we get
x - 2 (20) - 0 = 0
or x - 40 - 10 = 0
or x = 50

Answer:

Let units digit = x
Tens digit = y
∴ Required Number = 10y + x
According to 1st condition,
x + y = 9 ...(1)
On reversing
Units digit = y
Tens digit = x
∴ Number = 10x + y
According to 2nd condition,
9[10y + x] = 2[10x + y]
or 90y + 9x = 20x + 2y
or 90y + 9x - 20x - 2y =0
or - 11x + 88y =0
or x - 8y =0 ...(2)
Now, (2) - (1) gives

Substitute this value of y in (2), we get
x - 8 × 1 = 0
or x = 8
Hence, required number = 10y + x
= 10 × 1 + 8 = 18

Answer:

Let Meena received x notes of ₹ 50
also, Meena received y notes of ₹ 100
According to 1st condition,
x + y = 25 ...(1)
According to 2nd condition,
50x + 100y = 2000
or x + 2y = 40 ...(2)
Now, (2) - (1) gives
x + 2y = 40
x + y = 25

Substitute this value of y in (1), we get
x + 15 = 25
or x = 25 - 15 = 10
Hence, Meena received 10 notes of ₹50 and
15 notes of ₹100.

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Given pair of linear equations is
x - 3y - 3 = 0
and 3x - 9y - 2= 0

Answer:

Given pair of linear equations:
2x + y = 5
and 3x + 2y = 8
or 2x + y - 5 = 0
and 3x + 2y - 8= 0
Here a₁ = 2, b₁ = 1, c₁ = - 5
a₂ = 3, b₂ = 2, c₂ = - 8

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Given pair of linear equations is
2x + 3y = 7
and (a - b) x + (a + b) y = 3a + b - 2
or 2x + 3y - 7 = 0
and (a - b) x + (a + b) y - (3a + b - 2) = 0

or 9a + 3b - 6 = 7a + 7b
or 2a - 4b - 6 = 0
or a - 2b - 3 = 0
Substitute the value of a from (1) in above, we
get:
9b - 4 - 2b - 3 = 0
or 7b - 7 = 0
or 7b = 7
or b = 1
Substitute this value of b in (1), we get
a = 9 × 1 - 4 = 9 - 4
a = 5
Hence a = 5 and b = 1

Answer:

Given pair of linear equations is:
3x + y = 1
and (2k - 1) x + (k - 1) y = 2k + 1
or 3x + y - 1 = 0
and (2k - 1) x + (k - 1) y - (2k + 1) = 0
Here, a₁ = 3, b₁ = 1, c₁ = - 1
a₂ = (2k - 1), b₂ = k - 1, c₂ = -(2k + 1)
∴ System of equations have no solution, if

Answer:

Given pair of linear equation is
8x + 5y = 9 ...(1)
3x + 2y = 4 ...(2)
Substitution method
From (2), 2y = 4 - 3x

Answer:

Let monthly fixed hostel charges = ₹x
and cost of food per day = ₹y
According to 1st condition
x + 20y = 1000 ...(1)
According to 2nd condition
x + 26y = 1180 ...(2)

Answer:

Let numerator of fraction = x
Denominator of fraction = y

Answer:

Let, number of right questions attempted by
Yash = x
and number of wrong questions attempted by
Yash = y
According to 1st condition,
3x - y = 40
or 3x - y - 40 = 0 ...(1)
According to 2nd condition,
4x - 2y = 50
or 4x - 2y - 50 = 0 ...(2)

Answer:

Let speed of car at place A = x km/hour and speed of car at place B = y km/hour Distance between places A and B = 100 km
In case of 5 hours
Distance covered by car A = 5x km
[∵ Distance = Speed × Time]
Distance covered by car B = 5y km
According to 1st condition,
5x - 5y = 100
or x - y = 20
or x - y - 20 = 0 ...(1)
In case of one hour Distance covered by car A = x km
[∵ Distance = Speed × Time]
Distance covered by car B = y km
According to 2nd condition,
x + y = 100
or x + y - 100 = 0 ...(2)

