The graph shows an arrow beginning on 5 with an open circle and pointing to the left, thus that portion of the graph says, all real numbers less than 5. There is a second arrow beginning on 9 with a closed circle and pointing to the right, representing all real numbers greater than or equal to 9. Since we are joining the two parts of the graph, we have a compound inequality utilizing the "or" statement. So our answer is \(\displaystyle \small x< 5 \quad or \quad x \ge 9\).   

"The product of seven and a number " is \(\displaystyle 7x\). "Twenty subtracted from the product of seven and a number" is \(\displaystyle 7x - 20\) . "Exceeds one hundred" means that this is greater than one hundred, so the correct inequality is

\(\displaystyle 7x - 20 > 100\)

"The sum of a number and sixteen" is translates to \(\displaystyle x + 16\); twice that sum is \(\displaystyle 2(x+16)\). " Is no less than sixty" means that this is greater than or equal to sixty, so the desired inequality is

 \(\displaystyle 2(x+16) \geq 60\).

"The sum of a number and sixteen" translates to \(\displaystyle x + 16\); twice that sum is \(\displaystyle 2 (x+16)\). "Does not exceed eighty" means that it is less than or equal to eighty, so the desired inequality is

\(\displaystyle 2 (x+16) \leq 80\)

In terms of inequalities, we know two things. \(\displaystyle x\) is greater than or equal to 14 (which is the same as \(\displaystyle x\) being greater than 13). And we know \(\displaystyle x\) is less than or equal to 22 (which is the same as \(\displaystyle x\) being less than 23).

So, we have to find the correct statement where we can find the two inequalities. We see we need  \(\displaystyle 14\leq x \text{ or } 13< x\)  and \(\displaystyle x\leq 22 \text{ or }x< 23\).

Therefore the only answer that fits our needs is \(\displaystyle 13< x\leq 22\).

On the number line, the graph starts at –1 and ends at 3.

The line runs between –1 and 3, so we know our inequality involves only values of x that fall between these two numbers. The open circle at –1 indicates that –1 is not included, while the shaded circle on 3 indicates that 3 is included.

\(\displaystyle x>-1\)

\(\displaystyle x\leq3\)

Combining these two inequalities into one give us our answer.

\(\displaystyle -1< x\leq3\)

\(\displaystyle 6x + 22 \leq 97\)

\(\displaystyle 6x + 22 -22 \leq 97-22\)

\(\displaystyle 6x \leq 75\)

\(\displaystyle 6x \div 6 \leq 75\div 6\)

\(\displaystyle x \leq 12.5\)

or, in interval notation, \(\displaystyle (-\infty , 12.5]\)

\(\displaystyle 6x + 22 > 67\)

\(\displaystyle 6x + 22 -22 > 67 -22\)

\(\displaystyle 6x > 45\)

\(\displaystyle 6x \div 6 > 45\div 6\)

\(\displaystyle x > 7.5\)

or, in interval notation, \(\displaystyle \left ( 7.5, \infty \right )\)

The first step is to distribute (multiply) through the parentheses:

\(\displaystyle 3(x + 4) \leq x - 6\)

\(\displaystyle 3x + 12 \leq x - 6\)

Then subtract \(\displaystyle x\) from both sides of the inequality:

\(\displaystyle 2x + 12 \leq -6\)

Next, subtract the 12:

\(\displaystyle 2x \leq -18\)

Finally, divide by two:

\(\displaystyle x \leq -9\)

To solve \(\displaystyle 3x-4\leq 12x+16\), it is necessary to isolate the variable and the integers.

Subtract \(\displaystyle 3x\) and \(\displaystyle 16\) from both sides of the equation.

\(\displaystyle -20\leq 9x\)

Divide by nine on both sides.

\(\displaystyle -\frac{20}{9}\leq x\)

This answer is also the same as:  \(\displaystyle x\geq -\frac{20}{9}\)