
Solve and graph the inequality. Given answer in the interval notation.
\[x\in 15<4x+3\le 31\]
(a) (3, 7)
(b) [3, 7)
(c) (3, 7]
(d) [3, 7]




Answer
507.9k+ views
Hint: First of all consider each inequality separately that is 15 < 4x + 3 and \[4x+3\le 31\]. Now, solve them and find the values of x from each of them and take their intersection to get the interval of x and plot it in the given graph/number line.
Complete step-by-step answer:
In this question, we have to solve and graph the inequality
\[15<4x+3\le 31\]
Let us consider the inequality given in the question.
\[15<4x+3\le 31\]
Let us first solve, 15 < 4x + 3
By subtraction 3 from both the sides, we get,
15 – 3 < 4x
12 < 4x
\[x>\dfrac{12}{4}\]
\[x>3....\left( i \right)\]
Now, let us solve
\[4x+3\le 31\]
By subtracting 3 from both the sides, we get,
\[4x\le 31-3\]
\[4x\le 28\]
\[x\le \dfrac{28}{4}\]
\[x\le 7....\left( ii \right)\]
So, from (i) and (ii), we get,
\[\left( x>3 \right)\cap \left( x\le 7 \right)\]
\[x\in (3,7]\]
We have used ( ) bracket on the side of 3 because \[x\ne 3\] by x > 3 and we have used [ ] bracket on the side of 7 because x = 7 as well as x < 7. Let us plot this interval on the number line.
Hence, the option (c) is the right answer.
Note: In this question, almost all the options are looking the same, so students often get confused but if they look clearly, they would see that in some options, equality with 3 or 7 is shown by using [ ] brackets. As we have found that \[x\in (3,7]\] and this we can see only in option (c), so that is correct. So, in these questions, where options are so close, students must check the options twice to get the correct answer. Also, in the case of inequality, if we divide or multiply it by a negative number, its sign gets reversed. So, this must be taken care of.
Complete step-by-step answer:
In this question, we have to solve and graph the inequality
\[15<4x+3\le 31\]
Let us consider the inequality given in the question.
\[15<4x+3\le 31\]
Let us first solve, 15 < 4x + 3
By subtraction 3 from both the sides, we get,
15 – 3 < 4x
12 < 4x
\[x>\dfrac{12}{4}\]
\[x>3....\left( i \right)\]
Now, let us solve
\[4x+3\le 31\]
By subtracting 3 from both the sides, we get,
\[4x\le 31-3\]
\[4x\le 28\]
\[x\le \dfrac{28}{4}\]
\[x\le 7....\left( ii \right)\]
So, from (i) and (ii), we get,
\[\left( x>3 \right)\cap \left( x\le 7 \right)\]
\[x\in (3,7]\]
We have used ( ) bracket on the side of 3 because \[x\ne 3\] by x > 3 and we have used [ ] bracket on the side of 7 because x = 7 as well as x < 7. Let us plot this interval on the number line.

Hence, the option (c) is the right answer.
Note: In this question, almost all the options are looking the same, so students often get confused but if they look clearly, they would see that in some options, equality with 3 or 7 is shown by using [ ] brackets. As we have found that \[x\in (3,7]\] and this we can see only in option (c), so that is correct. So, in these questions, where options are so close, students must check the options twice to get the correct answer. Also, in the case of inequality, if we divide or multiply it by a negative number, its sign gets reversed. So, this must be taken care of.
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