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Section 2-1 : Solutions and Solution Sets

We will start off this chapter with a fairly short section with some basic terminology that we use on a fairly regular basis in solving equations and inequalities.

First, a solution to an equation or inequality is any number that, when plugged into the equation/inequality, will satisfy the equation/inequality. So, just what do we mean by satisfy? Let’s work an example or two to illustrate this.

Example 1 Show that each of the following numbers are solutions to the given equation or inequality.
  1. \(x = 3\) in \({x^2} - 9 = 0\)
  2. \(y = 8\) in \(3\left( {y + 1} \right) = 4y - 5\)
  3. \(z = 1\) in \(2\left( {z - 5} \right) \le 4z\)
  4. \(z = - 5\) in \(2\left( {z - 5} \right) \le 4z\)
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a \(x = 3\) in \({x^2} - 9 = 0\) Show Solution

We first plug the proposed solution into the equation.

\[\begin{align*}{3^2} - 9 & \mathop = \limits^? 0\\ 9 - 9 & = 0\\ 0 & = 0 \,\,\,\,{\mbox{OK}}\end{align*}\]

So, what we are asking here is does the right side equal the left side after we plug in the proposed solution. That is the meaning of the “?” above the equal sign in the first line.

Since the right side and the left side are the same we say that \(x = 3\) satisfies the equation.


b \(y = 8\) in \(3\left( {y + 1} \right) = 4y - 5\) Show Solution

So, we want to see if \(y = 8\) satisfies the equation. First plug the value into the equation.

\[\begin{align*}3\left( {8 + 1} \right) &\mathop = \limits^? 4\left( 8 \right) - 5\\ 27 & = 27\,\,\,\,{\mbox{OK}}\end{align*}\]

So, \(y = 8\) satisfies the equation and so is a solution.


c \(z = 1\) in \(2\left( {z - 5} \right) \le 4z\) Show Solution

In this case we’ve got an inequality and in this case “satisfy” means something slightly different. In this case we will say that a number will satisfy the inequality if, after plugging it in, we get a true inequality as a result.

Let’s check \(z = 1\).

\[\begin{align*}2\left( {1 - 5} \right) & \mathop \le \limits^? 4\left( 1 \right)\\ - 8 & \le 4 \,\,\,\,{\mbox{OK}}\end{align*}\]

So, -8 is less than or equal to 4 (in fact it’s less than) and so we have a true inequality. Therefore \(z = 1\) will satisfy the inequality and hence is a solution


d \(z = - 5\) in \(2\left( {z - 5} \right) \le 4z\) Show Solution

This is the same inequality with a different value so let’s check that.

\[\begin{align*}2\left( { - 5 - 5} \right) & \mathop \le \limits^? 4\left( { - 5} \right)\\ - 20 & \le - 20\,\,\,\,{\mbox{OK}}\end{align*}\]

In this case -20 is less than or equal to -20 (in this case it’s equal) and so again we get a true inequality and so \(z = - 5\) satisfies the inequality and so will be a solution.

We should also do a quick example of numbers that aren’t solution so we can see how these will work as well.

Example 2 Show that the following numbers aren’t solutions to the given equation or inequality.
  1. \(y = - 2\) in \(3\left( {y + 1} \right) = 4y - 5\)
  2. \(z = - 12\) in \(2\left( {z - 5} \right) \le 4z\)
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a \(y = - 2\) in \(3\left( {y + 1} \right) = 4y - 5\) Show Solution

In this case we do essentially the same thing that we did in the previous example. Plug the number in and show that this time it doesn’t satisfy the equation. For equations that will mean that the right side of the equation will not equal the left side of the equation.

\[\begin{align*}3\left( { - 2 + 1} \right) & \mathop = \limits^? 4\left( { - 2} \right) - 5\\ - 3 & \ne - 13\,\,\,\,{\mbox{NOT OK}}\end{align*}\]

So, -3 is not the same as -13 and so the equation isn’t satisfied. Therefore \(y = - 2\) isn’t a solution to the equation.


