Answer
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Hint:Solve each interval separately and solve for x. Take the first interval and then solve for X. doing this step, we will see that x is less than 4. Which means x can be anything less than the digit 4. Also, for the second interval, solving for x we will get that x is greater than 5. Which means x can be anything more than 5. And when we combine both the intervals, we will get x is less and 4 till negative infinity and x is greater than 5 till infinity.
Complete step by step solution:
The given interval we have here is:-
$3x - 2 > 13$ Or $3x - 2 > 13$
Let’s consider $3x - 2 > 13$ as 1 st interval and $3x - 2 > 13$ as 2 nd interval.
So,
Let’s solve the 1 st interval first
Adding 3 to both the sides, we will get:-
$
2x - 3 + 3 < 5 + 3 \\
\Rightarrow 2x < 8 \\
$
Diving both the sides of the 1 st interval, we will get:-
$
\dfrac{{2x}}{2} < \dfrac{8}{2} \\
\Rightarrow x < 4 \\
$
Therefore, we got out first constraint and that is $x < 4$
So if we want to show this in intervals. It will be all the real numbers less than 4. Which can be shown as:-
Interval=$\left( { - \infty ,4} \right)$
Similarly,
We will add 2 to both the sides of the 2 nd interval, we will get:-
$
3x - 2 + 2 > 13 + 2 \\
\Rightarrow 3x > 15 \\
$
Dividing both the sides of the 2 nd interval by 3, we will get:-
$\left( { - \infty ,4} \right) \cup \left( {5,\infty } \right)$
Therefore, we got our second constraint too, which is $x > 5$
So, if we want to show this in terms of interval, we will write it as all real numbers greater than 5. Mathematically, which can be written as:-
Interval=$\left( {5,\infty } \right)$
Hence, if you join both the intervals together to get the full answer, you will get:-
Final required interval= $\left( { - \infty ,4} \right) \cup \left( {5,\infty } \right)$
Note: We have used the union symbol, because for the answer both the intervals are satisfying. Therefore, to write in a continuous form, we will use the union symbol. Also, for the sides where the intervals are ending that are more than 4 and greater than 5 sides, we have not used square brackets because the numbers are not included.
Complete step by step solution:
The given interval we have here is:-
$3x - 2 > 13$ Or $3x - 2 > 13$
Let’s consider $3x - 2 > 13$ as 1 st interval and $3x - 2 > 13$ as 2 nd interval.
So,
Let’s solve the 1 st interval first
Adding 3 to both the sides, we will get:-
$
2x - 3 + 3 < 5 + 3 \\
\Rightarrow 2x < 8 \\
$
Diving both the sides of the 1 st interval, we will get:-
$
\dfrac{{2x}}{2} < \dfrac{8}{2} \\
\Rightarrow x < 4 \\
$
Therefore, we got out first constraint and that is $x < 4$
So if we want to show this in intervals. It will be all the real numbers less than 4. Which can be shown as:-
Interval=$\left( { - \infty ,4} \right)$
Similarly,
We will add 2 to both the sides of the 2 nd interval, we will get:-
$
3x - 2 + 2 > 13 + 2 \\
\Rightarrow 3x > 15 \\
$
Dividing both the sides of the 2 nd interval by 3, we will get:-
$\left( { - \infty ,4} \right) \cup \left( {5,\infty } \right)$
Therefore, we got our second constraint too, which is $x > 5$
So, if we want to show this in terms of interval, we will write it as all real numbers greater than 5. Mathematically, which can be written as:-
Interval=$\left( {5,\infty } \right)$
Hence, if you join both the intervals together to get the full answer, you will get:-
Final required interval= $\left( { - \infty ,4} \right) \cup \left( {5,\infty } \right)$
Note: We have used the union symbol, because for the answer both the intervals are satisfying. Therefore, to write in a continuous form, we will use the union symbol. Also, for the sides where the intervals are ending that are more than 4 and greater than 5 sides, we have not used square brackets because the numbers are not included.
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