Answer:
[tex]x\leq6\text{ or } x>30[/tex]
Step-by-step explanation:
We are given the compound inequality:
[tex]\displaystyle 9x+4\leq 58\text{ or } \frac{1}{6}x-2>3[/tex]
Note that this is an "or" inequality. In other words, we just have to solve each inequality individually and then combine the answers!
1)
We have:
[tex]9x+4\leq58[/tex]
Subtract 4 from both sides:
[tex]9x\leq54[/tex]
Divide both sides by 9:
[tex]x\leq6[/tex]
So, our first solution is all numbers less than or equal to 6.
2)
We have:
[tex]\displaystyle \frac{1}{6}x-2>3[/tex]
Add 2 to both sides:
[tex]\displaystyle \frac{1}{6}x>5[/tex]
To cancel out the fraction, let's multiply both sides by 6. So:
[tex]\displaystyle 6\left(\frac{1}{6}x\right)>6(5)[/tex]
The left side will cancel. Therefore:
[tex]x>30[/tex]
So, our second solution is all numbers greater than 30.
Therefore, our full solution will be:
[tex]x\leq6\text{ or } x>30[/tex]
And we're done!
