RD Sharma Solutions Chapter 3 Pair of Linear Equation in Two Variables Exercise 3.3 Class 10 Maths
Chapter Name  RD Sharma Chapter 3 Pair of Linear Equation in Two Variables 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 3.3 Solutions
Solve the following systems of equations:
1. 11x + 15y + 23 = 0
7x – 2y – 20 = 0
Solution
The given system of equation is
11x + 15y + 23 = 0 …(i)
7x – 2y – 20 = 0 …(ii)
From (ii), we get
2y = 7x – 20
⇒ y = (7x – 20)/2
Substituting y = (7x – 20)/2 in (i) we get
11x + 15[(7x – 20)/2] + 23 = 0
⇒ 11x + (105x – 300)/2 + 23 = 0
⇒ (22x + 105x – 300 + 46)/2 = 0
⇒ 127x – 254 = 0
⇒ 127x = 254
⇒ x = 254/127 = 2
Putting x = 2 in y = (7x – 20)/2 we get
⇒ y = (7×2 – 20)/2
= (14 – 20)/2
= 6/2 = 3
Hence, the solution of the given system of equation is x = 2, y = 3.
2. 3x – 7y + 10 = 0
y – 2x – 3 = 0
Solution
The given system of equation is
3x – 7y + 10 = 0 …(i)
y – 2x – 3= 0 …(ii)
From (ii), we get
y = 2x + 3
Substituting y = 2x + 3 in (i) we get
3x – 7(2x + 3) + 10 = 0
⇒ 3x + 14x – 21 + 10 = 0
⇒  11x = 11
⇒ x = 11/11 = 1
Putting x =  1 in y = 2x + 3, we get
⇒ y = 2×(1) + 3
= 2 + 3
= 1
⇒ y = 1
Hence, the solution of the given system of equations is x = 1, y = 1.
3. 0.4x + 0.3x = 1.7
0.7x + 0.2y = 0.8
Solution
The given system of equation is
0.4x + 0.3y = 1.7 …(i)
0.7x  0.2y = 0.8 …(ii)
Multiplying both sides of (i) and (ii), by 10, we get
4x + 3y = 17 …..(iii)
7x – 2y = 8 ….(iv)
From (iv), we get
7x = 8 + 2y
⇒ x = (8 + 2y)/7
Substituting x = (8 +2y)/7 in (iii), we get
4[(8+2y)/7] + 3y = 17
⇒ (32 + 8y)7 + 3y = 17
⇒ 32 + 29y = 17×7
⇒ 29y = 87
⇒ y = 87/29 = 3
Putting y = 3 in x = (8+2y)/7, we get
x = (8 + 2×3)/7
= (8 + 6)/7
= 14/7 = 2
Hence, the solution of the given system of equation is x = 2, y = 3.
4. x/2 +y = 0.8
Solution
x/2 + y = 0.8
And 7/(x+y/2) = 10
∴ x + 2y = 1.6 and (7×2)/(2x + y) = 10
x + 2y = 1.6 and 7 = 10x + 5y
Multiply first equation by 10
10x + 20y = 16 and 10x + 5y = 7
Subtracting the two equations
15y = 9
y = 9/15 = 3/5
x = 1.6 – 2(3/5) = 1.6 – 6/5 = 2/5
Solution is (2/5, 3/5)
5. 7(y+3 ) – 2(x + 3) = 14
4(y – 2) + 3(x – 3) = 2
Solution
The given system of equation is
7(y+3 ) – 2(x + 3) = 14 …(i)
4(y – 2) + 3(x – 3) = 2 …(ii)
From (i), we get
7x + 21 – 2x – 4 = 14
⇒ 7y = 14 + 4 – 21 + 2x
⇒ y = (2x – 3)/7
From (ii), we get
4y – 8 + 3x – 9 = 2
⇒ 4y + 3x – 17 – 2 = 0
⇒ 4y + 3x – 19 = 0 …(iii)
Substituting y = (2x – 3)/7 in (iii), we get
4[(2x – 3)/7] + 3x – 19 = 0
⇒ (8x – 12)/7 + 3x – 19 = 0
⇒ 8x – 12 + 21x – 133 = 0
⇒ 29x – 145 = 0
⇒ 29x = 145
⇒ x = 145/29 = 5
Putting x = 5 in y = (2x – 3)/7, we get
y = (2×5 – 3)/ 7
= 7/7 = 1
⇒ y = 1
Hence, the solution of the given system of equations is x = 5, y = 1.
