The Retired Professor Simplifies Algebra: Linear Inequalities

Before we start it is important to address whether there is a problem with math anxiety.  If there is a belief by the student that it is impossible to learn math, that impediment must be addressed first.  Fortunately, there are some excellent books written on the subject that can really help.

Linear inequalities are like equations, except that they each use one of the symbols <, >, ≤, or ≥.

They are solved much like equations, with two differences.  The first difference is when an inequality is multiplied or divided on both sides by a negative number, the direction of the inequality changes. 

To understand why the above is true take two numbers, say 2 and 4.  Observe, 2 < 4.  Next multiply each side by a negative number, say -3.  Now we get -6 and -12.  Observe, -6 > -12.

The other difference is that while it an equation with the variable solved for makes no difference which side of the equation the variable ends up on.  Observe, x = 3 and 3 = x are equally valid ways of expressing a solution.  In inequalities, the solution set x < 6 is the same as 6 > x.  Notice the direction of the inequality.  Most people prefer the variable on the left, so if you solve with the variable on the right be careful when switching sides.

Now, a few examples. 

Solve 5x < 10

Divide both sides by 5.

5x/5 < 10/5

x < 2

Notice any value greater than 2 is a solution, so we refer to the solutions of inequalities as solution sets.

Solve 3(x – 2) < 5x + 8.

Break down the symbols of grouping:  3x + (-6) < 5x + 8

Add 6 and -5x to each side.

3x + (-5x) + (-6) + 6 < 5x + (-5x) + 8 + 6

-2x < 14

Divide both sides by -2.

-2x/(-2) >  14/(-2)

x > -7

Solve for x, 2x + 3 < 2x + 1

Add -2x to each side. 

2x + (-2x) + 3 < 2x + (-2x) + 1

3 < 1 

Notice the variable has disappeared, and what is left is a false statement, so the inequality has no solution.

Solve for x, 2x + 3 > 2x + 1

Add -2x to each side. 

2x + (-2x) + 3 > 2x + (-2x) + 1

3 > 1 

Notice the variable has disappeared, and what is left is a true statement, so the inequality has all real numbers as a solution set.

For linear inequalities we will start with the special cases first, since the special cases give great insight in the process.

First, inequalities of the form │ax + b│ < c where c is 0 or any negative number, or │ax + b│ ≤ d, where d is any negative number and d is any negative number, have no solution.  An absolute value cannot ever be less than 0, so it cannot be less than a negative number.

Next, inequalities of the form │ax + b│ > d, or │ax + b│ ≥ d, where d is any negative number must always be true, so any value of x will solve the inequality.  Hence, the solution set for this case is all real numbers.

The above cases utilize an insight into what an absolute value really is.

Now we consider the more general case.  An absolute value being less than or being less than or equal to a non-negative number.  We get sets of possible solutions, and we must consider only those solutions that make both parts of the set of possible solutions valid.  Hence, we combine the two parts of the solution set with the word “and” which is the same as the intersection of the two sets or answers that we get when solving the inequalities.  Any solution to only one of the two inequalities makes the absolute value inequality less than or at worse equal to a negative value, which is impossible.

As an example, find the solution set for │x + 5│ < 3. 

(Many textbooks say to solve x + 5 < 3 and x + 5 > -3.  This does not make clear why the < becomes >.  Our approach makes this more understandable.

Inequality 1 is x + 5 < 3, with solutions x < -2 after subtracting 5 form each side of the inequality.

Inequality 2 is –(x + 5) < 3.  Multiplying each side by -1gives x + 5 > -3.  Notice when we multiplied by a negative number < became >.  So, solving by subtracting 5 from each side of the inequality we get solutions x > -8. 

Our solution set is -8 < x < -2.  This is the same as x > -8 “and” x < -2.

Notice x can be negative.  That is fine.

When the absolute value uses the > or ≥ symbol its two possible solution sets is combined using “or” and the solution set is any solution in ether possible solution set.

As another example consider │x - 9│ > 5.  Our two inequalities are –(x – 9) > 5 and x – 9 > 5.

Inequality 1 is x – 9 > 5, which is solved by we add 9 to each side and get x > 14.

Inequality 2 is –(x – 9) > 5.  First, multiply both sides by -1 to produce x – 9 < -5.  Adding 9 to each side we get the solutions x < 4.

The solution set is x < 4 or x > 14.

Note:  The absolute value must be isolated before starting.  Had │x - 9│ - 5 > 0 been the problem │x - 9│ > 5 it would have had to be manipulated into the second form before starting.

If a problem starts with │x - 9│ < -3 simply indicate the problem has no solution.

If a problem starts with │x - 9│ > -3 simply indicate the problem has a solution set of all real numbers.

If a problem starts with │x - 9│ - 8 > -3 add 8 to each side, and solve │x - 9│ > 5 as was done above.  Just seeing the negative number on the right side of the inequality is not enough to make a call of no solution or all real numbers are solutions, the absolute value must first be isolated on one side of the inequality.

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