Question

Solve the following pair of simultaneous equations.

`8/x + 5/y = -3, 6/x + 25/y = 2`

Answer 1

Given equations:

`8/x + 5/y = -3`    ...(1)

`6/x + 25/y = 2`     ...(2)

Let, `a = 1/x, b = 1/y,`

Then the equations will become:

8a + 5b = −3    ...(1)

6a + 25b = 2     ...(2)

Here, multiplying equation (1) by 5:

5(8a) + 5(5b) = 5(−3)

40a + 25b = −15     ...(3)

Now, subtracting equation (2) from equation (3):

(40a + 25b) − (6a + 25b) = −15 − 2

40a + 25b − 6a − 25b = −17

40a − 6a = −17

34a = −17

a = `(-17)/34`

∴ a = `-1/2`

Let’s substitute `a = -1/2` into equation (1):

`8(-1/2) + 5b = -3`

`(-8)/2 + 5b = -3`

−4 + 5b = −3

5b = − 3 + 4

5b = 1

∴ b = `1/5` 

Thus, finding x and y from `a = 1/x, b = 1/y,`

`a = 1/x = -1/2`

∴ x = −2

`b = 1/y = 1/5`

∴ y = 5