Binomial Theorem

Question
CBSEENMA11014364

Solve the following inequality:

negative 3 space less or equal than space 4 minus fraction numerator 7 straight x over denominator 2 end fraction less or equal than 18












Solution

negative 3 space less or equal than space 4 space minus space fraction numerator 7 straight x over denominator 2 end fraction less or equal than 18
Now,     negative 3 space less or equal than space 4 minus fraction numerator 7 straight x over denominator 2 end fraction rightwards double arrow negative 6 space less or equal than space 8 space minus 7 straight x space rightwards double arrow space 7 straight x space less or equal than space 8 space plus space 6 space rightwards double arrow space 7 straight x space less or equal than space 14 space rightwards double arrow space straight x space less or equal than space 2 ...(i)

Also,    space space space space space space space 4 minus fraction numerator 7 straight x over denominator 2 end fraction less or equal than 18 space rightwards double arrow space 8 minus 7 straight x less or equal than 36 space rightwards double arrow space minus 7 straight x less or equal than 36 minus 8
rightwards double arrow                space space space space space minus 7 straight x space less or equal than space 28 space rightwards double arrow space fraction numerator negative 7 straight x over denominator negative 7 end fraction greater or equal than fraction numerator 28 over denominator negative 7 end fraction rightwards double arrow space straight x greater or equal than space minus 4                                    ...(ii)
From (i) and (ii), we have negative 4 less or equal than straight x less or equal than 2 which is the required solution of the given inequality.
Hence, the solution is [-4, 2].

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