Pa.Linear Eq Ex 3.6 Solutions
Question - 1 : - Solve the following pairs of equations by reducing them to a pair of linear equations:
Answer - 1 : -
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Solution:┬а
Let us assume 1/x = m and 1/y = n┬а , then the equation will change as follows.
m/2 + n/3 = 2
тЗТ 3m+2n-12 = 0тАжтАжтАжтАжтАжтАжтАжтАж.(1)
m/3 + n/2 = 13/6
тЗТ 2m+3n-13 = 0тАжтАжтАжтАжтАжтАжтАжтАжтАж.(2)
Now, using cross-multiplication method, we get,
m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)
m/10 = n/15 = 1/5
m/10 = 1/5 and n/15 = 1/5
So, m = 2 and n = 3
1/x = 2 and 1/y = 3
x = 1/2 and y = 1/3
(ii) 2/тИЪx + 3/тИЪy = 2
4/тИЪx + 9/тИЪy = -1
Solution:
Substituting 1/тИЪx = m and 1/тИЪy = n in the given equations, we get
2m + 3n = 2 тАжтАжтАжтАжтАжтАжтАжтАжтАж..(i)
4m тАУ 9n = -1 тАжтАжтАжтАжтАжтАжтАжтАжтАж(ii)
Multiplying equation (i) by 3, we get
6m + 9n = 6 тАжтАжтАжтАжтАжтАжтАж.тАж..(iii)
Adding equation (ii) and (iii), we get
10m = 5
m = 1/2тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж.тАж(iv)
Now by putting the value of тАШmтАЩ in equation (i), we get
2├Ч1/2 + 3n = 2
3n = 1
n = 1/3
m =1/тИЪx
┬╜ = 1/тИЪx
x = 4
n = 1/тИЪy
1/3 = 1/тИЪy
y = 9
Hence, x = 4 and y = 9
(iii) 4/x + 3y = 14
3/x -4y = 23
Solution:
Putting in the given equation we get,
So, 4m + 3y = 14┬а ┬а ┬а=> 4m + 3y тАУ 14 = 0┬а тАжтАжтАжтАжтАж..тАж..(1)
3m тАУ 4y = 23┬а ┬а ┬а=> 3m тАУ 4y тАУ 23 = 0┬а тАжтАжтАжтАжтАжтАжтАжтАжтАж.(2)
By cross-multiplication, we get,
m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)
-m/125 = y/50 = -1/ 25
-m/125 = -1/25 and y/50 = -1/25
m = 5 and b = -2
m = 1/x = 5
So , x = 1/5
y = -2
(iv) 5/(x-1) + 1/(y-2) = 2
6/(x-1) тАУ 3/(y-2) = 1
Solution:
Substituting 1/(x-1) = m and 1/(y-2) = n┬а in the given equations, we get,
5m + n = 2 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(i)
6m тАУ 3n = 1 тАжтАжтАжтАжтАжтАжтАжтАжтАж.(ii)
Multiplying equation (i) by 3, we get
15m + 3n = 6 тАжтАжтАжтАжтАжтАжтАжтАж.(iii)
Adding (ii) and (iii) we get
21m = 7
m = 1/3
Putting this value in equation (i), we get
5├Ч1/3 + n = 2
n = 2- 5/3 = 1/3
m = 1/ (x-1)
тЗТ 1/3 = 1/(x-1)
тЗТ x = 4
n = 1/(y-2)
тЗТ 1/3 = 1/(y-2)
тЗТ y = 5
Hence, x = 4 and y = 5
(v) (7x-2y)/ xy = 5
(8x + 7y)/xy = 15
Solution:
(7x-2y)/ xy = 5
7/y тАУ 2/x = 5тАжтАжтАжтАжтАжтАжтАжтАжтАжтАж..(i)
(8x + 7y)/xy = 15
8/y + 7/x = 15тАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(ii)
Substituting 1/x =m in the given equation we get,
тАУ 2m + 7n = 5┬а ┬а ┬а=> -2 + 7n тАУ 5 = 0┬а тАжтАж..(iii)
7m + 8n = 15┬а ┬а ┬а=> 7m + 8n тАУ 15 = 0 тАжтАж(iv)
By cross-multiplication method, we get,
m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)
m/(-65) = n/(-65) = 1/(-65)
m/-65 = 1/-65
m = 1
n/(-65) = 1/(-65)
n = 1
m = 1 and n = 1
m = 1/x = 1┬а ┬а ┬а ┬а n = 1/x = 1
Therefore, x = 1 and y = 1
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
Solution:
6x + 3y = 6xy
6/y + 3/x = 6
Let 1/x = m and 1/y = n
=> 6n +3m = 6
=>3m + 6n-6 = 0тАжтАжтАжтАжтАжтАжтАжтАж.(i)
┬а ┬а
2x + 4y = 5xy
=> 2/y + 4/x = 5
=> 2n +4m = 5
=> 4m+2n-5 = 0тАжтАжтАжтАжтАжтАжтАжтАж..(ii)
3m + 6n тАУ 6 = 0
4m + 2n тАУ 5 = 0
By cross-multiplication method, we get
m/(-30 тАУ(-12)) = n/(-24-(-15)) = 1/(6-24)
m/-18 = n/-9 = 1/-18
m/-18 = 1/-18
m = 1
n/-9 = 1/-18
n = 1/2
m = 1 and n = 1/2
m = 1/x = 1 and n = 1/y = 1/2
x = 1 and y = 2
Hence, x = 1 and y = 2
(vii) 10/(x+y) + 2/(x-y) = 4
15/(x+y) тАУ 5/(x-y) = -2
Solution:
Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,
10m + 2n = 4┬а ┬а ┬а =>┬а 10m + 2n тАУ 4 = 0┬а ┬а ┬а тАжтАжтАжтАжтАжтАж..тАж..(i)
