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Section 3.2 Solving Inqualities

Subsection 3.2.1 Inequalities as Questions

Figure 3.2.1. Solving Inequalities

When I talked about equations with varibles in Section 1.2, I said that an equation with a variable was implicity a question. The queation

\begin{equation*} 4x - 7 = 12 \end{equation*}

was implicit the question: what number(s), when we multiply by 4 and then subtract 7, result in 12? An inequality is also a question. If we make this equality an inequality,

\begin{equation*} 4x - 7 \lt 12 \end{equation*}

we are asking the question: what numers, when we mulitply by 4 and then subtrat 7, are less than 12?

The major difference, now, is that we expect many more answers (in general). There might be only one number which satisfies the equation, but infinitely many numbers which satisfy the inequality. This raising the question of how to represent these large gets of numbers. For the purposes of this section, we'll just represent them using the inequalities themslves. It I want to talk about all the numbers larger than 5, I can just talk about all \(x\) such that \(x \gt 5\text{.}\) In Chapter 10, we'll talk about other notations for set of numbers, including set notation and interval notation.

Subsection 3.2.2

For now, sovling inequalities will be simply tryin go write the inequalities in a consise form. Let's consider the example we started with above.

\begin{equation*} 4x - 7 \lt 12 \end{equation*}

To write this in a more succint form, let's do some manipulations, keeping in mind the rules about what operations perserve and reverse inequalities. First, I'll add 7 to both side; addition and subtration were operations that preserved inequalities.

\begin{equation*} 4x \lt 19 \end{equation*}

Then I'll divide by sides by 4. Division by positive numbers is also an operation that preserve inequalities.

\begin{equation*} x \lt \frac{19}{4} \end{equation*}

This is a reasonable answer. The originaly inequality describes (is solved by!) all numbers which are strictly larger than \(\frac{21}{4}\text{.}\)

Here is another example.

\begin{equation*} 15 - 6x \geq 42 \end{equation*}

First, I'll subtract 15 from both sides. Subtraction preserves inequalities.

\begin{equation*} - 6x \geq 35 \end{equation*}

Then I'll divide both sides by \(-6\text{.}\) Dividing by a negative number reverses inequalities, so I'll reverse the notation.

\begin{equation*} x \leq \frac{35}{-6} = - \frac{35}{6} \end{equation*}

The set of number specified by this inequality is all number less than \(- \frac{35}{6}\text{.}\) That's a reasonable description of this set of numbers, so that is a solution to the inequality.

\begin{equation*} 9x - 4 \lt 17x + 5 \end{equation*}

I can add 4 to both sides of the inequality, which preserves the inequality.

\begin{equation*} 9x \lt 17x + 9 \end{equation*}

Now I can subtract \(17x\) from both sides of the inequality, which preserves the inequality.

\begin{equation*} -8x \lt 9 \end{equation*}

Now I can divide by \(-8\text{,}\) which reverses the inequality.

\begin{equation*} x \lt \frac{9}{-8} = - \frac{9}{8} \end{equation*}

This solves the inequality. The original expression is satisfied by all \(x\) which are less than \(-\frac{9}{8}\text{.}\)

\begin{equation*} \sqrt{6x-4} \geq 17 \end{equation*}

To get rid of the square root, we need to square both sides. That's an exponential operation, so it preserves inequalities only if both sides of the inequality are positive. Here, that is the case: 17 is positive and the square root is a positive number. We square both sides of the inequality.

\begin{equation*} 6x-4 \geq 289 \end{equation*}

Then we can add \(4\) to both sides, which preserve the inequality.

\begin{equation*} 6x \geq 293 \end{equation*}

Finally, we divide by \(6\text{,}\) which also preserves the inquality because \(6\) is positive.

\begin{equation*} x \geq \frac{293}{6} \end{equation*}

This gives us a solution to the inequality: all number greater than or equal to \(\frac{293}{6}\text{.}\) Before we finish this example, though, we have one subtlety to mention. We need the square root at the start of the problem to be possible, to the term in side the square root cannot be negative: \(6x-4 \geq 0\text{.}\) We can solve this inequality similarly to get \(x \geq \frac{4}{6} = \frac{2}{3}\text{.}\) In our solution of \(x \geq \frac{293}{6}\) is more restrictive, since \(\frac{293}{6}\) is much later than \(\frac{2}{3}\text{,}\) so the square root restrict is satisfied by our solution.

\begin{equation*} x^2 + 6 \geq 3 \end{equation*}

First, we can subtract 6 from both sides, which preserves the inequality.

\begin{equation*} x^2 \geq -3 \end{equation*}

If we were solving an equation, at this point we would try to square root both sides. That wouldn't work too well, since we can't take square roots of negative numbers. However, this is an inequality, so we don't need the precise equality. The right side is a square, which is always positive. All positive numbers are greater than \(-3\text{.}\) Therefore, we can conclude that this inequality is satisfied by all possible numbers.

This example hopefully makes the point that when solving inequalities, we can't be to mechanistic about it. We can't just walk through a series of mechanical steps; we have to stop and think about what the inequality means and what it would entail for numbers to satisfy it. A mechanical approach to this equation would try to take a negative square root, fail, and stop, missing the fact that all numbers already satisfy the inequality.

\begin{equation*} x^2 + 6 \geq 10 \end{equation*}

This is very similar to the previous example, but we avoid the negative square root when we start with \(10\) on the right side. The first step is the same: we can subtract 6 from both sides, which preserves the inequality.

\begin{equation*} x^2 \geq 4 \end{equation*}

Then we can square root both side. The square root is an exponent and both sides are positive, so it preserves the inequality.

\begin{equation*} x \geq 2 \end{equation*}

We see that all numbers greater than or equal to \(2\) satisfy the inequality.

However, here again there are subtleties that a mechanical approach might miss. When we took the square root, we didn't consider the possibility that \(x\) might be a negative number. This is more complciated, since the square of a negative becomes positive. Therefore, any slightly large negative numbers, such as \(-10\text{,}\) have squares larger than 4 (in this case, \((-10)^2 = 100\text{.}\) Since the negative square become positive, we have a reveral of inequalities. The solution includes the numbers above, which are larger than \(2\text{,}\) but also all negative numbers less or equal to \(-2\text{.}\) If \(x \leq -2\text{,}\) then its square must be at least 4, satisfying the original equality.

You might ask: why didn't I give a rule about squares, square roots and negative numbers. I could have given such a rule; however, if I followed in that direction, I'd end up with a monumental list of rules for manipulating inequalities in a variety of situations. This isn't really how most mathematics is done; instead, we build up some practice and intuition about inequalities and carefully think about what each individual inequality means. It's more conceptual and less mechanistic. That makes is trickier to learn and teach -- I can't just give you all the rules and algorithms -- but such is the nature of mathematics.