Answer:

Let length of rectangle = x units
and breadth of rectangle = y units
∴ Area of rectangle = xy sq. units
According to 1st condition,
(x - 5) (y + 3) = xy - 9
or xy + 3x - 5y - 15 = xy - 9

Answer:

Given pair of linear equation is

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Given pair of linear equations is

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Given pair of linear equations is

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5u + v = 2 ...(1)
and 6u - 3v = 1 ...(2)
Multiplying (1) by 3, we get
15u + 3v = 6 ...(3)
Now, (3) + (2) gives

Answer:

Given pair of linear equations is

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Given pair of linear equations is

or y = 2 Hence, x = 1 and y = 2.

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Given pair of linear equations is:

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Given pair of linear equations is

Substitute this value of x in (3), we get:
3 (1) + y = 4
or 3 + y = 4
or y = 4 - 3 = 1
Hence, x = 1 and y = 1

Answer:

Let the speed of Ritu in still water = x km/hour
and the speed of current = y km/hour
∴ Speed in upstream = (x - y) km/hour
and speed in downstream = (x + y) km/hour
Distance covered by Ritu in downstream in
2 hours
= Speed × Time
= (x + y) × 2
According to 1st condition,
2 (x + y) = 20
x + y = 10 ...(1)
Distance covered by Ritu in upstream
in 2 hours
= Speed × Time
= 2 (x - y) km
According to 2nd condition,
2 (x - y) = 4
x - y = 2 ...(2)
Now, (1) + (2) gives

Substitute this value of x in (1), we get
6 + y = 10
y = 10 - 6 = 4
Hence, Ritu's speed in still water
= 6 km/hour
and speed of current = 4 km/hour

Answer:

Let one woman can finish the work = x days
One man can finish the work = y days

Hence, one woman and one man alone can finish work in 18 days and 36 days respectively.

Answer:

Let speed of train = x km/hour
and speed of bus = y km/hour
Total distance = 300 km
Case I:
Time taken by train to cover 60 km

Answer:

Let Ani’s age = x years
and Biju’s age = y years
Dharam’s age = 2x years

Substitute this value of x in (1), we get
19 - y = 3
or - y = 3 - 19
or - y = - 16
or y = 16
Hence Ani’s age = 19 years
Biju’s age = 16 years.

Answer:

Let Capital of one friend = ₹x
and capital of 2nd friend = ₹y
According to 1st condition
x + 100 = 2 (y - 100)
or x + 100 = 2y - 200
or x - 2y = - 200 - 100
or x - 2y = - 300 ...(1)
According to 2nd condition
y + 10 = 6 (x - 10)
or y + 10 = 6x - 60
or 6x - y = 10 + 60
or 6x - y = 70 ...(2)
Multiplying (1) by 6, we get
6x - 12y = - 1800 ...(3)
Now, (3) - (2) gives
6x - 12y = - 1800
6x - y = 70

Answer:

Let speed of train = x km/hour
and time taken by train = y hour
∴ Distance covered by train
= (Speed) × (Time)
= (xy) km
According to 1st condition,
(x + 10) (y - 2)=xy
or xy - 2x + 10y - 20 =xy
or - 2x + 10y - 20 = 0
or x - 5y + 10 =0 ...(1)
According to 2nd condition,
(x - 10) (y + 3)=xy
or xy + 3x - 10y - 30 =xy
or 3x - 10y - 30 = 0 ...(2)
Multiplying (1) by 3, we get
3x - 15y + 30 = 0 ...(3)
Now, (3) - (2) gives
3x - 15y + 30 = 0
3x - 10y - 30 = 0

Answer:

