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NCERT Solutions for Class 10th Maths:Linear Equations Exercise 3.6




In this page we have NCERT Solutions for Class 10th Maths:Linear Equations for Exercise 3.6. Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) $\frac {1}{2x} + \frac {1}{3y}= 2$
$\frac {1}{3x}+ \frac {1}{2y}= \frac {13}{6}$

(ii) $\frac {2}{\sqrt {x}}+\frac {3}{\sqrt {y}}= 2$
$\frac {4}{\sqrt {x}}- \frac {9}{\sqrt {y}}= -1$

(iii) $\frac {4}{x}+ 3y= 14$
$\frac {3}{x}- 4y= 23$

(iv) $\frac {5}{x-1}+ \frac {1}{y-2} = 2$
$\frac {6}{x-1} - \frac {3}{y-2} = 1$

(v) $\frac {(7x-2y)}{xy}= 5$
$\frac {(8x+ 7y)}{xy}= 15$

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) $\frac {10}{(x+y)}+ \frac {2}{(x-y)}= 4$
$\frac {15}{(x+y)}- \frac {5}{(x-y)}= -2$

(viii) $\frac {1}{(3x+y)}+ \frac {1}{(3x-y)}= \frac {3}{4}$
$\frac {1}{2(3x-y)} - \frac {1}{2(3x-y)} = \frac {-1}{8}$
Answer
(i) $\frac {1}{2x} + \frac {1}{3y}= 2$
$\frac {1}{3x}+ \frac {1}{2y}= \frac {13}{6}$
Let 1/x = a and 1/y = b, then the equations changes as below:
a/2 + b/3 = 2
⇒ 3a + 2b =12 ... (i)
a/3 + b/2 = 13/6
⇒ 2a + 3b =13 ... (ii)
Multiplying (i) by 2 and (ii) by 3 and then subtracting
4b-9b=24 -39
Or b=3
Substituting b=3 in (i)
We get a =2
Now 1/x = 2 and 1/y = 3
Hence, x = 1/2 and y = 1/3
 
(ii) $\frac {2}{\sqrt {x}}+\frac {3}{\sqrt {y}}= 2$
$\frac {4}{\sqrt {x}}- \frac {9}{\sqrt {y}}= -1$
Let 1/√x = a and 1/√y = b, then the equations changes as below:
2a + 3b = 2 ... (i)
4a – 9b = -1 ... (ii)
Multiplying equation (i) by 3, then adding with (ii)
10a = 5
a = 1/2
Putting in equation (i), we get
2 × 1/2 + 3b = 2
b = 1/3
Now a = 1/√x = 1/2
√x = 2
x = 4
and
b = 1/√y = 1/3
√y = 3
y = 9
Hence, x = 4, y = 9

(iii) $\frac {4}{x}+ 3y= 14$
$\frac {3}{x}- 4y= 23$
Putting 1/x = a in the given equations, we get
4a + 3y = 14 -(i)
3a - 4y = 23   -(ii)
Multiplying (i) by 3 and (ii) by 4 and then subtracting
9y +16y=42-92
Or y =-2
Now putting this value in (i), we get
p=5
 
Also, p = 1/x = 5
⇒ x = 1/5
So, x = 1/5 and y = -2 is the solution.
 
(iv) $\frac {5}{x-1}+ \frac {1}{y-2} = 2$
$\frac {6}{x-1} - \frac {3}{y-2} = 1$
Putting 1/(x-1) = a and 1/(y-2) = b in the given equations, we obtain
5a + b = 2 ... (i)
6a – 3b = 1 ... (ii)
Now, by multiplying equation (i) by 3 and adding with equation (ii), we get
21a = 7
⇒ a = 1/3
Putting this value in equation (ii) we get,
 6×1/3 – 3b =1
  b = 1/3
Now,
a = 1/(x-1) = 1/3
 3 = x - 1
 x = 4
Also,
b = 1/(y-2) = 1/3
  3 = y-2
 y = 5
Hence, x = 4 and y = 5 is the solution.
(v) The give equation is
$\frac {(7x-2y)}{xy}= 5$
$\frac {(8x+ 7y)}{xy}= 15$
 Now
(7x-2y)/xy = 5
 ⇒ 7x/xy - 2y/xy = 5
 ⇒ 7/y - 2/x = 5 ...
8x+7y/xy = 15
 ⇒ 8x/xy + 7y/xy = 15
 ⇒ 8/y + 7/x = 15
Putting 1/x = a and 1/y = b, we get,
7b – 2a = 5 ... (i)
8b + 7a = 15 ... (ii)
Multiplying equation (ii) by 7 and multiplying equation (i) by 2 and adding, we get
49b + 16b = 35 + 30
65b = 65
b = 1
Putting the value of b in equation (ii)
8 + 7a = 15
a = 1
Now,
a = 1/x = 1
x = 1
b = 1 = 1/y
y = 1
Hence, x =1 and y = 1 is the solution.
(vi) The given equation is
6x + 3y = 6xy
2x + 4y = 5xy
 
