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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
1. Two linear equations in the same two variables are called a pair of linear equations
in two variables. The most general form of a pair of linear equations is
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
where, a1, a2, b1, b2, c1, c2 are real numbers, such that
a12 + b12 ≠ 0, a 22 + b22 ≠ 0
2. A pair of linear equations in two variables can be represented, and solved, by the:
(i) graphical method
(ii) algebraic method
3. Graphical Method :
The graph of a pair of linear equations in two variables is represented by two lines.
(i) If the lines intersect at a point, then that point gives the unique solution of the
two equations. In this case, the pair of equations is consistent.
(ii) If the lines coincide, then there are infinitely many solutions — each point on the
line being a solution. In this case, the pair of equations is dependent (consistent).
(iii) If the lines are parallel, then the pair of equations has no solution. In this case,
the pair of equations is inconsistent.
4. Algebraic Methods : We have discussed the following methods for finding the solution(s) of a pair of linear equations :
(i) Substitution Method
(ii) Elimination Method
(iii) Cross-multiplication Method
5. If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
then the following situations can arise :
(i)
a1 b1
≠
a 2 b2
In this case, the pair of linear equations is consistent.
(ii)
a1 b1 c1
=
≠
a 2 b2 c 2
In this case, the pair of linear equations is inconsistent.
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(iii)
a1 b1 c1
=
=
a 2 b2 c 2
In this case, the pair of linear equation is dependent and consistent.
6. There are several situations which can be mathematically represented by two
equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.
EXERCISE 1
1. Aftab tells his daughter, “Seven years ago, I was seven times as old as
you were then. Also, three years from now, I shall be three times as old as
you will be.” Represent this situation algebraically and graphically.
Answer: Let us assume that seven years ago Aftab’s age was x years and his
daughter’s age was y.
So, as per the question
x = 7y
⇒ x − 7 y = 0 --------------------------------- (i)
The equation gives us the following table:
X
7
14
21
28
35
42
49
Y
1
2
3
4
5
6
7
Now, three years from now means 10 years from 7 years back
Aftab age will be x + 10
Daughter’s age will be y +
As
⇒
⇒
⇒
10
per question x + 10 = 3( y + 10)
x + 10 = 3 y + 30
x = 3 y + 20
x − 3 y − 20 = 0
---------------------------- (ii)
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The equation gives us following table:
X
23
26
29
32
35
38
41
Y
1
2
3
4
5
6
7
60
Father's Age
50
⇒ x − 3 y − 20 = 0
40
30
20
⇒ x − 7y = 0
10
0
1
2
3
4
5
6
7
Daughter's Age
2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later,
she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Answer: Let us assume number of bats = x and that of balls = y. As per the question we get following equations:
3 x + 6 y = 3900
⇒ 3x = 3900 − 6 y
⇒ x = 1300 − 2 y ---------------------- (i)
x + 2 y = 1300
⇒ x = 1300 − 2 y ---------------------- (ii)
As both equations are similar so there coincident lines in the graph, meaning there
can be infinite number of solutions to this problem.
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3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs
160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300.
Represent the situation algebraically and geometrically.
Answer: Let us assume that cost of apple = x and that of grapes = y
As per the question we get following equations:
2 x + y = 160
⇒ y = 160 − 2 x --------------------------- (i)
The equation gives us following table:
X
1
2
3
4
5
Y
158
156
154
152
150
4 x + 2 y = 300
⇒ 2 y = 300 − 4 x
⇒ y = 150 − 2 x --------------------------- (ii)
The equation gives us following table:
X
1
2
3
4
5
Y
148
146
144
142
140
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160
155
Grapes
150
y=160-2x
145
y=150-2x
140
135
130
1
2
3
4
5
Apples
Since we get parallel lines in this graph so there will be no solution as equations are
inconsistent.
EXERCISE 2
1. Form the pair of linear equations in the following problems, and find their
solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of
girls is 4 more than the number of boys, find the number of boys and girls
who took part in the quiz.
