Solve each inequality and graph the solution:

(a) $4x+7\geq 2x-3$

(b) $x^{2}-4x+3>0$

(c) $x^{2}+x+1 < 0$

(d) $\dfrac{1+x}{1-x}>0$

Solution (a) $$\begin{array}{@{\hspace*{-4pt}}rcll@{\quad}l} 4x+7&\geq& 2x-3 & & \\ 4x&\geq& 2x-10 & & {\color{#0066A7}{\hbox{Subtract 7 from both sides.}}} \\ 2x&\geq& -10 & & {\color{#0066A7}{\hbox{Subtract 2\(x\) from both sides.}}} \\ x&\geq& -5 & & {\color{#0066A7}{\hbox{Divide both sides by 2.}}}\quad \\ &&&&{\color{#0066A7}{\hbox{(The direction of the inequality symbol is unchanged.)}}} \end{array}$$

Figure 1 $x\geq 5$

The solution using interval notation is $[ -5,\infty )$. See Figure 1 for the graph of the solution.

(b) This is a quadratic inequality. The related quadratic equation $x^{2}-4x+3= ( x-1) ( x-3) =0$ has two solutions, $1$ and $3$. We use these numbers to partition the number line into three intervals. Now select a test number in each interval, and determine the value of $x^{2}-4x+3$ at the test number. See Table 2.

Table 2

Figure 2 $x<1$ or $x>3$

We conclude that $x^{2}-4x+3>0$ on the set $( -\infty ,1) \cup ( 3,\infty)$. See Figure 2 for the graph of the solution.

(c) The quadratic equation $x^{2}+x+1=0$ has no real solution, since its discriminant is negative. [See Example 4(b)]. When this happens, the quadratic inequality is either positive for all real numbers or negative for all real numbers. To see which is true, evaluate $x^{2}+x+1$ at some number, say, $0$. At $0,$ $x^{2}+x+1=1,$ which is positive. So, $x^{2}+x+1>0$ for all real numbers $x$. The inequality $x^{2}+x+1 < 0$ has no solution.

(d) The only solution of the rational equation $\dfrac{1+x}{1-x} =0$ is $x=-1;$ also, the expression $\dfrac{1+x}{1-x}$ is not defined for $x=1.$ We use the solution $-1$ and the value $1$, at which the expression is undefined, to partition the real number line into three intervals. Now select a test number in each interval, and evaluate the rational expression $\dfrac{1+x}{1-x}$ at each test number. See Table 3.

Table 3

Figure 3 $-1<x<1$

We conclude that $\dfrac{1+x}{1-x}>0$ on the interval $( -1,1)$. See Figure 3 for the graph of the solution.