In Precalculus, it's essential that you can easily and efficiently solve sentences like
‘$\,\frac{3x}{2}1 \ge \frac 15  7x\,$’
and ‘$\,x^2 \ge 3\,$’ .
This web exercise gives you practice with the first type of sentence, which is a linear inequality in one variable.
The next web exercise gives you practice with the second type of sentence, which is a nonlinear inequality in one variable.
This page gives an inanutshell review of essential concepts and skills for solving linear inequalities in one variable.
If you're needing more, a thorough review is also outlined.
All links to web exercises open in a new window, so you don't lose this page.
SOLVING LINEAR INEQUALITIES
The sentence ‘$\,\frac{3x}{2}1 \ge \frac 15  7x\,$’ is an example of a linear inequality in one variable.

It's an inequality because of the ‘$\,\ge\,$’.
An inequality is a mathematical sentence that uses at least one of the following:
 $\le$ (less than or equal to)
 $\lt$ (less than)
 $\ge$ (greater than or equal to)
 $\gt$ (greater than)

It's an inequality in one variable because it uses only a single variable (in this case, $\,x\,$).
That is, it doesn't use (say) both $\,x\,$ and $\,y\,$.
The single variable can appear any number of times: in the sentence
‘$\displaystyle\,\frac{3\color{blue}{x}}{2}1 \ge \frac 15  7\color{blue}{x}\,$’, it appears twice.

It's linear since every occurrence of the variable is as simple as possible.
For example, there are no $\,x^2\,$ terms, no variables inside trig functions or under square roots, and no variables in denominators.
All you have are numbers times $\,x\,$ to the first power (i.e., terms of the form $\,kx\,$, where $\,k\,$ is a
real number).
Only extremely basic tools are needed to solve linear inequalities in one variable:
 Similar to the Addition Property of Equality,
you can add/subtract the same number to/from both sides of the inequality.
 Similar to the Multiplication Property of Equality,
you can multiply/divide both sides of the inequality by a nonzero number.
However, if you multiply/divide by a negative number, then you
must change the direction of the inequality symbol.
Why is this? Why do you need to flip the inequality when you multiply or divide by a negative number?
Let's look at an example to understand this situation.
In the sketch below,
‘$\,a\lt b\,$’ is true,
because
$\,a\,$ lies to the left of
$\,b\,$.
Multiplying both sides by
$\,1\,$
sends $\,a\,$ to its opposite ($\,a\,$) ,
and sends $\,b\,$ to its opposite ($\,b\,$) .
Now, the opposite of $\,a\,$ is to the right of the opposite of
$\,b\,$.
That is,
$\, a\gt b\,$.
So, multiplying two numbers by $\,1\,$ changes their left/right positioning relative to each other!
For example, $\,1 < 2\,$ (one is to the left of two), but $\,1 > 2\,$ (negative one is to the right of negative two).
That's why you need to change the inequality symbol!
So, solution of the sample sentence could
look something like this:
$$
\eqalign{
\frac{3x}{2}1 \ &\ge\ \frac 15  7x \qquad\qquad\qquad&\text{original inequality}\cr\cr
10\bigl(\frac{3x}{2}1\bigr) \ &\ge\ 10(\frac 15  7x) &\text{to clear fractions, multiply both sides by 10, which is the least common multiple of 2 and 5}\cr\cr
15x  10 \ &\ge\ 2  70x &\text{after multiplying out, all fractions are gone}\cr\cr
10 \ &\ge\ 2 85x &\text{subtract $\,15\,x$ from both sides}\cr\cr
12 \ &\ge\ 85x &\text{subtract $\,2\,$ from both sides}\cr\cr
\frac{12}{85} \ &\le\ \frac{85x}{85} &\text{divide both sides by $\,85\,$, change direction of inequality symbol}\cr\cr
\frac{12}{85} \ &\le\ x &\text{simplify}\cr\cr
x \ &\ge\ \frac{12}{85} &\text{write in more conventional form, with the variable on the left}
}
$$
By the way, head up to
wolframalpha.com and type in
3x/2 1 >= 1/5  7x (you can cutandpaste). Voila!
For a much more thorough review of solving linear inequalities, do this:
Now, practice solving some fairly complicated linear inequalities by clickclickclicking below.
All steps in the solution process are shown.
Each problem is accompanied by a graph, which offers additional insights into the
solution set.
If the information contained in the graph isn't meaningful to you, first study the next section,
Solving Nonlinear Inequalities in One Variable (graphical concepts),
and then come back!