6. x/7 + y/3 = 5
x/2 – y/9 = 6
Sol: The given system of equation is
x/7 + y/3 = 5 …(i)
x/2 – y/9 = 6 …(ii)
From (i), we get
(3x +7y)/21 = 5
⇒ 3x + 7y = 105
⇒ 3x = 105 – 7y
⇒ x = (105 – 7y)/3
From (ii), we get
(9x – 2y)/18 = 6
⇒ 9x – 2y = 108 ...(iii)
Substituting x = (105 – 7y)/3 in (iii), we get
9[(105 – 7y)/3] – 2y = 108
⇒ (948 – 63y)/3 – 2y = 108
⇒ 945 – 63y – 6y = 108×3
⇒ 945 – 69y = 324
⇒ 945 – 324 = 69y
⇒ 69y = 621
⇒ y = 621/69 = 9
Putting y = 9 in x = (1105 – 7y)/3, we get
x = (105 – 7×9)/3 = (105 – 63)/3
⇒ x = 42/3 = 14
Hence, the solution of the given system of equations is x = 14, y = 9.
7. x/3 + y/4 = 11
5x/6 – y/3 = 7
Solution
The given system of equations is
x/3 + y/4 = 11 …(i)
5x/6 – y/3 = 7 …(ii)
From (i) , we get
(4x + 3y)/12 = 11
⇒ 4x + 3y = 132 …(iii)
From (ii), we get
(5x + 2y)/6 = 7
⇒ 5x – 2y =  42 …(iv)
Let us eliminate y from the given equations. The coefficients of y in the equations (iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.
Multiplying (iii) by 2 and (iv) by 3, we get
8x + 6y = 264 …(v)
15x – 6x = 126 …(vi)
Adding (v) and (vi), we get
8x + 15x = 264 – 126
⇒ 23x = 138
⇒ x = 138/23 = 6
Substituting x = 6 in (iii) , we get
4 ×6 + 3y = 132
⇒ 3y = 132 – 24
⇒ 3y = 108
⇒ y = 108/3 = 36
Hence, the solution of the given system of equations is x = 6, y= 36.
8. 4u + 3y = 8
6u – 4y = 5
Solution
Taking 1/x = u, then given equations become
4u + 3y =8 ...(i)
6u – 4y =  5 …(ii)
From (i) we get
4u = 8 – 3y
⇒ u = (8 – 3y)/4
Substitutig u = (8 – 3y) /4 in (ii), we get
From (ii), we get
6[(8 – 3y)/4] – 4y = 5
⇒ 3(8 – 3y)/2 – 4y =  5
⇒ (24 – 9y)/2 – 4y =  5
⇒ (24 – 9y – 8y)/2 = 5
⇒ 24 – 17y = 10
⇒ 17y = 10 – 24
⇒ 17y = 34
⇒ y = 34/17 = 2
Putting y = 2, in u = (8 – 3y)/4, we get
u = (8 – 3×2)/4 = (86)/4= 2/4= ½
Hence, x = 1/u = 2
So, the solution of the given system of equation is x = 2, y = 2.
9. x + y/2 = 4
x/3 + 2y = 5
Solution
The given system of equation is
x + y/2 = 4 …(i)
x/3 + 2y = 5 …(ii)
From (i) , we get
(2x + y)/2 = 4
2x + y = 8
y = 8 – 2x
From (ii), we get
x + 6y = 15 …(iii)
Substituting y = 8 – 2x in (iii), we get
x + 6 (8 – 2x) = 15
⇒ x + 48 – 12x = 15
⇒ 11x = 33
⇒ x = 33/11 = 3
Putting x = 3, in y = 8 – 2x, we get
y = 8 – 2×3
= 8 – 6 = 2
⇒ y = 2
Hence, solution of the given system of equation is x= 3 , y = 2.
10. x + 2y = 3/2
2x + y = 3/2
Solution
The given system of equation is
x + 2y = 3/2 …(i)
2x + y = 3/2 …(ii)
Let us eliminate y from the given equations. The Coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.
