- Find the solution set for \(x\) given that the following three relations for \(x\), \(y\), where \(x, y \in \mathbb{R}\), are simultaneously true: \[\begin{equation*} y < x + 1, \quad y + 6x < 20, \quad x = 5y - 7. \end{equation*}\]

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 \(\frac{1}{2} < x < 3\).

- 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:

\(x<-3\) | \(-3<x<3\) | \(3<x<5\) | \(x>5\) | |
---|---|---|---|---|

\(x+3\) | \(-\) | \(+\) | \(+\) | \(+\) |

\(x-5\) | \(-\) | \(-\) | \(-\) | \(+\) |

\(x-3\) | \(-\) | \(-\) | \(+\) | \(+\) |

\(\dfrac{(x+3)(x-5)}{x-3}\) | \(-\) | \(+\) | \(-\) | \(+\) |

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.