The Retired Professor Simplifies Algebra: Linear Equations

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.

Any equation has two expressions that are equal.  So, whatever is done to one side of the equation, if also done to the other, does not change the fact that the two expressions are equal.

The objective is to preform operations so as to cause the variable to be alone on one side of the equation, and not appear on the other side.

We will build complexity slowly.  In the end you will have seen different examples that will help with your understanding.

Solve for x:  x + 2 = 9

In order to get the variable alone on one side of the equation subtract 2 from each side.

x + 2 – 2 = 9 – 2, which gives x = 7.

Solve for x: 3x = 9

In order to get the variable alone on one side of the equation divide both sides by 3.

3x/3 = 9/3, which gives x = 3.

Solve for x: 2x + 1 = 4x - 7

Think that x stands for a number, so it can be treated as you would treat a number.

You must decide which way you want to send the variable, and send the number terms the other way.

2x – 2x + 1 + 7 = 4x – 2x – 7 + 7

8 = 2x

Now divide each side by 2.

8/2 = 2x/2, which gives 4 = x.

Note:  It really does not matter which side of the equation the variable ends up on.

Solve for x:  2(x + 3) = -x – 9

Always remove the parentheses, or other symbol of grouping, first.  This requires the distributive property, a(b + c) = ab + ac.

Eliminating the parentheses gives:  2x + 6 = -x – 9

Continue to solve.  2x + x + 6 + 9 = -x + x – 9 + 9

3x = 15

Divide by 3:  3x/3 = 15/3, which solves as x = 5.

Sometimes the variable disappears.  If the result is a true statement all values for the variable are solutions, or all real numbers solve the equation.  If the result is a false statement, the equation has no solution.

Solve for x:  2x + 5 = 3x + 5 – x

Collect like terms on each side of the equation.

2x + 5 = 2x + 5

Subtract 2x from each side of the equation.

2x – 2x + 5 = 2x – 2x + 5, or 5 = 5 (TRUE)  So, all real numbers are solutions.

Solve for x:  2x + 5 = 3x + 6 – x

Collect like terms on each side of the equation.

2x + 5 = 2x + 6

Subtract 2x from each side of the equation.

2x – 2x + 5 = 2x – 2x + 6, or 5 = 6 (FALSE)  So, the equation has no solution.

If we think about an absolute value, say │7│ or │-7│, we get the non-negative number the size of what is between the absolute value bars.  In this case either expression is equal to 7. 

Now, we must allow a variable in the expression we wish to take the absolute value of.

Many textbooks solve with the two signs placed on the other side of the equation. 

│x│ = 6 is written x = 6 and x = -6.  While it may seem like a wasted extra step at this point, soon the reason for changing this to -x = 6 and x = 6.  The first equation can be easily solved for x as x = -6. 

As an example, solve │2x + 5│ = 3.

There will be two equations, both linear equations.

-(2x + 5) = 3 and 2x + 5 = 3

First equation:  -(2x + 5) = 3 

Multiply each side of the equation by -1.

(-1)( -(2x + 5)) = (-1)3 

2x + 5 = -3

2x + 5 - 5 = -3 – 5

2x = -8

2x/2 = -8/2

x = -4

Second equation:

2x + 5 = 3 

2x + 5 - 5 = 3 – 5

2x = -2

2x/2 = -2/2

x = -1

There are two solutions, -1 and -4.

Since an absolute value can never be negative, we can immediately write “no solution” for any absolute value that is equal to a negative value.

(In the above the negative values are for the variable, not the absolute value.)

As an example, suppose you are asked to solve the absolute value │x + 3│ = -4

Immediately write down “No solution exists.”  Solving the two equations is misleading, you will get solutions to those equations that will not solve the absolute value equation.

Now for the harder examples.

Solve │x + 3│ = │x + 4│

Here, we get four equations with all variations of positive and negative expressions. 

x + 3 = x + 4, -(x + 3) = x + 4, x + 3 = -(x + 4), and –(x + 3) = -(x + 4)

If you multiply the third equation by -1 on each side it becomes the second equation.  Likewise, multiplying the fourth equation by -1 on both sides gives the first equation.  So, there are only two equations to consider.

x + 3 = x + 4

Subtracting x from both sides gives

x – x + 3 = x – x + 4

3 = 4, a false statement, which indicates this equation produces no solutions.

-(x + 3) = x + 4

Using the distributive law, we get

-x + (-3) = x + 4

-x + x + (-3) – 4 = x + x + 4 – 4

-7 = 2x

Dividing by 2 we get

-7/2 = 2x/2

-7/2 = x, which is a solution of the absolute value equation.

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