The method of cross-multiplication is widely discussed in this exercise. Solving the systems of equations by the method of cross-multiplication is clearly explained in the RD Sharma Solutions Class 10. It’s a valuable resource developed by experts at BYJUâ€™S for the purpose of clarifying doubts and help students focus on their weaker areas. Further, students can also make use of the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.4 PDF given below.

## RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.4 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.4

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

**1. x + 2y + 1 = 0**

** 2x – 3y â€“ 12 = 0 **

**Solution: **

The given system of equations is

**x + 2y + 1 = 0**

**2x – 3y â€“ 12 = 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

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

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

** 2x + y + 10 = 0**

**Solution: **

The given system of equations is

**3x + 2y + 25 = 0**

**2x + y + 10 = 0**

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

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

**3. 2x + y = 35, 3x + 4y = 65**

**Solution: **

The given system of equations can be written as

**2x + y â€“ 35 = 0**

**3x + 4y â€“ 65 = 0**

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 15 and y = 5.

**4.** **2x â€“ y = 6, x â€“ y = 2**

**Solution: **

The given system of equations can be written as

**2x â€“ y â€“ 6 = 0**

**x â€“ y â€“ 2 = 0**

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 4 and y = 2.

**5. (x + y)/ xy = 2**

** (x – y)/ xy = 6 **

**Solution: **

The given system of equations is

(x + y)/ xy = 2 â‡’ 1/y + 1/x = 2â€¦â€¦.. (i)

(x – y)/ xy = 6 â‡’ 1/y â€“ 1/x = 6â€¦â€¦… (ii)

Let 1/x = u and 1/y = v, so the equation becomes

u + y = 2â€¦.. (iii)

u â€“ y = 6â€¦â€¦(iv)

**For cross multiplication we use, **

**Comparing the above two equations (iii) & (iv) with the general form, we get**

Hence, the solution for the given system of equations is x = -1/2 and y = 1/4.

**6. ax + by = a-b**

** bx â€“ ay = a+b**

**Solution: **

The given system of equations can be written as

**ax + by â€“ (a-b) = 0 **

**bx â€“ ay â€“ (a+b) = 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 1 and y = -1.

**7. x + ay = b**

** ax + by = c**

**Solution: **

The given system of equations can be written as

x + ay â€“ b = 0

ax + by â€“ c = 0

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = (b^{2} + ac)/(a^{2} + b^{2})

and y = (-c^{2} + ab)/(a^{2} + b^{2}).

**8. ax + by = a ^{2}**

** bx + ay = b ^{2}**

**Solution: **

The given system of equations can be written as

**ax + by â€“ (a ^{2}) = 0 **

**bx + ay â€“ (b ^{2}) = 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = (a^{2} + ab + b^{2})/(a + b)

and y = -ab / (a+ b).

**9. 5/(x + y) â€“ 2/(x -y) = -1**

** 15/(x + y) + 7/(x – y) = 10**

**Solution:**

Letâ€™s substitute 1/(x + y) = u and 1/(x – y) = v, so the given equations becomes

5u – 2v = -1

15u + 7v = 10

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 3 and y = 2.

**10. 2/x + 3/y = 13**

** 5/x â€“ 4/y = -2**

**Solution: **

Let 1/x = u and 1/y = v, so the equation becomes

2u + 3y = 13 â‡’ 2u + 3y â€“ 13 = 0

5u â€“ 4y = -2 â‡’ 5u â€“ 4y + 2 = 0

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 1/2 and y = 1/3.

**11. 57/(x + y) + 6/(x – y) = 5**

** 38/(x + y) + 21/(x – y) = 9**

**Solution: **

Letâ€™s substitute 1/(x + y) = u and 1/(x – y) = v, so the given equations becomes

57u + 6v = 5 â‡’ 57u + 6v â€“ 5 = 0

38u + 21v = 9 â‡’ 38u + 21v â€“ 9 = 0

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = 11 and y = 8.

**12. xa â€“ yb = 2**

** ax â€“ by = a ^{2}-b^{2}**

**Solution: **

The given system of equations can be written as

xa â€“ yb â€“ 2 **= 0 **

**ax â€“ by â€“ (a ^{2}-b^{2}) = 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = a and y = b.

**13. x/a + y/b = a + b**

** x/a ^{2 }+ y/b^{2 }= 2**

**Solution: **

The given system of equations can be written as

x/a + y/b â€“ (a + b)** = 0 **

x/a^{2 }+ y/b^{2 }â€“ 2 **= 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = a^{2} and y = b^{2}.

**14. x/a = y/b **

** ax + by = a ^{2} + b^{2} **

**Solution: **

The given system of equations can be written as

x/a – y/b** = 0 **

ax + by â€“ (a^{2} + b^{2}) **= 0 **

**For cross multiplication we use, **

**Comparing the above two equations with the general form, we get**

Hence, the solution for the given system of equations is x = a and y = b.