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Answers
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Solution:
Let us assume 1/x = m and 1/y = n , then the equation will change as follows.
m/2 + n/3 = 2
⇒ 3m+2n-12 = 0…………………….(1)
m/3 + n/2 = 13/6
⇒ 2m+3n-13 = 0……………………….(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/√x + 3/√y = 2
4/√x + 9/√y = -1
Solution:
Substituting 1/√x = m and 1/√y = n in the given equations, we get
2m + 3n = 2 ………………………..(i)
4m – 9n = -1 ………………………(ii)
Multiplying equation (i) by 3, we get
6m + 9n = 6 ………………….…..(iii)
Adding equation (ii) and (iii), we get
10m = 5
m = 1/2…………………………….…(iv)
Now by putting the value of ‘m’ in equation (i), we get
2×1/2 + 3n = 2
3n = 1
n = 1/3
m =1/√x
½ = 1/√x
x = 4
n = 1/√y
1/3 = 1/√y
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 => 4m + 3y – 14 = 0 ……………..…..(1)
3m – 4y = 23 => 3m – 4y – 23 = 0 ……………………….(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) – 3/(y-2) = 1
Solution:
Substituting 1/(x-1) = m and 1/(y-2) = n in the given equations, we get,
5m + n = 2 …………………………(i)
6m – 3n = 1 ……………………….(ii)
Multiplying equation (i) by 3, we get
15m + 3n = 6 …………………….(iii)
Adding (ii) and (iii) we get
21m = 7
m = 1/3
Putting this value in equation (i), we get
5×1/3 + n = 2
n = 2- 5/3 = 1/3
m = 1/ (x-1)
⇒ 1/3 = 1/(x-1)
⇒ x = 4
n = 1/(y-2)
⇒ 1/3 = 1/(y-2)
⇒ y = 5
Hence, x = 4 and y = 5
(v) (7x-2y)/ xy = 5
(8x + 7y)/xy = 15
Solution:
(7x-2y)/ xy = 5
7/y – 2/x = 5…………………………..(i)
(8x + 7y)/xy = 15
8/y + 7/x = 15…………………………(ii)
Substituting 1/x =m in the given equation we get,
– 2m + 7n = 5 => -2 + 7n – 5 = 0 ……..(iii)
7m + 8n = 15 => 7m + 8n – 15 = 0 ……(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 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 = 0…………………….(i)
2x + 4y = 5xy
=> 2/y + 4/x = 5
=> 2n +4m = 5
=> 4m+2n-5 = 0……………………..(ii)
3m + 6n – 6 = 0
4m + 2n – 5 = 0
By cross-multiplication method, we get
m/(-30 –(-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) – 5/(x-y) = -2
Solution:
Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,
10m + 2n = 4 => 10m + 2n – 4 = 0 ………………..…..(i)
15m – 5n = -2 => 15m – 5n + 2 = 0 ……………………..(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 …………………………………………(iii)
n = 1/(x-y) = 1
x-y = 1……………………………………………(iv)
Adding equation (iii) and (iv), we get
2x = 6 => x = 3 …….(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) – 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 …………………………….…… (1)
m/2 – n/2 = -1/8
m – n = -1/4 …………………………..…(2)
Adding (1) and (2), we get
2m = 3/4 – 1/4
2m = 1/2
Putting in (2), we get
1/4 – n = -1/4
n = 1/4 + 1/4 = 1/2
m = 1/(3x+y) = 1/4
3x + y = 4 …………………………………(3)
n = 1/( 3x-y) = 1/2
3x – y = 2 ………………………………(4)
Adding equations (3) and (4), we get
6x = 6
x = 1 ……………………………….(5)
Putting in (3), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1.......
Answer:
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