### 4.1 - Introduction to Equations

 Definition: An equation is an algebraic statement that two expressions are equal. An equation consists of: a left-hand-side expression (abbreviated LHS), a right-hand-side expression (abbreviated RHS), and an equal sign between them.

Here are three examples:
• 5 + 3 = 8

• 5 + 3 = 9

• x + 4 = 10
The first example is an equation that is always true; 5 plus 3 always equals 8. This is also known as an identity. The second example is an equation that is never true; 5 plus 3 never equals 9. This is also known as a contradiction. The third example is known as a conditional equation. It is only true under the condition that the variable , x, equals 6 since only 6 plus 4 equals 10. No other value of x makes the equation true. In this book we are primarily interested in conditional equations.

### Solutions

Finding all the values of the variable x that make the left-hand-side of an equation equal the right-hand-side is called solving the equation for x. When we set out to find these values, x is called the unknown. Once we have found the values they are called the solutions or roots of the equation. We say that they make the equation true. We also say that these values of x satisfy the equation.

Equations can have any number of solutions, as these examples show.

Example: The equation: has one solution, namely x = 3. We can verify that this solution satisfies the equation by substituting it back into the equation. This gives: Since both the left-hand-side and the right-hand-side simplify to 1/6, the solution x = 3 is verified.

Example: The equation:

20 x 3 + 25 x 2 + 5 x = 0
has three solutions: x = 0, x = −¼ and x = −1. This is called a solution set. We can verify that all three values of x satisfy the equation by substituting each of them into the equation and getting the true statement that 0 = 0.

Example: Some equations have no solutions over the real numbers but do have solutions over the complex numbers. An example is the equation:
x 2 = −9.
There is no real number which when squared will yield −9 but there are two complex numbers, namely 3 i and −3 i.

Example: Some equations have no solutions whatsoever. They are called contradictions. An example is the equation:
x = x + 1.
There is no number that is unchanged when you add 1 to it.

### Checking the solutions

Once you have found a possible solution for an equation you must substitute it into the original equation and check that it does indeed satisfy the equation. There are two reasons: (1) you may have made an error in solving the equation, and (2) the algebra process that we will describe below for solving an equation can produce so-called “extraneous solutions” (essentially garbage solutions!) You can’t tell that you have an extraneous solution until you substitute the solution in and discover that it doesn’t satisfy the original equation. Extraneous solutions must be rejected. Click here to learn more about extraneous solutions.

### Literal equations and formulas

Many of the equations that we will study will contain only numbers and one letter, namely the variable that is the unknown to be solved for. This letter will usually be x, the universal letter that represents the “unknown” (think x-rays, The x-Files, etc.) But we will also learn how to solve literal equations and formulas:
• Literal equations are equations that contain more than one letter. An example is a x + b = 10. One letter, usually x, is considered to be the variable or unknown to be solved for and the other letters are considered to be constants.

• Formulas are equations that also contain more than one letter. The letters are chosen to suggest certain physical quantities and any one of them can be considered to be the unknown. An example is the formula A = ½ b h, which gives the area A of a triangle having base b and height h.
Some notes on using the AutoSolve feature of the Algebra Coach:
• Don’t forget to set the AutoSolve unknown to whatever the equation’s unknown is (by using either the Tools, Options menu or by clicking on the status panel at the bottom of the program). By default it is x.

• If the equation contains only numbers and one letter (the unknown) then the solution will be a number and the Algebra Coach will check the solution. However if the equation is a literal equation or formula then the solution will contain letters and the Algebra Coach will not check the solution.

### Basic procedures for solving equations

There is one basic rule to follow when solving equations:

 Perform the same mathematical operation on both sides of the equation.

The reason is that this maintains the equality of the equation. The only question now is “which operation”? As we go through this book we will answer that question for ever more complicated types of equations. We begin, in this section, by explaining how to solve two of the simplest types of equations:
After that we will manipulate every equation we encounter, no matter how complicated, until we get it into one of these two forms.

### Solving equations in which the unknown occurs only once

Let’s assume that the unknown is x. The goal is to get x alone on one side of the equation. To do this, look at what operations have been applied to x to get the expression that contains x. Then:

 Invert (undo) the operations that were applied to x in the reverse order in which they were applied.

Every operation that is performed on the left-hand-side of the equation is also performed on the right-hand-side. Also, after every algebraic operation, both sides are simplified as much as possible.

Example: Solve the equation:
x − 5 = 13.
Solution: In the expression on the left-hand-side of the equation, x has had 5 subtracted from it so we must add 5 back on to get x alone. (Adding 5 is the inverse operation of subtracting 5.) Adding 5 to both sides gives:
x − 5 + 5 = 13 + 5.
Simplifying both sides then gives:
x = 18.
This last equation is just a manipulated form of the original equation, but because it has x alone on the left-hand-side and a number on the right-hand-side, it is the solution. We must still check the solution by substituting it into the original equation. This gives 18 − 5 = 13 or 13 = 13, so the left-hand-side equals the right-hand-side and the solution checks out.

