Solve the System of Linear Equations Using Cross-Multiplication Method

Video Explanation

Watch the video below to understand the complete solution step by step:

Solution

Question: Solve the following system of equations using cross-multiplication method:

5ax + 6by = 28  …… (1)

3ax + 4by = 18  …… (2)

Step 1: Write Equations in Standard Form

5ax + 6by − 28 = 0  …… (1)

3ax + 4by − 18 = 0  …… (2)

Step 2: Compare with ax + by + c = 0

From equation (1): a₁ = 5a, b₁ = 6b, c₁ = −28

From equation (2): a₂ = 3a, b₂ = 4b, c₂ = −18

Step 3: Apply Cross-Multiplication Formula

x / (b₁c₂ − b₂c₁) = y / (a₂c₁ − a₁c₂) = 1 / (a₁b₂ − a₂b₁)

Substitute values:

x / [ 6b(−18) − 4b(−28) ] = y / [ 3a(−28) − 5a(−18) ] = 1 / [ 5a(4b) − 3a(6b) ]

x / ( −108b + 112b ) = y / ( −84a + 90a ) = 1 / ( 20ab − 18ab )

x / 4b = y / 6a = 1 / 2ab

Step 4: Find the Values of x and y

x / 4b = 1 / 2ab

⇒ x = 2 / a

y / 6a = 1 / 2ab

⇒ y = 3 / b

Final Answer

∴ The solution of the given system of equations is:

x = 2/a and y = 3/b

Conclusion

Thus, by using the cross-multiplication method, we find that the solution of the given system of linear equations is (2/a, 3/b).