In this page Cross Multiplication Method Question4 we are going to
see solution for first problem of the worksheet cross multiplication
method.

While solving a pair of linear equations in two unknowns x and y using elimination method,we have to think about how to eliminate any one of the variable in two equations to get value of one unknown and we have to plug it in the second to get the value of other variable. There is another method called cross multiplication which simplifies the procedure.

**Solve the following system of equations using cross multiplication method**

(iv) (5/x) - (4/y) = -2 , (2/x) + (3/y) = 13

**Solution:**

First we have to make the given equations in the form of a₁ x + b₁ y + c₁ = 0,a₂ x + b₂ y + c₂ = 0.

Let 1/x = a and 1/y = b

5 a - 4 b = - 2

2 a + 3 b = 13

5 a - 4 b + 2 = 0 ----- (1)

2 a + 3 b - 13 = 0 ----- (2)

x/(52-6) = y/(4 -(-65)) = 1/(15-(-8))

x/46 = y/(4 +65) = 1/(15+8)

x/(46) = y/(69) = 1/(23)

x/(46) = 1/(23) y/(69) = 1/(23)

x = 46/23 y = 69/23

x = 2 y = 3

Therefore solution is (2,3).

**Verification:**

Now let us apply the answer that we got in the first or second equation to check whether we got correct answer or not.

5 a - 4 b + 2 = 0

5(2) - 4(3) + 2 = 0

10 - 12 + 2 = 0

-12 + 12 = 0 cross multiplication method question4 cross multiplication method question4

- Back to worksheet
- Elimination method
- Elimination method worksheet
- Cross Multiplication Method
- Synthetic division
- L.C.M
- G.C.F
- Factoring
- Types of matrix