2.9.3 Solving Linear Inequalities: Part II
- Divide by a positive constant, keep the inequality.
- Divide by a negative constant, switch the inequality.
In the last section we were able to turn the equation .69x - 3000 > 2000 into .69x > 5000, by adding 3000 to both sides. Now let's try to get rid of the .69. An equation that involves decimals and big numbers can be tough to get your head around, so first, let's try something simpler, in the hopes that we can then develop a general method. For example, let's begin with the inequality 2x>6 and figure out how we can get rid of the 2. If this were a linear equality, we would just multiply both sides by 1/2.
Can we do that here? Let's say that x represents the distance to school. If the distance to school and back is longer than 6 miles, is the one-way distance to school more than 3 miles? In other words, does 2x>6 imply that x>3? Sure! So it looks like if we multiply both sides by a positive number, such as the ½ used in this example, then we've performed a legal operation.
Further, since multiplying by ½ is the same thing as dividing by 2, we can draw the conclusion that whenever we have a positive number, we can legally multiply or divide by that number. We now have a general technique to solve ax>b when a>0; a technique which we developed by thinking about a specific real world example.
If a>0 andax>b then we can divide both sides by a and keep the same inequalitiy.
If 3x>9 then
If 2x<-8 then
We can now solve our original question: If .69x>5000 then . (In writing your final answer, remember that the x represents the number of Apps)
For example, can we solve -x>-2 by just multiplying both sides by -1 to get that x>2? You would probably imagine that the answer is yes, but to be sure, let's investigate this example a bit.
Negatives and inequalities together are tough to conceptualize, so let's study an easier problem first. This is a great problem solving technique: if something is too abstract and difficult, make it more specific and simple. In this case, numbers are more specific than variables.
So, let's use an x that actually makes the statement true, such as x=1. We then have that -1>-2, which is a true statement. Multiplying both sides by -1, we get the inequality 1>2. But this is not a true statement! In order to make the inequality true, it looks like we'll have to reverse the order of the inequality.
To better understand why this just happened, let's study the expression -1>-2 a little bit further. If we add 2 to both sides, we get that -1+2>0. In other words, the quantity on the left is bigger than 0. Now let's multiply both sides by -1. Since 0×-1 is still 0, the right hand side doesn't change. After multiplying the left hand side by -1, we get -(-1+2) = 1-2, which is now a negative number. Since just the left hand side changes sign, the inequality will now only be true if we switch the inequality from a greater than inequality to a less than inequality: 1-2<0. Adding the 2 back to the right hand side we get that 1<2. Our inequality has changed its direction!
Futher, since -1 is equal to 1/-1, we can see that dividing by negative numbers has the same exact effect as multiplying. We can summarize what we've learned in this section as:
If a>0andax>bthen we can divide both sides by a and keep the same inequalitiy.
If a< 0andax>bthen we can divide both sides by a and switch the inequalitiy.
If -3x>9 then
If -2x > -8 then
We now have a deep understanding of single linear relationships. We know what they look like and now have the algebraic skills that we can use to solve both equalities and inequalities. In the next section, we can begin to study what happens when we have more than one relationship in a system of linear equations.