1. $3x-5<6-2x$

   $$\begin{array}{rcll}5x-5& <& 6& \mathbf{\text{Using the addition principle for inequalities}}\mathbf{;}\mathbf{\text{ adding }}\mathbf{2}\mathit{x}\\ 5x& <& 11& \mathbf{\text{Using the addition principle for inequalities; adding }}\mathbf{5}\\ x& <& \frac{11}{5}& \mathbf{\text{Using the multiplication principle for inequalities}}\mathbf{;}\\ & & & \mathbf{\text{multiplying by }}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{,}\mathbf{\text{ or dividing by }}\mathbf{5}\end{array}$$

   Any number less than $\frac{11}{5}$ is a solution. The solution set is $\left\{x|x<\frac{11}{5}\right\}$, or $\left(-\infty ,\frac{11}{5}\right)$. The graph of the solution set is shown below.

   

2. $13-7x\ge 10x-4$

   $$\begin{array}{rcll}13-17x& \ge & -4& \mathbf{\text{Subtracting }}\mathbf{10}\mathit{x}\\ -17x& \ge & -17& \mathbf{\text{Subtracting }}\mathbf{13}\\ x& \le & 1& \mathbf{\text{Dividing by }}\mathbf{-}\mathbf{17}\mathbf{\text{ and reversing the inequality sign}}\end{array}$$

   The solution set is $\left\{x\right|x\le 1\}$, or $(-\infty ,1]$. The graph of the solution set is shown below.

   

1. The radicand, $x-6$, must be greater than or equal to 0. We solve the inequality $x-6\ge 0:$

   $$\begin{array}{rcl}x-6& \ge & 0\\ x& \ge & 6.\end{array}$$

   The domain is $\left\{x\right|x\ge 6\}$, or $[6,\infty )$.

2. Any real number can be an input for x in the numerator, but inputs for x must be restricted in the denominator. We must have $3-x\ge 0$ and $\sqrt{3-x}\ne 0$. Thus, $3-x>0$. We solve for x:

   $$\begin{array}{rcll}3-x& >& 0& \\ -x& >& -3& \mathbf{\text{Subtracting }}\mathbf{3}\\ x& <& 3.& \mathbf{\text{Multiplying by }}\mathbf{-}\mathbf{1}\mathbf{\text{ and reversing the inequality sign}}\end{array}$$

   The domain is $\left\{x\right|x<3\}$, or $(-\infty ,3)$.

We have

The solution set is $\left\{x\right|-4<x\le 1\}$, or $(-4,1]$. The graph of the solution set is shown below.

We have

The solution set is $\left\{x\right|x\le -1\mathit{\text{ or }}x>3\}$. We can also write the solution set using interval notation and the symbol $\cup$ for the union, or inclusion, of both sets: $(-\infty ,-1]\cup (3,\infty )$. The graph of the solution set is shown below.

1. Familiarize. Suppose that a job takes 20 hr. Then $n=20$, and under plan A, Natália would earn $\mathrm{\$250}+\mathrm{\$10}\cdot 20$, or $\mathrm{\$250}+\mathrm{\$200}$, or $450. Her earnings under plan B would be $\mathrm{\$20}\cdot 20$, or $400. This shows that plan A is better for Natália if a job takes 20 hr. If a job takes 30 hr, then $n=30$, and under plan A, Natália would earn $\mathrm{\$250}+\mathrm{\$10}\cdot 30$, or $\mathrm{\$250}+\mathrm{\$300}$, or $550. Under plan B, he would earn $\mathrm{\$20}\cdot 30$, or $600, so plan B is better in this case. To determine all values of n for which plan B is better for Natália, we solve an inequality. Our work in this step helps us write the inequality.

2. Translate. We translate to an inequality:

   

3. Carry out. We solve the inequality:

   $$\begin{array}{cccc}\hfill 20n& >& 250+10n\hfill & \\ \hfill 10n& >& 250\hfill & \mathbf{\text{Subtracting }}\mathbf{10}\mathit{n}\mathbf{\text{ on both sides}}\hfill \\ \hfill n& >& 25.\hfill & \mathbf{\text{Dividing by }}\mathbf{10}\mathbf{\text{ on both sides}}\hfill \end{array}$$

4. Check. For $n=25$, the income from plan A is $\mathrm{\$250}+\mathrm{\$10}\cdot 25$, or $\mathrm{\$250}+\mathrm{\$250}$, or $500, and the income from plan B is $\mathrm{\$20}\cdot 25$, or $500. This shows that for a job that takes 25 hr to complete, the income is the same under either plan. In the Familiarize step, we saw that plan B pays more for a 30-hr job. Since $30>25$, this provides a partial check of the result. We cannot check all values of n.

5. State. For values of n greater than 25 hr, plan B is better for Natália.