b \(z = - 12\) in \(2\left( {z - 5} \right) \le 4z\) Show Solution

This time we’ve got an inequality. A number will not satisfy an inequality if we get an inequality that isn’t true after plugging the number in.

\[\begin{align*}2\left( { - 12 - 5} \right) & \mathop \le \limits^? 4\left( { - 12} \right)\\ - 34\require{cancel} & \bcancel{ \le } - 48\,\,\,\,{\mbox{NOT OK}}\end{align*}\]

In this case -34 is NOT less than or equal to -48 and so the inequality isn’t satisfied. Therefore \(z = - 12\) is not a solution to the inequality.

Now, there is no reason to think that a given equation or inequality will only have a single solution. In fact, as the first example showed the inequality \(2\left( {z - 5} \right) \le 4z\) has at least two solutions. Also, you might have noticed that \(x = 3\) is not the only solution to \({x^2} - 9 = 0\). In this case \(x = - 3\) is also a solution.

We call the complete set of all solutions the solution set for the equation or inequality. There is also some formal notation for solution sets although we won’t be using it all that often in this course. Regardless of that fact we should still acknowledge it.

For equations we denote the solution set by enclosing all the solutions is a set of braces, \(\left\{ {} \right\}\). For the two equations we looked at above here are the solution sets.

\[\begin{align*}3\left( {y + 1} \right) & = 4y - 5 & \hspace{0.25in} & {\mbox{Solution Set }} :\,\,\,\left\{ 8 \right\}\\ {x^2} - 9 & = 0 & \hspace{0.25in} & {\mbox{Solution Set }} :\,\,\,\left\{ { - 3,3} \right\}\end{align*}\]

For inequalities we have a similar notation. Depending on the complexity of the inequality the solution set may be a single number or it may be a range of numbers. If it is a single number then we use the same notation as we used for equations. If the solution set is a range of numbers, as the one we looked at above is, we will use something called set builder notation. Here is the solution set for the inequality we looked at above.

\[\left\{ {z|z \ge - 5} \right\}\]

This is read as : “The set of all \(z\) such that \(z\) is greater than or equal to -5”.

Most of the inequalities that we will be looking at will have simple enough solution sets that we often just shorthand this as,

\[z \ge - 5\]

There is one final topic that we need to address as far as solution sets go before leaving this section. Consider the following equation and inequality.

\[\begin{align*}{x^2} + 1 & = 0\\ {x^2} & < 0\end{align*}\]

If we restrict ourselves to only real solutions (which we won’t always do) then there is no solution to the equation. Squaring \(x\) makes \(x\) greater than equal to zero, then adding 1 onto that means that the left side is guaranteed to be at least 1. In other words, there is no real solution to this equation. For the same basic reason there is no solution to the inequality. Squaring any real \(x\) makes it positive or zero and so will never be negative.

We need a way to denote the fact that there are no solutions here. In solution set notation we say that the solution set is empty and denote it with the symbol : \(\emptyset \). This symbol is often called the empty set.

We now need to make a couple of final comments before leaving this section.

In the above discussion of empty sets we assumed that we were only looking for real solutions. While that is what we will be doing for inequalities, we won’t be restricting ourselves to real solutions with equations. Once we get around to solving quadratic equations (which \({x^2} + 1 = 0\) is) we will allow solutions to be complex numbers and in the case looked at above there are complex solutions to \({x^2} + 1 = 0\). If you don’t know how to find these at this point that is fine we will be covering that material in a couple of sections. At this point just accept that \({x^2} + 1 = 0\) does have complex solutions.

Finally, as noted above we won’t be using the solution set notation much in this course. It is a nice notation and does have some use on occasion especially for complicated solutions. However, for the vast majority of the equations and inequalities that we will be looking at will have simple enough solution sets that it’s just easier to write down the solutions and let it go at that. Therefore, that is what we will not be using the notation for our solution sets. However, you should be aware of the notation and know what it means.