Question

Find the largest value of $x$ for which $2(x - 1) \leqslant 9 - x $ and $ x \in W $.

Answer 1

Hint: The given question is related to linear inequalities. The linear inequalities are the type of mathematical expression in which instead of the equal to sign ( $ = $ ) we just define the nature of the variable using the greater than sign or the lesser than sign $ ( < {\text{ or > )}} $ . The question will be solved by first simplifying the inequality by rearrangement and then finding the value which is maximum.

Complete step-by-step answer:
The given inequality will be solved by rearranging the terms by taking variables on one side and constants on the other side of the inequality,
 $ \Rightarrow 2(x - 1) \leqslant 9 - x $
 $ \Rightarrow 2x - 2 \leqslant 9 - x $
 $ \Rightarrow 2x + x - 2 \leqslant 9 - x $
 $ \Rightarrow 3x \leqslant 9 + 2 $
Solving it further we get,
 $ \Rightarrow 3x \leqslant 11 $
 $ \Rightarrow x \leqslant \dfrac{{11}}{3} $
 $ \Rightarrow x \leqslant 3.67 $
The given inequality has been written to say that the variable belongs to the set of whole numbers and therefore the decimal values will be ignored as well as the negative values. The value of the variable will be less than $ 3.67 $ , but since the variable belongs only to the whole number set,
We get values of $ x $ as,
 $ x = \{ 0,1,2,3\} $
The question asks about the largest value of the variable so the largest value of the set is $ 4 $ , which is the correct answer.

Note: The rules of rearrangement in inequalities are similar to the rules of rearrangement in normal equations that we solve. The only caveat in the inequalities is the rule that when a negative number is multiplied to both the sides of the inequality the sign of the inequality reverses i.e. greater than sign becomes less than sign and vice-versa.