Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math Exercise 3.4

Exercise 3.4

1. Solve each of the following systems of equations by the method of cross-multiplication:-

1. x + 2y + 1 = 0
2x - 3y - 12 = 0

Solution

The given system of equation is
Hence, the solution of the given system of equations is  x = 3, y = -2.

2. 3x + 2y + 25 = 0
2x + y + 10 = 0

Solution

The given system of equation is
Hence, x = 5, y = -20 is the solution of the given system of equations.

3. 2x + y - 35 = 0
3x + 4y - 65 = 0

Solution

The given system of equations may be written as 
Hence,  x = 15, y = 5 is the solution of the given system of equations.

4. 2x - y - 6 = 0
x - y - 2 = 0 

Solution

The given system of equations may be written as
 
Hence , x = 4, y = 2 is the solution of the given system of the equations.

5. (x+y)/xy = 2
(x-y)/xy = 6

Solution

The given system of equations is

6. ax + by = a - b
bx - ay = a + b

Solution

The given system of equations is
Hence, x = 1, y = 1 is the solution of the given system of the equations.

7. x + ay - b = 0
ax - by - c = 0

Solution

The given system of equations may be written as
8. ax + by = a2
bx + ay = b2

Solution

The system of the given equations may be written as

9. 5/(x+y) - 2/(x-y) = -1
15/(x+y) + 7/(x-y) = 10
where x ≠ 0 and y ≠ 0

Solution


10. 2/x + 3/y = 13
5/x - 4/y = -2
where x ≠ 0 and y ≠ 0

Solution

The given system of equation is

11. 57/(x+y) + 6/(x-y) = 5
38/(x+y) + 21/(x-y) = 9

Solution

Now adding eq.(3) and (4) we get x = 11
And after substituting the value of x in eq. (4) we get y = 8
Hence we get the value of x = 11 and y = 8 .

12. x/a + y/b = 2
ax-by = a2 - b2

Solution

The system of the given equations may be written as .

13. x/a + y/b = a+b
x/a2 + y/b2 = 2

Solution


Hence we get the value = a2 and y = b2

14. x/a = y/b
ax + by = a2 + b2 

Solution



15. 2ax + 3by = a + 2b
3ax + 2by = 2a + b

Solution

 The given system of equations is


16. 5ax + 6by = 28
3ax + 4by - 18 = 0

Solution

The given system of equation is

17. (a+2b)x + (2a-b)y = 2
(a-2b)x + (2a+b)y = 3

Solution

The given system of equations may be written as

18. x[a-b + ab/(a-b)] = y[a+b - ab/(a+b)]
x+y = 2a2

Solution

The given system of equation is


19. bx + cy = a + b
ax[1/(a-b) - 1/(a+b)] + cy[1/(b-a) - 1/(b+a) = 2a/(a+b)

Solution

The given system of equation is




20. (a-b)x + (a+b)y = 2a2 - 2a2
(a+b) (x+y) = 4ab

Solution

The given system of equation is




21. a2x + a2y = c2
b2x + a2y = d2

Solution 

The given system of equations may be written as
22. 57/(x+y) + 6/(x-y) = 5
38/(x+y) + 21/(x-y) = 9

Solution



23. 2(ax-by) + a + 4b = 0
2(bx+ay) + b - 4a = 0

Solution

The given system of equation may be written as
 
24. 6(ax + by) = 3a + 2b
6(bx - ay) = 3b - 2a

Solution



25. a2/x - b2/y = 0
a2b/x + c2a/y = a + b, x,y ≠ 0

Solution



26. mx - my = m2 + n2
x + y = 2m

Solution

The given system of equations may be written as

27. ax/b - by/a = a+b
ax - by = 2ab

Solution

The given system of equation may be written as


28. bx/a + ay/b - (a2 + b2) = 0
x + y - 2ab = 0

Solution

The given system of equation may be written as
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