Multiplying (i) by 1 and (ii) by 2, we get
x + 2y = 3/2 …(iii)
4x + 2y = 3 …(iv)
Subtracting (iii) from (iv), we get
4x – x + 2y – 2y = 3 – 3/2
⇒ 3x = (6 – 3)/2
⇒ 3x = 3/2
⇒ x =3/2×3
⇒ x = ½
Putting x = ½, in equation (iv) , we get
4×1/2 + 2y = 3
⇒ 2 + 2y = 3
⇒ 2y = 3 2
⇒ y = ½
Hence, solution of the given system of equation is x = 1/2, y = 1/2.
11. √2x + √3y = 0
√3x  √8 y = 0
Solution
√2x + √3y = 0 ...(i)
√3x  √8y = 0 …(ii)
From equation (i), we obtain:
x = −√3y/√2 …..(iii)
Substituting this value in equation (ii), we obtain:
√3(−√3y/√2)  √8y = 0
3y/√2  2√2y = 0
y(3/√2  2√2) = 0
y = 0
Substituting the value of y in equation (iii), we obtain:
x = 0
∴ x = 0, y = 0
12. 3x – (y+7)/11 + 2 = 10
2y + (x+ 11)/7 = 10
Solution
The given systems of equation is
3x – (y+7)/11 + 2 = 10 …(i)
2y + (x + 11)/7 = 10 …(ii)
From (i), we get
(33x – y – 7 + 22)/11 = 10
⇒ 33x – y + 15 = 10×11
⇒ 33x + 15 – 110 = y
⇒ y = 33x – 95
From (ii) we get
(14y + x + 11)/7 = 109
⇒ 14y + x + 11 = 10×7
⇒ 14y +x + 11 = 70
⇒ 14y + x = 70 – 11
⇒ 14y + x = 59 …(iii)
Substituting y = 33x – 95 in (iii), we get
14(33x – 95) +x = 59
⇒ 462x – 1330 + x = 59
⇒ 463x = 59 + 1330
⇒ 463x = 1389
⇒ x = 1389/463 = 3
Putting x = 3, in y = 33x – 95 , we get
y = 33×3 – 95
⇒ y = 99 – 95 = 4
⇒ y = 4
Hence, solution of the given system of equation is x = 3, y = 4.
13. 2x – 3/y = 9
3x + 7/y = 2 y≠0
Solution
The given system of equation is
2x – 3/y = 9 …(i)
3x + 7/y = 2, y ≠ 0 …(ii)
Taking 1 /y = u, the given equations becomes
2x – 3u = 9 …(iii)
3x + 7u = 2 …(iv)
From (iii), we get
2x = 9 + 3u
⇒ x = (9 + 3u)/2
Substituting x = (9+3u)/2 in (iv) , we get
3(9 +3u)/2 + 7u = 2
⇒ (27+ 9u + 14u)/2 = 2
⇒ 27 + 23u = 2×2
⇒ 23u = 4 – 27
⇒ u = 23/23 =  1
Hence, y = 1/u = 1/1 =  1
Putting u = 1 in x = (9+ 3u)/2, we get
x = (9 + 3×1)/2 = (9 3)/2 = 6/2 = 3
⇒ x = 3
Hence, solution of the given system of equation is x = 3, y = 1.
14. 0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Sol : The given systems of equations is
0.5x + 0.7y = 0.74 …(i)
0.3x + 0.5y = 0.5 …(ii)
Multiplying (i) and (ii) by 100, we get
50x + 70y = 74 …(iii)
30x + 50y = 50 …(iv)
From (iii), we get
50x = 74 – 70y
x = (74 – 70y)/50
Substituting x = (74 – 70y)/50 in equation (iv), we get
30(74 – 70y)/50 + 50y = 50
⇒ 3(74 – 70y)/5 + 50y = 50
⇒ (222 – 210y)/5 + 50y = 50
⇒ 222 – 210y + 250y = 250
⇒ 40y = 250 – 222
⇒ 40y = 28
⇒ y = 28/40 = 14/20 = 7/10 = 0.7
Putting y = 0.7 in x = (74 – 70y)/50 , we get
x = (74 – 70×0.7)/50
= (74 – 49 )/50
= 25/50 = 1/2 = 0.5
Hence, solution of the given system of equation is x = 0.5, y = 0.7
15. 1/7x + 1/6y = 3
1/2x – 1/3y = 5
Solution
1/7x + 1/6y = 3 …(i)
1/2x – 1/3y = 5 …(ii)
Multiplying (ii) by 1/2, we get
1/7x + 1/6y = 3
⇒ (4+7)/28x = (6+5)/2
⇒ 11/28x = 11/2
⇒ x = (11×2)/(28×11) = 1/14
When x = 1/14, we get
1/7(1/14) + 1/6y = 3 (Using (i)]
⇒ 2 + 1/6y = 3
⇒ 1/6y = 3 – 2 = 1
⇒ y = 1/6
Thus , the solution of given equation is x = 1/14 and y = 1/6.
