Question

One third of a number decreased by five is at least three. Two-thirds of a number plus eight is greater than twelve. Solve and graph each inequality

Answer 1

The inequalities are solved algebraically, resulting in the solution x ≥ 24. Graphically, this is represented by a line on a number line starting at 24 extending to the right, with a closed circle at 24.

Step-by-step explanation:

We have two inequalities to solve:

1. One third of a number decreased by five is at least three:
   (1/3)x - 5 ≥ 3
2. Two-thirds of a number plus eight is greater than twelve:
   (2/3)x + 8 > 12

To solve the first inequality, we can start by adding 5 to both sides to isolate the term involving x:

(1/3)x ≥ 3 + 5
(1/3)x ≥ 8

Multiplying both sides by 3 to solve for x, we have:

x ≥ 24

For the second inequality, we subtract 8 from both sides:

(2/3)x > 12 - 8
(2/3)x > 4

Now, multiplying both sides by 3/2 to solve for x:

x > 6

To find the common solution set for both inequalities, we take the intersection of x ≥ 24 and x > 6, which is x ≥ 24. The graphical representation will be a line on the number line starting at 24 and extending to the right, with a closed circle at 24 to indicate that 24 is included in the solution set.