Modules for Prerequisite Mathematics

First I want to isolate the absolute value, so I'll subtract \(4\) from both sides.

The first case is for \(h \geq 5\text{,}\) when the term in the absolute value is positive.

In this case, we solve and get \(h = 13\text{,}\) which is a valid solution in the range for this case. The second case is \(h \lt 5\text{,}\) where the term in the absolute value is negative.

Then we proceed to solve by multiplying by \((-1)\) and adding \(5\) to get \(h = -3\text{.}\) This solution is in the approprate range for this case. We conclude that there are two solutions: \(h = 13\) and \(h = -3\text{.}\) We can also observe that these are exactly the two number which are distance \(8\) away from 5, which is a reasonable interpretation of the original equation.

First I want to isolate the absolute value, so I'll subtract \(2\) from both sides.

Then I'll divide both sidse by \(-8\text{.}\)

Then we work in cases. The term in the absolute value is positive when \(r \geq -8\text{,}\) so that is our first case. In this case, we can just drop the absolute value.

Then we can solve, finding that \(r = \frac{-31}{4}\text{.}\) This is greater (slightly) than \(-8\text{,}\) so it is a valid solution for the first case. We move on to the second case, where the term in the absolute value is negative. This happens when \(r \lt -8\text{.}\) We multiply the absolute value term by \((-1)\text{.}\)

Then we solve. We multiply both sides by \((-1)\) and then subtract \(8\) from both sides to sinde \(r = \frac{-33}{4}\text{.}\) This is less (slightly) than \(-8\text{,}\) so it is a valid solution for the second case. In total, we have two solution, \(r = \frac{-31}{8}\) and \(r = \frac{-33}{8}\text{.}\)

Here there are two terms which involve \(|w + 5|\text{.}\) I would like to only have one such term. To get there, I can subtract \(|w + 5|\) from both sides of the equation. On the right side, we have \(4\) of this term; subtracting one of them leaves, of course, \(3\) of the term.

Then I can divide both sides of the equation by \(3\text{.}\)

Now I can work with the two cases that apply to an absolute value. In the first case, \(w \geq -5\) and I can just drop the absolute value.

We solve this to get \(w = -1\text{,}\) which is a valid solution in our case. In the second case, \(w \lt -5\) and I multiply by \((-1)\) for the absolute value.

I can solve this (multiply by \((-1)\) and subtract 5) to get \(w = -9\text{.}\) This is a value solution in the range for the second case. In total, we conclude that we have two solutions: \(w = -1\) and \(w = -9\text{.}\)

I can split this into two inequalities. For the negative case, I have to reverse the inequality.

Then we can solve these inequalities by adding 3 to both sides.

First I'll divide each side by \(5\text{.}\)

Then we can work in the two cases required for absolute value. For the first case, \(b \geq -10\) and we can drop the absolute value.

We can subtract \(b\) from both sides.

Then I can divide by \(2\text{.}\)

We get the range \(b \gt 5\text{,}\) which is included in this case, so we can accept the entire range as part of the solution. Now we move on to the second case, where \(b \lt -10\text{.}\) In this case, we multiply the absolute value term by \((-1)\text{.}\)

Then we solve. We start by multiplying by \((-1)\text{,}\) which reverses the inequality.

Then we can subtract \(b\) from both sides.

Finally, we can divide by \(-4\text{,}\) which again reverses the inequality.

We get a range of \(b \gt -\frac{5}{2}\text{.}\) However, this case is only consider \(b \lt -10\text{.}\) There is no overlap between these ranges, so there are no solution in this case. We conclude that the original inequality is solved simply by the range \(b \gt 5\text{.}\)

We work in our two cases. In the first case, we want the term in the absolute value to be positive: \(k^2 + 9 \gt 0\) or \(k^2 \gt -9\text{.}\) But square are always positive, so this inequality is always satisfied. In this situation, the first case covers all number and there is not second case. Since the term in the absolute value is positive, we can just drop the absolute value and solve.

We subtract \(9\) from both sides.

Here I follow Subsection 3.1.4 to solve this inequality. One piece of the solution is the direct square root of the inequality: \(k \gt \sqrt{5}\text{.}\) But I also have to consider negative \(k\) which become positive when squares. This gives me the second part of the solution: \(k \lt -\sqrt{5}\text{.}\) The solution to the original inequality consists of both these ranges.