Question

Solve the inequality $ 3x + 2 > - 16,2x - 3 \leqslant 11 $
 $ A)( - 6,7] $
 $ B)[ - 6,7) $
 $ C)( - 6,7) $
 $ D)[ - 6,7] $

Answer 1

Hint: First, we need to know about the concept of open brackets and closed brackets.
Where the open brackets represented as $ () $ and the range of the values from strictly less than $ < $ and the closed brackets represented as $ [] $ and the range of the values are less than or equal to $ \leqslant $.
We need to solve the given linear equation further to approach the solution.

Complete step by step answer:
Since given that we have two inequalities, $ 3x + 2 > - 16,2x - 3 \leqslant 11 $
Now take the first inequality and solve further,
Let $ 3x + 2 > - 16 $ then subtracted the values from $ 2 $ then we get, $ 3x + 2 - 2 > - 16 - 2 \Rightarrow 3x > - 18 $
Now by the division operation, we get, $ 3x > - 18 \Rightarrow x > - 6 $
Thus, we get the first inequality range as $ - 6 < x $
Now take the second inequality $ 2x - 3 \leqslant 11 $ then add both values by the number $ 3 $ then we get $ 2x - 3 + 3 \leqslant 11 + 3 \Rightarrow 2x \leqslant 14 $
By the division operation we get; $ 2x \leqslant 14 \Rightarrow x \leqslant 7 $
Thus, we get the second inequality as $ x \leqslant 7 $
Compare both the range values we get, $ - 6 < x \leqslant 7 $
Since $ < $ represents the open bracket and $ \leqslant $ represents the closed bracket, then we get $ x \in ( - 6,7] $ is the range of the given two inequalities.

So, the correct answer is “Option A”.

Note: The range $ A)( - 6,7] $ is also called as open bracket $ - 6 $ and closed bracket $ 7 $
Since we used addition and subtraction operation in the above solution,
The addition is the sum or adding the given two or more numbers, or values, or variables and in addition, if we sum the two or more numbers a new frame of the number will be found, also in subtraction which is the minus of two or more than two numbers or values but here comes with the condition that in subtraction the greater number sign will stay constant.