15m тАУ 5n = -2┬а ┬а ┬а=>┬а ┬а15m тАУ 5n + 2 = 0┬а ┬а тАжтАжтАжтАжтАжтАжтАжтАж..(ii)
Using cross-multiplication method, we get,
m/(4-20) = n/(-60-(20)) = 1/(-50 -30)
m/-16 = n/-80 = 1/-80
m/-16 = 1/-80 and n/-80 = 1/-80
m = 1/5 and n = 1
m = 1/(x+y) = 1/5
x+y = 5 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(iii)
n = 1/(x-y) = 1
x-y = 1тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(iv)
Adding equation (iii) and (iv), we get
2x = 6┬а ┬а=> x = 3 тАжтАж.(v)
Putting the value of x = 3 in equation (3), we get
y = 2
Hence, x = 3 and y = 2
(viii) 1/(3x+y) + 1/(3x-y) = 3/4
1/2(3x+y) тАУ 1/2(3x-y) = -1/8
Solution:
Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,
m + n = 3/4 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж.тАжтАж (1)
m/2 тАУ n/2 = -1/8
m тАУ n = -1/4┬а тАжтАжтАжтАжтАжтАжтАжтАжтАжтАж..тАж(2)
Adding (1) and (2), we get
2m = 3/4 тАУ 1/4
2m = 1/2
Putting in (2), we get
1/4 тАУ n = -1/4
n = 1/4 + 1/4 = 1/2
m = 1/(3x+y) = 1/4
3x + y = 4┬а тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(3)
n = 1/( 3x-y) = 1/2
3x тАУ y = 2 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж(4)
Adding equations (3) and (4), we get
6x = 6
x = 1 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж.(5)
Putting in (3), 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:
Answer - 2 : -
(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 the 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.
(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.
Solutions:
(i) Let us consider,
Speed of Ritu in still water = x km/hr
Speed of Stream = y km/hr
Now, speed of Ritu during,
Downstream = x + y km/h
Upstream = x тАУ y km/h
As per the question given,
2(x+y) = 20
Or x + y = 10тАжтАжтАжтАжтАжтАжтАжтАжтАж.(1)
And, 2(x-y) = 4
Or x тАУ y = 2тАжтАжтАжтАжтАжтАжтАжтАжтАж(2)
Adding both the eq.1 and 2, we get,
2x = 12
x = 6
Putting the value of x in eq.1, we get,
y = 4
Therefore,
Speed of Ritu rowing in still water = 6 km/hr
Speed of Stream = 4 km/hr
(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.
Solutions:
(ii) Let us consider,
Number of days taken by women to finish the work = x
Number of days taken by men to finish the work = y
Work done by women in one day = 1/x
Work done by women in one day = 1/y
As per the question given,
4(2/x + 5/y) = 1
(2/x + 5/y) = 1/4
And, 3(3/x + 6/y) = 1
(3/x + 6/y) = 1/3
Now, put 1/x=m and 1/y=n, we get,
2m + 5n = 1/4 => 8m + 20n = 1тАжтАжтАжтАжтАжтАжтАж(1)
3m + 6n =1/3 => 9m + 18n = 1тАжтАжтАжтАжтАжтАжтАж.(2)
Now, by cross multiplication method, we get here,
m/(20-18) = n/(9-8) = 1/ (180-144)
m/2 = n/1 = 1/36
m/2 = 1/36
m = 1/18
m = 1/x = 1/18
or x = 18
n = 1/y = 1/36
y = 36
Therefore,
Number of days taken by women to finish the work = 18
Number of days taken by men to finish the work = 36.
(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 the 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.
Solutions:
(iii) Let us consider,
Speed of the train = x km/h
Speed of the bus = y km/h
According to the given question,
60/x + 240/y = 4 тАжтАжтАжтАжтАжтАжтАж(1)
100/x + 200/y = 25/6 тАжтАжтАжтАжтАж.(2)
Put 1/x=m and 1/y=n, in the above two equations;
60m + 240n = 4тАжтАжтАжтАжтАжтАжтАжтАж..(3)
100m + 200n = 25/6
600m + 1200n = 25 тАжтАжтАжтАжтАжтАжтАж.(4)
Multiply eq.3 by 10, to get,
600m + 2400n = 40 тАжтАжтАжтАжтАжтАжтАжтАж(5)
Now, subtract eq.4 from 5, to get,
1200n = 15
n = 15/1200 = 1/80
Substitute the value of n in eq. 3, to get,
60m + 3 = 4
m = 1/60
m = 1/x = 1/60
x = 60
And y = 1/n
y = 80
Therefore,
Speed of the train = 60 km/h
Speed of the bus = 80 km/h