Solving Inequalities

As with solving equations, the goal of solving inequalities is to find the values of the variable or variables that make the inequality true. For instance, for what values of x is $$(x - 1)$$ less than 4? Well, whenever $$(x - 1)$$ is less than 4, x is less than 5. Check a couple of values, if you need to convince yourself.

• Take a value less than 5, say, 3. Is 3 – 1 less than 4? Yes, 2 is less than 4.
• Try one bigger than 5, say, 10. Is 10 – 1 less than 4? No, 9 is greater than 4.

Thus, the solution when solving for x in the inequality $$x - 1 < 4$$ , is $$x < 5$$.

In general, when solving inequalities, we can perform many of the same actions that we can when solving equations:

1. We can add or subtract the same thing from both sides of an inequality and the inequality still holds. So, for example:

and so, adding 7 to each side:

$$6 + 7 < 9 + 7$$ is true, since we get $$13 < 16$$.

we can subtract 4 from each side, as we do with an equation, and the inequality still holds:

Just as with equations , perhaps even more so, it is always good to check your answer when solving inequalities. To do this, first try y = 12. When $$
y = 12$$, you should get an equality: $$12 + 4 = 16$$. In this case, the inequality is true, since we have a ‘$$ \leq$$’, but even if it were strictly less than (<), you can still check for equality here. This just lets you make sure that your manipulations were correct.

Next, you should try a number smaller than 12. Let’s pick 0, since that is generally a nice number to work with. Since y = 0 is one of the solutions to our inequality, when we replace y with 0 in the original inequality, it should be true. And it is:

2. We can multiply or divide both sides of the inequality by a positive number and the inequality still holds. For example:

So, we can solve the inequality, $$4z \leq 24$$ by dividing each side by the same positive number, namely, 4:

Check this answer. First set z = 6 in the inequality $$4z \leq 24$$. That makes the true statement that $$4 \cdot 6 \leq 24$$. Then set z equal to anything greater than or less than 6. If you pick something smaller than 6, then the original inequality should be true, but if you pick something greater than 6, your original inequality will not hold.

This brings us to the one thing that you need to be very careful about when solving inequalities: MULTIPLYING and DIVIDING a negative number to each side of the inequality.

The action of multiplying or dividing a negative number by both sides of an inequality does not preserve the inequality. Let’s look at an example of why. We know that:

This is NOT a true statement. However, notice that $$4 > 2$$ is a true statement. It turns out that if we multiply or divide a negative number on both sides of an inequality, we must reverse the inequality to so that it remains equivalent to the original inequality. Let’s try an example:

So, we divide both sides by -4, and since we are dividing by a negative, we reverse the inequality from “greater than or equal to” ($$ \geq $$) to “less than or equal to” ($$ \leq $$):

$${\Large\frac{3}{4}} \leq k$$

Equivalently, we can bring the k to the left side, but make sure that your inequality still indicates that $${\Large\frac{3}{4}}
$$ is less than or equal to k (in other words, k is greater than or equal to $${\Large\frac{3}{4}}$$.

Example:

Solve the inequality for x (that is, find the values of x that make the inequality true):

1. First, use the distributive property to multiply the polynomial in the parentheses by –2:

2. Next, bring all x’s to one side and all constants to the other side. One way we can do this is by adding 12 to each side, and then subtracting 6x from each side. Remember that this does not change the inequality: we are just adding and subtracting the same thing to each side.

3. Now, we need to divide both sides by -14. Remember that we have to reverse the inequality, since we are dividing by a negative number:

Solve for the variable (that is, find the values of the variable that make the equation true):

Solution: $$x > 3$$

This may not look like a “problem” to be solved, but for what values of x is x greater than 3? For all values where: $$x > 3$$. So, in a sense it was already “solved”, but it is good to realize that when we “solve” an inequality, our answer should tell us which values make the equation true. In this case, the inequality is telling us directly which values will make it true: all those greater than 3.

4. $${\Large\frac{{5z\left( {12 - 2z} \right)}}{6}} \leq - {\Large\frac{5}{3}}z^2 + z$$