Review of Inequalities

1. A. \(h = 12\)

Let's plug in \(h=12\) and see if the inequality is true.

\(\begin{aligned}

6 & \geq \frac{h}{2} \\

6 & \stackrel {?}{\geq} \frac{12}{2} \\

6 & \geq 6

\end{aligned}\)

Yes, \(h =12\) is a solution!

Now let's try \(h =14\).

\(\begin{aligned}

&6 \geq \frac{h}{2} \\

&6 \stackrel {?}{\geq} \frac{14}{2} \\

&6 \not {\geq} 7

\end{aligned}\)

No, \(h=14\) is not a solution.

Let's try \(h =16\).

\(\begin{aligned}

6 & \geq \frac{h}{2} \\

6 & \stackrel {?}{\geq} \frac{16}{2} \\

6 & \not{\geq} 8

\end{aligned}\)

No, \(h = 16\) is not a solution.

The following \(h\)-value satisfies the inequality \(6 \geq \frac{h}{2}\):

• \(h = 12\)

2. B.\(z =9\), C. \(z =10\)

Let's plug in \(z=8\) and see if the inequality is true.

\(\begin{aligned}

&7 < \frac{z}{2}+3 \\

&7  \stackrel {?}{ < } \frac{8}{2} + 3 \\

&7 \stackrel{?}{ < } 4+3 \\

&7 \nless 7 \\

\end{aligned}\)

No, \(z = 8\) is not a solution.

Now let's try \(z=9\).

\(\begin{aligned}

&7 < \frac{z}{2}+3 \\

&7 \stackrel {?}{ < } \frac{9}{2} + 3 \\

&7 \stackrel{?}{ < } 4.5 +3 \\

&7 < 7.5 \\

\end{aligned}\)

Yes, \(z =9\) is a solution!

Let's try \(z =10\).

\(\begin{aligned}

&7 < \frac{z}{2}+3 \\

&7 \stackrel {?}{ < } \frac{10}{2} + 3 \\

&7 \stackrel{?}{ < } 5 +3 \\

&7  < 7.5 \\

\end{aligned}\)

Yes, \(z =10\) is a solution!

The following \(z\)-values satisfy the inequality \(7 < \frac{z}{2}+3\):

• \(z =9\)
• \(z= 10 \)

3. A. \(f=4\)

Let's plug in \(f=4\) and see if the inequality is true.

\(\begin{aligned}

f+8 & \leq 12 \\

4+8 & \stackrel{?}{\leq} 12 \\

12 & \leq 12

\end{aligned}\)

Yes, \(f=4\) is a solution.

Now let's try \(f=5\).

\(\begin{aligned}

f+8 & \leq 12 \\

5+8 & \stackrel {?} { \leq } 12 \\

13 & \leq 12

\end{aligned}\)

No, \(f=5\) is not a solution.

Let's try \(f=6\).

\(\begin{aligned}

f+8 & \leq 12 \\

6+8 & \stackrel {?} { \leq } 12 \\

14 & \leq 12

\end{aligned}\)

No, \(f=6\) is not a solution.

The following \(f\)-values satisfy the inequality \(f+8 \leq 12:\)

• \(f=4\)

4. A. \(n=10\)

Let's plug in \(n=10\) and see if the inequality is true.

\(\begin{aligned}

3 n-7 & < 26 \\

3(10)-7 & \stackrel{?}{ < } 26 \\

30-7 & \stackrel{?}{ < } 26 \\

23 & < 26

\end{aligned}\)

Yes, \(n=10\) is a solution!

Now let's try \(n=11\).

\(\begin{aligned}

3 n-7 & < 26 \\

3(11)-7 & \stackrel{?}{ < } 26 \\

33-7 & \stackrel{?}{ < }26 \\

23 & \nless 26

\end{aligned}\)

No, \(n=11\) is not a solution.

Let's try \(n=12\).

\( \begin{aligned}

3 n-7 & < 26 \\

3(12)-7 & \stackrel{?}{ < } 26 \\

36-7 & \stackrel{?}{ < } 26 \\

29 & \nless 26

\end{aligned}\)

No, \(n=12\) is not a solution.

The following \(n\)-values satisfy the inequality \(3 n-7 < 26\):

• \(n=10\)