Question

Solve the following pair of simultaneous equations.

`4x + 3/y = 18, 3x + 2/y = 13`

Answer 1

Given equations:

`4x + 3/y = 18`     ...(1)

`3x + 2/y = 13`     ...(2)

Let, z = `1/y`

Then the equations will become:

4x + 3z = 18     ...(1)

3x + 2z = 13     ...(2)

Here, multiplying equation (1) by 3:

3(4x) + 3(3z) = 3(18)

12x + 9z = 54     ...(3)

And, multiplying equation (2) by 4:

4(3x) + 4(2z) = 4(13)

12x + 8z = 52     ...(4)

Now, subtracting equation (3) from equation (4):

(12x + 9z) − (12x + 8z) = 54 − 52

12x + 9z − 12x − 8z = 2

9z − 8z = 2

∴ z = 2

Let’s substitute z = 2 into equation (2):

3x + 2(2) = 13

3x + 4 = 13

3x = 13 − 4

3x = 9

x = `9/3`

∴ x = 3

Thus, finding y from z = `1/y`,

`1/y` = 2

∴ y = `1/2`