Answer: Let us assume number of girls to be x and that of boys to be y. Then we
get following equations:
x = y + 4 -------------------- (i)
The equation will give following table:
X
4
5
6
7
8
9
Y
0
1
2
3
4
5
x + y = 10
⇒ x = 10 − y --------------- (ii)
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The equation will give following table:
X
10
9
8
7
6
5
Y
0
1
2
3
4
5
12
10
Girls
8
x=y+4
6
x=10-y
4
2
0
1
2
3
4
5
6
Boys
As both lines intersect at 7, so Number of Girls= 7 and that of Boys=3
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens
together cost Rs 46. Find the cost of one pencil and that of one pen.
5 x + 7 y = 50
⇒ 5 x = 50 − 7 y ------------------ (i)
Answer:
The equation will give following table:
y
X
-15
-10
-5
0
5
10
15
8
9
10
31
24
17
10
3
-4
-11
-1.2
-2.6
-4
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7 x + 5 y = 46
⇒ 7 x = 46 − 5 y ----------------- (ii)
The equation gives following table
y
X
-9
-2
5
12
19
13
8
3
-2
-7
35
30
25
20
Pen
15
5x=50-7y
10
7x=46-5y
5
0
-5
1
3
5
7
9
11 13
15 17
19 21
23 25
27
29 31
-10
-15
Pencil
Both lines are intersecting at 3, so x= 3
Putting the value of x in either of the equations we can get the value of y, which is
equal to 5. Also, 5 is the only value of y which gives similar values for x in both the
tables.
So cost of one pencil = Rs. 3 and one pen = 5
5 x + 7 y = 50
⇒ 5 × 3 + 7 × 5 = 15 + 35 = 50
Check:
2. On comparing the ratios
a1 b1
c
, and 1 , find out if the lines representing
a 2 b2
c2
the following pairs of linear equations intersect at a point, are parallel or coincident:
(i)
5x – 4y + 8 = 0
7x + 6y – 9 = 0
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a1 5 x 5
=
=
a2 7 x 7
b1 − 4 y
4
=
= −
b2
6y
6
c1
8
8
=
= −
c2 − 9
9
As it is clear that
a1 b1
≠
, so lines will be intersecting.
a 2 b2
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
a1
9 1
=
=
a 2 18 2
b1 3 1
= =
b2 6 2
c1 12 1
=
=
c 2 24 2
It is clear that
a1 b1 c1
=
=
, so lines are coincidental.
a 2 b2 c 2
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
a1 6 3
= =
a2 2 1
b1 − 3 3
=
=
b2 − 1 1
c1 10
=
c2
9
It is clear that
a1 b1 c1
=
≠
, so lines will be parallel
a 2 b2 c 2
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3. On comparing the ratios
a1 b1
c
, and 1 find out whether the following pair
a 2 b2
c2
of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
a1 3
=
a2 2
b1 − 3
3
=
= −
b2
2
2
Here,
a1 b1
≠
, so the pair of linear equations is consistent.
a 2 b2
(ii) 2x – 3y = 8 ; 4x – 6y = 9
a1 2 1
= =
a2 4 2
b1 − 6 1
=
=
b2 − 3 2
c1 − 8 8
=
=
c2 − 9 9
Here,
a1 b1 c1
=
≠
, so the pair of linear equations is inconsistent.
a 2 b2 c 2
3
5
x + y = 7 , 9x – 10y = 14
2
3
a1
3
1
=
=
a2 2 × 9 6
b1
5
1
=
= −
b2 3 × − 10
6
(iii)
Here,
a1 b1
≠
, so the pair of linear equations is consistent.
a 2 b2
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(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a1
5
1
=
= −
a 2 − 10
2
b1
3
1
=
= −
b2 − 6
2
c1
11
1
=
= −
c 2 − 22
2
Here,
(v)
a1
a2
b1
b2
c1
c2
a1 b1 c1
=
=
, so the pair of linear equations is dependent and consistent.
a 2 b2 c 2
4
x + 2 y = 8 ; 2x + 3y = 12
3
4
2
=
=
3× 2 3
2
=
3
−8 2
=
=
− 12 3
In this case,
a1 b1 c1
=
=
, so the pair of linear equation is dependent and consista 2 b2 c 2
ent.