Now,
6x + 3y = 6xy
6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5
Putting 1/x = a and 1/y = b , we get,
6b + 3a = 6   -(i)
2b + 4a =5   -(ii)
Multiplying (ii) by 3 and then subtracting from (i)
3a -12a = 6-15
a=1
Now substituting these values in (i)
6b+3=6
b=1/2
Now, a = 1/x = 1 and b = 1/y = 1/2
Hence, x = 1 and y = 2
(vii) The given equation is
$\frac {10}{(x+y)}+ \frac {2}{(x-y)}= 4$
$\frac {15}{(x+y)}- \frac {5}{(x-y)}= -2$
Putting 1/x+y = a and 1/x-y = b in the given equations, we get:
10a + 2b = 4 -(i)
15a – 5b = -2 -(ii)
Multiplying (i) by 5 and (ii) by 2 and then adding
50a +30 a=20 -4
a=1/5
Substituting the value in (i)
b=1
Now p = 1/(x+y) = 1/5 and q = 1/(x-y) = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
This is again a   pair of simultaneous equation
Adding equation (iii) and (iv), we get
2x = 6
x = 3
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2
(viii) The given equations are
$\frac {1}{(3x+y)}+ \frac {1}{(3x-y)}= \frac {3}{4}$
$\frac {1}{2(3x-y)} - \frac {1}{2(3x-y)} = \frac {-1}{8}$
Putting 1/(3x+y) = a and 1/(3x-y) = b in the given equations, we get
a + b = 3/4 ... (i)
a/2 - b/2 = -1/8
a - b = -1/4 ... (ii)
Adding (i) and (ii), we get
2a = 3/4 - 1/4
2a = 1/2
a = 1/4
Substituting  the value in equation (ii), we get
1/4 - b = -1/4
b= 1/4 + 1/4 = 1/2
Now
p = 1/3x+y = 1/4
3x + y = 4 ... (iii)
q = 1/3x-y = 1/2
3x - y = 2 ... (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1
Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
 
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
 
 
Answer
i) Let the speed of Ritu in still water and the speed of stream be x km/h
and y km/h respectively.
Speed of Ritu while rowing Upstream = (x - y) km/h
Speed of Ritu while rowing Downstream = (x + y) km/h
According to question,
2(x + y) = 20
⇒ x + y = 10 ... (i)
2(x - y) = 4
⇒ x - y = 2 ... (ii)
Adding equation (i) and (ii), we get
Putting this equation in (i), we get
y = 4
Hence, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.
 
ii) Let the number of days taken by a woman and a man be x and y respectively.
Therefore, work done by a woman in 1 day = 1/x
work done by a man in 1 day = 1/y
According to the question,
4(2/x + 5/y) = 1
Or 2/x + 5/y = 1/4
3(3/x + 6/y) = 1
Or 3/x + 6/y = 1/3
Putting 1/x = a and 1/y = b in these equations, we get
2a + 5b = ¼   -(i)
3a + 6b= 1/3 -(ii)
Multiplying (i) by 3 and (ii) by 2 and then subtracting
15b -12 b= ¾ - 2/3
b=1/36
Substituting the values in (i)
a=1/18
 
Now, x=1/a and y =1/b
So, x = 18 and y = 36
Therefore, number of days taken by a woman = 18 and number of days taken by a man = 36

iii) Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
60/u + 240/v = 4
100/u + 200/v = 25/6
Putting 1/u = a and 1/v = b in the equations, we get
60a + 240b = 4 ... (i)
100a + 200b = 25/6
600a + 1200b = 25 ... (ii)
Multiplying equation (i) by 10, and subtracting (ii) from it, we get
1200b = 15
b = 15/200 = 1/80
Putting equation (i), we get
60a + 3 = 4
60a = 1
a = 1/60
Now, a= 1/u = 1/60 and b = 1/v = 1/80
u = 60 and v = 80
Therefore, speed of train = 60 km/h and speed of bus = 80 km/h.
 

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Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

  1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
  2. Mathematics for Class 10 by R D Sharma
  3. Pearson IIT Foundation Maths Class 10
  4. Secondary School Mathematics for Class 10
  5. Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.


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Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20