Question

Solve the following pair of simultaneous equations.

43x + 31y = 98, 31x + 43y = 50

Answer 1

Given equations:

43x + 31y = 98     ...(1)

31x + 43y = 50     ...(2)

Here, multiplying equation (1) by 31:

31(43x) + 31(31y) = 31(98)

1333x + 961y = 3038     ...(3)

And, multiplying equation (2) by 43:

43(31x) + 43(43y) = 43(50)

1333x + 1849y = 2150    ...(4)

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

(1333x + 961y) − (1333x + 1849y) = 3038 − 2150

1333x + 961y − 1333x − 1849y = 888

961y − 1849y = 888

−888y = 888

y = `888/(-888)`

∴ y = −1

Let’s substitute y = −1 into equation (1):

43x + 31(−1) = 98

43x − 31 = 98

43x = 98 + 31

43x = 129

x = `129/43`

∴ x = 3