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
a1 1
=
a2 2
b1 1
=
b2 2
c1
−5 1
=
=
c 2 − 10 2
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Both equations will give the same table as follows:
y
x
x'
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
-1
-2
-3
-4
10
9
8
7
6
5
4
3
2
1
0
-1
-2
-3
-4
12
10
8
y
6
4
x+y=5
2
2x+2y=10
0
-2
1
2
3 4
5 6
7
8 9 10 11 12 13 14 15
-4
-6
x
As the lines are coincident so there can be infinitely many solutions.
In this case,
a1 b1 c1
=
=
, so the pair of linear equation is dependent and consista 2 b2 c 2
ent.
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(ii) x – y = 8, 3x – 3y = 16
a1 1
=
a2 3
b1 1
=
b2 3
c1
−8 1
=
=
c 2 − 16 2
a1 b1 c1
=
≠
Here,
, so the pair of linear equations is inconsistent.
a 2 b2 c 2
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a1 1
=
a2 2
b1 1
=
b2 2
c1 − 6 3
=
=
c2 − 4 2
a1 b1 c1
=
≠
Here,
, so the pair of linear equations is inconsistent.
a 2 b2 c 2
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a1 2 1
= =
a2 4 2
b1 1
=
b2 2
c1 2
=
c2 5
a1 b1 c1
=
≠
Here,
, so the pair of linear equations is inconsistent.
a 2 b2 c 2
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5. Half the perimeter of a rectangular garden, whose length is 4 m more
than its width, is 36 m. Find the dimensions of the garden.
Answer: Let us assume Length = x and breadth =y
Then x=y+4
Perimeter = 2(x+y)
⇒ Pr imeter = x + y = x + x + 4 = 36
2
⇒ 2 x + 4 = 36
⇒ 2 x = 36 − 4 = 32
⇒ x = 32 ÷ 2 = 16
Hence, y=16-4=12
So, Length = 16 m and Breadth = 12 m
6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation
in two variables such that the geometrical representation of the pair so
formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines
Answer: (i) As you know that if
a1 b1
≠
, then the lines will be intersecting, so let
a 2 b2
us put a number for a2 and b2 in such a way which satisfies this condition.
Next possible equation can be 3 x + 2 y − 7 = 0
(ii) As you know that if
a1 b1 c1
=
≠
, then the lines will be parallel, so the possible
a 2 b2 c 2
equation can be as follows:
4x + 6 y − 8 = 0
(iii) As you know that if
a1 b1 c1
=
=
, then the lines will be coincidental, so the posa 2 b2 c 2
sible equation can be as follows:
4 x + 6 y − 16 = 0
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EXERCISE 3
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14, x-y=4
x + y = 14
⇒ x = 14 − y
Answer:
Substituting this value of x in the second equation we get
14 − y − y = 4
⇒ 14 − 2 y = 4
⇒ 2 y = 12
⇒ y= 6
⇒ x= 8
(ii) s – t = 3,
s t
+ = 6
3 2
Answer: s = 3+t
Substituting the value of s in the second equation we get
3+ t t
+ = 6
3
2
6 + 2t + 3t
⇒
= 6
6
⇒ 6 + 5t = 36
⇒ 5t = 30
⇒ t= 6
Hence, s = 3+6=9
(iii) 3x – y = 3, 9x – 3y = 9
Answer: y=3x-3
Putting this value of y in the second equation we get
9x-3(3x-3)=9
⇒ 9x − 9x + 9 = 9
It is clear that this equation will have infinitely many solutions. It is important to
note that here
a1 b1 c1
=
=
a 2 b2 c 2
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(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
Answer: 2x+3y=13 (after multiplying the equation by 10)
3y=13-2x
Or, y =
13 − 2 x
3
Putting this value of y in the second equation we get
 13 − 2 x 
4 x + 5
 = 23 (equation 2 is also multiplied like equation 1)
3 

⇒ 12 x + 65 − 10 x = 69
⇒ 2 x + 65 = 69
⇒ 2x = 4
⇒ x= 2
13 − 4
⇒ y=
= 3
3
(v) 2 x + 3 y = 0 , 3 x − 8 y = 0
Answer:
2x = − 3y ⇒ x = −
3
y
2
Putting this value of x in equation 2 we get
3× −
3
2
y−
8y = 0 ⇒ −
3
2
y−
8y = 0
− 3 y − 16 y
2
= 0⇒
− 3y − 4 y
2
= 0
⇒ − 3y − 4 y = 0 ⇒ − 3y = 4 y
Since coefficients of y are different on both LHS and RHS, so only possible value of
y=0
Hence, x=0
2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for
which y = mx + 3.