16. 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Solution
Let 1/x = u and 1/y = v, the given equations become
u/2 + v/3 = 2
⇒ (3u + 2v)/6 = 2
⇒ 3u + 2v = 12 …(i)
And, u/3 + v/2 = 13/6
⇒ (2u + 3v)/6 = 13/6
⇒ v = 6/2 = 3
Hence, x = 1/u = 1/2 and y = 1/v = 1/3
So, the solution of the given system equation is x = 1/2, y = 1/3.
17. (x +y)/xy = 2
(x – y)/xy = 6
Solution
The given system of equation is
(x + y)/xy = 2
⇒ x/xy + y/xy = 2
⇒ 1/y + 1/x = 2 ….(i)
And, (x – y)/xy = 6
⇒ x/xy – y/xy = 6
⇒ 1/y – 1/x = 6 ….(ii)
Taking 1/y = v and 1/x = u, the above equations become
v + u = 2 ….(iii)
v – u = 6 ….(iv)
Adding equation (iii) and equation (iv) , we get
v+ u + v – u = 2 + 6
⇒ 2v = 8
⇒ v = 8/2 = 4
Putting v = 4 in equation (iii), we get
4 + u = 2
⇒ u = 2 – 4 =  2
Hence, x = 1/u = 1/  2 = 1/2 and y = 1/v = 1/4
So, the solution of the given system of equation is x = 1/2, y = 1/4
18. 15/u + 2/v = 17
Solution
Let 1/u = x and 1/v = y, then the given system of equations become
15x + 2y = 17 ….(i)
x + y = 36/5 …(ii)
From (i), we get
2y = 17 – 5x
⇒ y = (17 – 15x)/2
Substituting y = (17 – 15x)/2 in equation (ii), we get
x + (17 – 15x)/2 = 36/5
⇒ (2x + 17 – 15x)/2 = 36/5
⇒ (13x + 17)/2 = 36/5
⇒ 5(13x + 17) = 36×2
⇒ 65x + 85 = 72
⇒ 65x = 72 – 85
⇒ 65x = 13
⇒ x = 13/65 = 1/5
Putting x = 1/5 in equation (ii), we get
1/5 + y = 36/5
⇒ y = 36/5 – 1/5
⇒ y = 35/5 = 7
Hence u = 1/x = 5 and v = 1/y = 1/7 .
So, the solution of the given system of equation is u = 5, v = 1/7.
19. 3/x – 1/y =  9
2/x + 3/y = 5
Solution
Let 1/x = u and 1/y = v, Then, the given system of equations becomes
3u – v = 9 ...(i)
2u + 3v = 5 …(ii)
Multiplying equation (i) by 3 and equation (ii) by 1, we get
9u – 3v = 27 …(iii)
2u + 3v = 5 …(iv)
Adding equation (i) and equation (ii), we get
9u + 2u – 3v + 3v = 27 + 5
⇒ 11u = 22
⇒ u = 22/11 = 2
Putting u = 2 in equation (iv), we get
2×(2) + 3v = 5
⇒ 4 + 3v = 5
⇒ 3v = 5 + 4
⇒ v = 9/3 = 3
Hence, x = 1/u = 1/2 = 1/2 and y = 1/v = 1/3.
So, the solution of the given system of equation is x = 1/2, y = 1/3.
20. 2/x + 5/y = 1
60/x + 40y = 19, x ≠0, y ≠0
Solution
Taking 1/x = u and 1/y = v, the given becomes
2u + 5v = 1 …(i)
60u + 40u = 19 …(ii)
Let us eliminate ‘u’ from equation (i) and (ii), multiplying equation (i) by 60 and equation (ii) by 2, we get
120u +300v = 60 …(iii)
120u + 80v = 38 …(iv)
Subtracting (iv) from (iii), we get
300v – 80v = 60 – 38
⇒ 220v = 22
⇒ v = 22/220 = 1/10
Putting v = 1/10 in equation (i), we get
2u + 5×1/10 = 1
⇒ 2u + 1/2 = 1
⇒ 2u = 1 – 1/2
⇒ 2u = (21)/2 = 1/2
⇒ 2u = 1/2
⇒ u = 1/4
Hence, x = 1/u = 4 and y = 1/v = 10
So, the solution of the given system of equation is x = 4 , y = 10.
