Can we solve $12/(x-3) < x+1$?

Can we solve $12/(x-3) < x+1$?

Question

1. Find the solution set for $x$ given that the following three relations for $x$, $y$, where $x, y \in \mathbb{R}$, are simultaneously true:

$$\begin{equation*} y < x + 1, \quad y + 6x < 20, \quad x = 5y - 7. \end{equation*}$$

1. Find the solution set of the inequality

$$\begin{equation*} \frac{12}{x-3} < x + 1, \quad (x \in \mathbb{R}, x \ne 3). \end{equation*}$$

Suggestion

We might get some useful insight into what is happening here by drawing a sketch.

Again, a sketch might be helpful here, to help check that our answer is sensible. Remember that $x-3$ might be positive or negative.

Solution

It helps to sketch a diagram; here, the regions included by the inequalities are shaded.

From the sketch, we can see that the solution set for $x$ consists of the $x$-coordinates of points lying on the line between $A$ and $B$.

We can find the coordinates of $A$ by solving the simultaneous equations $y=x+1\quad\text{and}\quad x=5y-7.$ Substituting the second equation into the first gives $y=5y-6$ so that $y=\frac64=\frac32$. Thus $x=y-1=\frac12$, and $A$ has coordinates $(\frac12, \frac32)$.

Likewise, the coordinates of $B$ are found by solving $y+6x=20\quad\text{and}\quad x=5y-7.$ Again, substitution gives $y+30y-42=20$ which yields $y=2$ and hence $x=3$, so $B$ has coordinates $(3,2)$.

Thus the solution set for $x$ is $\frac{1}{2} < x < 3$.

1. Find the solution set of the inequality

$$\begin{equation*} \frac{12}{x-3} < x+1, \quad (x \in \mathbb{R}, x \ne 3). \end{equation*}$$

We might be tempted to multiply through by $x-3$, but then we have to worry about its sign, and whether or not we need to change the direction of the inequality.

A sketch might again help us to follow what is happening.

So $\dfrac{12}{x-3}<x+1$ when we are between $A$ and $x=3$ and when we are to the right of $B$.

We can find the $x$-coordinates of $A$ and $B$ by solving the equation $\dfrac{12}{x-3}=x+1.$ Multiplying both sides by $x-3$ gives $12=(x+1)(x-3)$, which rearranges to $x^2-2x-15=0$.

We can factorise this to get $(x-5)(x+3)=0$, so $A$ has $x$-coordinate $-3$ and $B$ has $x$-coordinate $5$.

Thus the solution set is $-3<x<3$ or $x>5$.

An alternative approach is to manipulate the inequality directly. But we cannot simply multiply by

$x-3$

, as this may be positive or negative. Instead, we can rearrange the inequality to get zero on one side, by subtracting one side from the other:

$$\begin{align*} &&x+1&>\frac{12}{x-3}\\ \iff&& x+1-\frac{12}{x-3}&>0\\ \iff&& \frac{(x+1)(x-3)-12}{x-3}&>0\\ \iff&& \frac{(x+3)(x-5)}{x-3}&>0. \end{align*}$$

Now the expression on the left changes sign at $x = -3$, $x = 3$ and $x = 5$, and so we can make a table showing the sign of the expression:

Thus the solution set (where the expression is positive) is $-3<x<3$ or $x>5$.

Another technique is to multiply both sides of the inequality by $(x-3)^2$, as this is never negative; the resulting inequality can be solved in essentially the same way as we have done here.