Answer: 2 x = 11 − 3 y
Substituting in equation 2 we get
11 − 3 y − 4 y = − 24
⇒ 11 − 7 y = − 24
⇒ 7 y = 11 + 24 = 35
⇒ y= 5
⇒ 2 x = 11 − 3 × 5 = − 4
⇒ x= −2
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Now putting values of x and y in equation 3 we get
5 = m× − 2 + 3
⇒ − 2m = 5 − 3 = 2
⇒ m= −1
3. Form the pair of linear equations for the following problems and find their
solution by substitution method.
(i) The difference between two numbers is 26 and one number is three
times the other. Find them.
Answer: Let us assume one of the numbers = x and another number = y
Then as per question x − y = 26 ---------------- (i)
x = 3 y -------------------------------------------- (ii)
Substituting the value of x in equation (i)
3 y − y = 26
⇒ 2 y = 26
⇒ y = 13
⇒ x = 13 × 3 = 39
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
x + y = 180° ------------- (i)
x − y = 18° -------------- (ii)
⇒ x = y + 18°
Substituting in equation (i) we get
y + y + 18° = 180°
⇒ 2 y = 180° − 18° = 162°
⇒ y = 81°
⇒ x = 81° + 18° = 99°
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later,
she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each
ball.
7 x + 6 y = 3800 ----------- (i)
3 x + 5 y = 1750 --------------------- (ii)
⇒ 3x = 1750 − 5 y
1750 − 5 y
⇒ x=
3
Answer:
Substituting in equation (i) we get
7(1750 − 5 y )
+ 6 y = 3800
3
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⇒
⇒
⇒
⇒
12250 − 35 y + 18 y = 11400
12250 − 17 y = 11400
17 y = 12250 − 11400 = 850
y = 50
1750 − 250
⇒ x=
= 500
3
Hence, Cost of Bat = Rs. 500
Cost of Ball = Rs. 50
(iv) The taxi charges in a city consist of a fixed charge together with the
charge for the distance covered. For a distance of 10 km, the charge paid is
Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the
fixed charges and the charge per km? How much does a person have to pay
for travelling a distance of 25 km?
Answer: Let us assume the fixed charge = x
And variable charge = y
As per question:
x + 10 y = 105 ------------------- (i)
x + 15 y = 155 ------------------- (ii)
⇒ x = 155 − 15 y
Substituting in equation (i) we get
155 − 15 y + 10 y = 105 ⇒ 155 − 5 y = 105 ⇒ 5 y = 155 − 105 = 50 ⇒ y = 10
⇒ x = 155 − 15 × 10 = 5
So for 25 km the fair = 5 + 25 × 10 = 255
(v) A fraction becomes
9
, if 2 is added to both the numerator and the de11
nominator. If, 3 is added to both the numerator and the denominator it becomes
5
. Find the fraction.
6
Answer: Let us assume the numerator = x and denominator = y
As per question:
x+ 2 9
=
y + 2 11
⇒ 11x + 22 = 9 y + 18
⇒ 9 y − 11x = 4 --------------- (i)
x+ 3 5
=
y+ 3 6
⇒ 6 x + 18 = 5 y + 15
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⇒ 5 y − 6 x = 3 --------------- (ii)
⇒ 5 y = 6x + 3
6x + 3
⇒ y=
5
Substituting in equation (i) we get
54 x + 27
− 11x = 4
5
⇒ 54 x + 27 − 55 x = 20
⇒ 27 − x = 20
⇒ x= 7
6× 7 + 3
⇒ y=
= 9
5
Hence, required fraction =
7
9
(vi) Five years hence, the age of Jacob will be three times that of his son.
Five years ago, Jacob’s age was seven times that of his son. What are their
present ages?