21. 1/5x + 1/6x = 12
1/3x – 3/7y = 8, x ≠0, y≠0
Solution
Taking 1/x = u and 1/y = v, the given equations become
u/5 + v/6 = 12
⇒ (6u + 5v)/30 = 12
⇒ 6u + 5v = 360 …(i)
And u/3 – 3v/7 = 8
⇒ (7u + 9v)/21 = 8
⇒ 7u – 9v = 168 …(ii)
Let us eliminate ‘v’ from equation (i) and (ii), multiplying equation (i) by 9 and equation (ii) by 5, we get
54u + 45v = 3240 …(iii)
35u – 45v = 840 …(iv)
Adding equation (i) adding equation (ii), we get
54u + 35u = 3240 + 840
⇒ 89u = 4080
⇒ u = 4080/89
Putting u = 4080/89 in equation (i) , we get
6× 4080/89 + 5v = 360
⇒ 24480/89 + 5v = 360
⇒ 5v = 360 – 24480/89
⇒ 5v = (32040 24480)/89
⇒ 5v = 7560/89
⇒ v = 7560/5×89
⇒ v = 1512/89
Hence, x = 1/u = 89/4080 and y = 1/v = 89/1512
So, the solution of the given system of equation is x = 89/4080, y = 89/1512
22. 2/x + 3/y = 9/xy
4/x + 9/y = 21/xy , where x ≠0, y ≠ 0
Solution
The system of given equation is
2/x + 3/y = 9/xy …(i)
4/x + 9/y = 21/xy , where x ≠0, y = 0 …(ii)
Multiplying equation (i) adding equation (ii) by xy, we get
2y + 3x = 9 …(iii)
4y + 9x = 21 …(iv)
From (iii), we get
3x = 9 – 2y
⇒ x = (9 – 2y)/3
Substituting x = (9 – 2y)/3 in equation (iv) , we get
4x + 9(92y)/3 = 21
⇒ 4y + 3(9 2y) = 21
⇒ 4y + 27 – 6y = 21
⇒ 2y = 21 – 27
⇒ 2y = 6
⇒ y =3
Putting y = 3 in x = (92y)/3 , we get
x = (9 – 2×3)/3
= (96)/3
= 3/3 = 1
Hence, solution of the system of equation is x = 1, y = 3
23. 6/(x+y) = 7/(x – y) + 3
1/2(x+ y) = 1/3(x – y), where x + y ≠ 0 and x – y ≠ 0
Solution
Let 1/x + y = u and 1/x – y = v. Then the given system of equation becomes
6u = 7v + 3
⇒ 6u – 7v = 3 …(i)
And, u/2 =v/3
⇒ 3u = 2v
⇒ 3u – 2v = 0 …(ii)
Multiplying equation (ii) by 2, and equation (i) by 1, we get
6u – 7v = 3 …(iii)
6u – 4v = 0 …(iv)
Subtracting equation (iv) from equation (iii), we get
 7 + 4v = 3
⇒ 3v = 3
⇒ v = 1
Putting v = 1 in equation (ii), we get
3u – 2×(1) = 0
⇒ 3u + 2 = 0
⇒ 3u = 2
⇒ u = 2/3
Now, u = 2/3
⇒ 1/(x +2) = 2/3
⇒ x + y = 3/2 …(v)
And, v =  1
⇒ 1/(x –y) = 1
⇒ x –y = 1 …(vi)
Adding equation (v) and equation (vi), we get
2x = 3/2 – 1
⇒ 2x = (32)/2
⇒ 2x = 5/2
⇒ x =5/4
Putting x = 5/4 in equation (vi) , we get
5/4 – y = 1
⇒ 5/4 + 1 = y
⇒ (5+4)/4 = y
⇒ 1/4 = y
⇒ y = 1/4
Hence, solution of the system of equation is x = 5/4, y = 1/4.
