Question

Solve the following equation by the trial-and-error method: z − 3 = 2z − 5

Answer 1

z − 3 = 2z − 5
 We try several values of z till we get the L.H.S. equal to the R.H.S.

z	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	1 − 3 = −2	2 ×× 1 − 5 = −3	No
2	2 − 3 = −1	2 ×× 2 − 5 = −1	Yes

∴ z = 2