Reiny - the problem with the different inequality signs was typed from the textbook. I will notify instructor Thanks
Another problem
3<5x-27<53
1st part
3<5x-27
27+3<5x-27+27
30/5 < 5x/5
6<x
2nd part
5x-27<53
5x-27+27<53+27
5x/5 < 80/5
x<16
combine 6<x<16
Am I correct?
you can do both at once
3 < 5x-27 < 53
30 < 5x < 80
6 < x < 16
the nice thing about having the orientation of the inequality signs going the same way is that you can just perform the solution as it stands
3<5x-27<53
you are trying to get just "x" in the middle, so
add 27 to all parts
30 < 5x < 80
divide by +5
6 < x < 16 , same as yours
Now wasn't that easy, just took 2 steps.
That is why the orientation of the signs going the same way is so important.
Thanks
Yes, you are correct. The steps you have shown to solve the inequality are correct.
In order to solve the compound inequality "3 < 5x - 27 < 53", you can solve the two separate inequalities independently and then combine their solutions.
For the first inequality, "3 < 5x - 27", you added 27 to both sides of the inequality to isolate the term with x on one side. This resulted in "30 < 5x". Then, you divided both sides of the inequality by 5 to solve for x and found that "6 < x".
For the second inequality, "5x - 27 < 53", you added 27 to both sides of the inequality to isolate the term with x on one side. This resulted in "5x < 80". Then, you divided both sides of the inequality by 5 to solve for x and found that "x < 16".
Finally, combining the solutions from both inequalities, you have "6 < x < 16". This means that x is greater than 6 but less than 16.
Great job on solving the compound inequality correctly!