Let number of students in each row = x
and number of rows = y
Total number of students in the class = xy
According to 1st condition,
(x + 3) (y - 1)=xy
or xy - x + 3y - 3 = xy
or - x + 3y - 3=0
or x - 3y + 3 =0 ...(1)
According to 2nd condition,
(x - 3) (y + 2) = xy
or xy + 2x - 3y - 6 = xy
or 2x - 3y - 6 = 0 ...(2)
Now, (2) - (1) gives

Substitute this value of x in (1), we get
9 - 3y + 3 = 0
or - 3y + 12 = 0
or - 3y = - 12

∴ Number of students in each row = 9
and number of rows = 4
Hence, total number of students in the class =
9 × 4 = 36

Answer:

From II and III, we get
3$\angle$B = 2 ($\angle$A + $\angle$B)
or 3$\angle$B = 2$\angle$A + 2$\angle$B
or 3$\angle$B - 2$\angle$B = 2$\angle$A
or $\angle$B = 2$\angle$A ....(1)
From I and II, we get
$\angle$C = 3$\angle$B
or $\angle$C = 3(2$\angle$A) [Using (1)]
or $\angle$C = 6$\angle$A ...(2)
Sum of three angles of a triangle is 180$°$.
$\angle$A + $\angle$B + $\angle$C = 180$°$
or $\angle$A + 2$\angle$A + 6$\angle$A = 180$°$
or 9$\angle$A = 180$°$

Answer:

Given pair of linear equations is:
5x - y = 5 and 3x - y = 3
Consider,

Plotting A (1, 0); D (0, - 3); E (2, 3) on the graph,
we get the equation of line 3x - y = 3.

From the graph, it is clear that given lines
intersect at A (1, 0). Triangle formed by these lines
and y axis are shaded in the graph i.e., $∆$ABD.
Coordinates of the vertices of $∆$ABD are A (1, 0);
B (0, - 5) and D (0, - 3).

Answer:

Given pair of linear equations is
px + qy = p - q ...(1)
and qx - py = p + q ...(2)
Multiplying (1) by q and (2) by p, we get

Answer:

Given pair of linear equations is
ax + by=c
and bx + ay=1 + c
or ax + by - c=0
and bx + ay - (1 + c)=0

Answer:

Given pair of linear equations is:

Answer:

Given pair of linear equations is

Answer:

Given pair of linear equations is
152x - 378y = - 74
and - 378x + 152y = - 604
or 76x - 189y + 37 = 0 .....(1)
and - 189x + 76y + 302 = 0 .....(2)

Answer:

In cyclic quadrilateral ABCD,
$\angle$A = (4y + 20); $\angle$B = 3y - 5;
$\angle$C = 4x and $\angle$D = 7x + 5
Sum of opposite angles of a cyclic quadrilateral
are of measure 180$°$.
∴ $\angle$A + $\angle$C = 180$°$
or 4y + 20 + (4x) = 180$°$
or 4x + 4y = 180$°$ - 20
or 4x + 4y = 160
or x + y = 40
or y = 40 - x ...(1)
and $\angle$B + $\angle$D = 180$°$
or 3y - 5 + (7x + 5) = 180$°$
or 3y - 5 + 7x + 5 = 180$°$
or 7x + 3y = 180$°$ ...(2)
Substitute the value of y from (1) in (2), we get
7x + 3 (40 - x)= 180$°$
or 7x + 120 - 3x= 180$°$
or 4x = 180$°$ - 120$°$
or 4x = 60

Substitute this value of x in (1), we get
y = 40 - 15 = 25
∴ $\angle$A = 4y + 20 = 4 × 25 + 20 = 120$°$
$\angle$B = 3y - 5 = 3 × 25 - 5 = 70$°$
$\angle$C = 4x = 4 × 15 = 60$°$
$\angle$D = 7x + 5 = 7 × 15 + 5 = 110$°$
Hence, $\angle$A = 120$°$, $\angle$B = 70$°$; $\angle$C = 60$°$
and $\angle$D = 110$°$