Solving a linear inequality via equations

You can also solve a linear inequality by

first replacing the inequality sign by an equal sign,
then solving this equation, and
finally, determining the sign of the inequality for point to the left and to the right of the solution of the equation.

Determine the exact solution of the inequality \[x + 5 \ge 5x -1\] via equations.

\(x \le {{3}\over{2}}\)

We follow the following roadmap:

Find out whether the solutions are on the number line to the left or to the right of \({{3}\over{2}}\).
- First calculate the left- and right-hand sides of the inequality \(x + 5 \ge 5x -1\) when you substitute a value of \(x\) less than or equal to \({{3}\over{2}}\). For example, when you fill in \(x=-10\), then you get \(-5 \ge -51\) and this is a true statement. Any other value of \(x\) less than or equal to \({{3}\over{2}}\) may be used too, and you still get a true statement.
- Then calculate the left- and right-hand sides of the inequality \(x + 5 \ge 5x -1\) when you substitute a value of \(x\) greater than or equal to \({{3}\over{2}}\). For example, when you fill in \(x=10\), then you get \(15 \ge 49\) and this is a false statement. Any other value of \(x\) greater than or equal to \({{3}\over{2}}\) may be used too, and you still get a false statement.
- From these two numeric examples follows that solutions \(x\) of \(x + 5 \ge 5x -1\) must satisfy \(x \le {{3}\over{2}}\).
  The points where the inequality holds are shown in green in the number line below. An open circle around \(x={{3}\over{2}}\) indicates that we are dealing with an inequality of the type \(\lt\) or \(\gt\), where in this case the point itself is not a solution. A closed circle indicates an inequality of the type \(\le\) or \(\ge\), and then the point marked on the number line is element of the solution set.