24. xy/(x+y) = 6/5
xy/(yx) = 6
Solution
25. 22/(x+y) + 15/(x y) = 5
55/(x +y) + 45/(x y) = 14
Solution
26. 5/(x+y)  2/(xy) = 1
15/(x+ y) + 7/(x y) = 10
Solution
Hence, solution of the given system of equation is x = 3, y = 2 .
27. 3/(x+y) + 2/(xy) = 2
9/(x+y)  4/(xy) = 1
Solution
28. 1/2(x+2y) + 5/3(3x2y) = 3/2
5/4(x+2y)  3/5(3x2y) = 61/60
Solution
29. 5/(x+1)  2/(y1) = 1/2
10/(x+1) + 2/(y1) = 5/2, where x ≠ 1 and y ≠ 1
Solution
30. x+y = 5xy
3x + 2y = 13xy
Solution
31. x+ y = 2xy
(xy)/xy = 6 x ≠ 0, y ≠ 0
Solution
32. 2(3u  v) = 5uv
2(u + 3v) = 5uv
Solution
33. 2/(3x + 2y) + 3/(3x  2y) = 17/5
5/(3x + 2y) + 1/(3x 2y) = 2
Solution
34. 4/x + 3y = 14
3/x 4y = 23
Solution
35. 99x + 101y = 499
101x + 99y = 501
Solution
36. 23x  29y = 98
29x  23y = 110
Solution
The given system of equation is
23x  29y = 98 ...(i)
29x  23y = 110 ...(ii)
37. x  y +z = 4
x  2y 2z = 9
2x + y + 3z = 1
Solution
We have,
x  y +z = 4 ...(i)
x  2y 2z = 9 ...(ii)
2x + y + 3z = 1 ...(iii)
From equation (i) , we get
z = 4  x + y
⇒ z = x + y + 4
Subtracting the value of z in equation (ii), we get
38. x  y +z = 4
x + y + z = 2
2x + y  3z = 0
Solution
We have,
x  y +z = 4 ...(i)
x + y + z = 2 ...(ii)
2x + y  3z = 0 ...(iii)
From equation (i), we get
39. 44/(x+y) + 30/(xy) = 4
55/(x+y) + 40/(xy) = 13
Solution
Let 1/(x+ y) = u and 1/(x y) = v.
Then, the system of the given equations becomes
40. 4/x + 15y = 21
3/x + 4y = 5
Solution
41. 2(1/x) + 3(1/y) = 13
5(1/x)  4(1/y) = 2
Solution
42. 5/(x1) + 1/(y2) = 2
6/(x1)  3/(y2) = 1
Solution
5/(x1) + 1/(y2) = 2 ...(i)
6/(x1)  3/(y2) = 1 ...(ii)
Let 1/(x  1) = u, 1/(y2) = v
So, our equations become
5u+ v = 2 ...(iii)
6u  3v = 1 ...(iv)
From equation (iii),
5u + v = 2
v = 2  5u ...(v)
Putting value of v in (iv)
6u  3v = 1
6u  3(2  5u) = 1
6u  6 + 15u = 1
21u = 1+6
u = 7/21 = 1/3
Putting u = 1/3 in equation (v)
v = 2  5u
= 2  5×1/3
= 2  5/3
= (65)/3 = 1/3
Now, 1/(x1) = u
1/(x 1) = 1/3
⇒ x 1 = 3
⇒ x = 4
and , 1/(y  2) = v
⇒ 1/(y2) = 1/3
⇒ y  2 = 3
⇒ y = 5
Hence, solution of the given system of equation is x = 4, y = 5.
43. 10/(x+y) + 2/(xy) = 4
15/(x+y)  5/(xy) = 2
Solution
44. 1/(3x+y)+ 1/(3xy) = 3/4
1/2(3x+y)  1/2(3x y) = 1/8
Solution
45. 2/√x + 3/√y = 2
4/√x  9/√y = 1
Solution
46. (7x  2y)/xy = 5
(8x + 7y)/xy = 15
Solution
47. 152x  378y = 74
378x + 152y = 604
Solution
152x  378y = 74 ...(i)
378x + 152y = 604 ...(ii)
Adding the equations (i) and (ii), we obtain:
226x  226y = 678
⇒ x + y = 3 ...(iii)
Subtracting the equation(ii) from equation (i) , we obtain
530x  530y = 530
⇒ x  y = 1 ...(iv)
Adding equations (iii) and (iv), we obtain:
2x = 4
x = 2
Substituting the value of x in equation (iii) , we obtain:
y = 1