Answer: Let us assume five years ago son’s age = x and Jacob’s age = y
Then y = 7 x ------------- (i)
Son’s age five years from now = x+10 and Jacob’s age will be = y+10
So, y + 10 = 3( x + 10) = 3 x + 30
⇒ y = 3 x + 20 -------------- (ii)
Substituting in equation (i)
3 x + 20 = 7 x
⇒ 4 x = 20
⇒ x= 5
⇒ y = 35
Present age of Jacob = 40 years
Present age of son = 10 years
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EXERCISE 4
1. Solve the following pair of linear equations by the elimination method
and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
Answer: Multiplying equation (i) by 2 and subtracting equation (ii) from this as follows:
2 x + 2 y = 10
2x − 3y = 4
0 + 5y = 6
6
⇒ y = , substituting in equation (i)
5
6
x+ = 5
5
25 − 6 19
⇒ x=
=
5
5
(ii) 3x + 4y = 10 and 2x – 2y = 2
Answer: Multiplying equation (i) by 2 and equation (ii) by 3 and subtracting latter
from former as follows:
6 x + 8 y = 20
6x − 6 y = 6
0 + 14 y = 14
⇒ y = 1 , substituting in equation (i)
3 x + 4 = 10
⇒ 3x = 6
⇒ x= 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Answer: Multiplying equation (i) by 3 and subtracting equation (ii) as follows:
9 x − 15 y − 4 = 0
9x − 2 y − 7 = 0
0 − 10 y + 3 = 0
⇒ 10 y = 3 ⇒ y =
3 , substituting in equation (i)
10
3
− 4= 0
2
6x − 3 − 8
⇒
= 0
2
3x −
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⇒ 6 x = 11
11
⇒ x=
6
(iv)
x 2y
y
+
= − 1 and x − = 3
2 3
3
Answer: Simplify equation (i)
x 2y
+
= −1
2 3
3x + 4 y
⇒
= −1
6
⇒ 3x + 4 y = − 6 ----------------- (iii)
Simplify equation (ii)
y
= 3
3
⇒ 3x − y = 9 ------------------- (iv)
x−
Subtracting equation (iv) from (iii) as follows:
3x + 4 y = − 6
3x − y = 9
0 + 5 y = − 15
⇒ y = − 3 , substituting in equation (iii)
3 x − 12 = − 6
⇒ 3x = 6
⇒ x= 2
2. Form the pair of linear equations in the following problems, and find their
solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a
fraction reduces to 1. It becomes
1
, if we only add 1 to the denominator.
2
What is the fraction?
x+ 1
=1
y+ 1
⇒ x+ 1= y+ 1
⇒ x = y ------------ (i)
Answer:
x
1
=
y+ 1 2
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⇒ 2 x = y + 1 ------------ (ii)
Substituting value of x from equation (i) in equation (ii)
2y = y + 1
⇒ y=1
⇒ x= 1
(ii) The sum of the digits of a two-digit number is 9. Also, nine times this
number is twice the number obtained by reversing the order of the digits.
Find the number.
Answer: Let two digits of the number are x and y.
x+ y = 9
⇒ x = 9 − y ---------------- (i)
Then First number = 10 x + y ( x is at 10s place and y is at unit’s place)
Reversing the number we get a number which can be written as 10 y + x
As per question:
9(10 x + y ) = 2(10 y + x) ⇒ 90 x + 9 y = 20 y + 2 x ⇒ 90(9 − y ) + 9 y = 20 y + 2(9 − y )
⇒ 810 − 90 y + 9 y = 20 y + 18 − 2 y ⇒ 810 − 81y = 18 y + 18 ⇒ 99 y = 810 − 18 = 792
⇒ y = 792 ÷ 99 = 8
⇒ x = 9− 8= 1
Smaller number = 18 and larger number = 81
18 × 9 = 81 × 2 = 162
(iii) Meena went to a bank to withdraw Rs 2000. She asked the cashier to
give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how
many notes of Rs 50 and Rs 100 she received.
x + y = 25
⇒ x = 25 − y ------------ (i)
50 x + 100 y = 2000
⇒ 50(25 − y ) + 100 y = 2000
⇒ 1250 − 50 y + 100 y = 2000
⇒ 50 y = 2000 − 1250 = 750
⇒ y = 15 Number of hundred rupees notes
⇒ x = 10 Number of fifty rupees notes.
Answer:
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