Solve each compound inequality. Give the solution set in interval form

Question

3x+5<8 and 2x+5≥11

Answer 1

The word "and" tells us that we are looking for all values of x that satisfy BOTH inequalities simultaneously.  

 

3x + 5 < 8     and    2x + 5 ≥ 11

 

      3x < 3      and           2x ≥ 6

 

        x < 1       and            x ≥ 3

 

It is impossible for a number to be both less than 1 and greater than or equal to 3,  so the compound inequality has NO SOLUTION.  

Answer 2

3x + 5 < 8

What are all the possible values that, when substituted into x, will make the expression 3x + 5 less than 8?

We first rearrange the inequality similarly to how we work with equation: get x by itself through operations.

 

3x + 5 < 8

Subtract 5 from both sides:

3x < 8 - 5

3x < 3

Divide both sides by 3:

x < 1

 

We have x by itself on the left side and a number on the right side. What we've determined here is that any value of x less than 1 will make the originally inequality (3x + 5 < 8) true. 

 

Note than if we set x equal to 1, the original inequality is not true: 3*1 + 5 = 8, and 8 is not less than 8. It is 8.

 

The other inequality can be solved in a similar way.

 

Now, to put this into interval form, we have to pick the highest and lowest possible values of x. We know x has to be less than 1, so the highest possible value of x is something just a teeny bit less than 1. We don't actually get to include 1, as we talked about above. So the interval notation would be:

 

(-∞,1) 

 

The parentheses say: don't include these values. We can't include -∞ because it's infinity. We can't include 1 because it wouldn't make the original inequality true.