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