Example: Solve the equation: Solution: In the expression on the left-hand-side of the equation, x has had this sequence of operations applied to it in this order:
• subtraction of 8
• square root
• multiplication by 2
To get x alone on the left-hand-side, we must invert these operations in the reverse order in which they were applied. Specifically, we must perform this sequence of operations in this order:
• division by 2 (this is the inverse of the multiplication by 2)
• square (this is the inverse of the square root)
• addition of 8 (this is the inverse of the subtraction of 8)
 Why these operations and why this order? Here is an analogy. Picture how you get dressed in the morning (to be polite we will only talk about socks and shoes): You pull up your socks. You tie your shoes. In the evening you get undressed like this: You untie your shoes. You pull down your socks. Note that untying is the inverse of tying and pulling down is the inverse of pulling up. Also note that since your shoes were put on last they have to come off first.

Therefore start with and perform this sequence of operations:
• Divide both sides by 2 and simplify: • Square both sides and simplify (note that on the left-hand-side the square and square root are inverse operations): • Add 8 to both sides and simplify: This final equation is just a manipulated form of the original equation, but because it has x alone on the left-hand-side and a number on the right-hand-side, it is the solution. We must still check the solution by substituting it into the original equation. This gives which simplifies to 6 = 6, so the left-hand-side equals the right-hand-side and the solution checks out. Example: The cosine law, ,
is a formula that is used to solve oblique triangles. It is a relation between the lengths of sides a, b and c and the angle C which is opposite side c. Given any three of these quantities, the fourth can be found. Solve the cosine law for angle C.

Solution: On the left-hand-side the variable C has had these operations applied to it in this order:
• application of the cosine function
• multiplication by −2 · a · b
• the addition of a 2 + b 2
To get C alone on the left-hand-side, we must perform these operations in this order:
• the subtraction of a 2 + b 2
• division by −2 · a · b
• application of the arccos function (this is the inverse of the cosine function)

Therefore start with and perform this sequence of operations:
• Subtract a 2 + b 2 from both sides and simplify: • Divide both sides by −2 · a · b and simplify: • Apply the arccos function to both sides and simplify (note that cos and arccos are inverse functions): This final equation is just a manipulated form of the original equation, but because it has C alone on the left-hand-side and an expression that does not contain C on the right-hand-side, it is the solution.

 Algebra Coach Exercises

### Solving equations of the form a · b = 0

 Suppose that a and b are any expressions. Then the following statement is true: If a · b = 0 then either a = 0, or b = 0, or both. In other words a product can equal zero only if one of its factors is zero. This fact can be used to replace a “complicated” equation a · b = 0 by two simpler equations a = 0 and b = 0.

Note:
• It must be a product on the left-hand-side. Addition, subtraction or division will not work. For example a + b = 0 does not imply that a or b have any particular values.

• It must be zero on the right-hand-side. Any non-zero number will not work. For example a · b = 6 does not imply that a or b have any particular values.

Example: Solve the following equation for x:
(x − 1)(x + 2) = 0.
Solution: There are two factors on the left-hand-side of the equation:
• x − 1, and
• x + 2.
Any value of x that makes either of these factors equal to zero will make the left-hand-side of the equation equal to zero and is therefore a solution. To find these values we set each of the factors equal to zero and solve the resulting equations for x. Doing this gives these two equations:
• x − 1 = 0, and
• x + 2 = 0.
Both of these equations are of the type where the unknown occurs just once. To solve the first equation we add 1 to both sides and simplify and to solve the second equation we sutract 2 from both sides and simplify. The result is these two solutions:
• x = 1
• x = −2
We should still verify these solutions:
• Substitute 1 in for x in the original equation. This gives (1 − 1)(1 + 2) = 0, which simplifies to 0 = 0 so this solution checks out.

• Substitute −2 in for x in the original equation. This gives (−2 − 1)(−2 + 2) = 0, which simplifies to 0 = 0 so this solution checks out.

Example: Solve the following equation for x:
5 x (x + 1)(4 x + 1) = 0.
Solution: There are four factors on the left-hand-side of the equation:
• 5,
• x,
• x + 1, and
• 4 x + 1.
Any value of x that makes any one of these factors equal to zero will make the left-hand-side of the equation equal to zero and is therefore a solution. To find these values we set each of the factors equal to zero and solve the resulting equations for x. There are four corresponding equations:
• 5 = 0,
• x = 0,
• x + 1 = 0, and
• 4 x + 1 = 0.
The corresponding solutions are:
• this equation is a contradiction; it has no solution
• there is nothing to do - this equation is already solved; the solution is x = 0
• subtract 1 from both sides; this equation has the solution x = −1
• subtract 1, then divide by 4 on both sides; this equation has the solution x = −¼
Thus the solution set consists of three solutions: x = 0, x = −1 and x = −¼, and you can verify that all three check out.

Notes:
• Since the first factor, 5, obviously can’t equal zero you could simply ignore it and consider only the other three factors.

• Another approach is to begin this problem by dividing both sides of the original equation by 5. This gives the following equation in which the factor of 5 doesn’t appear:
x (x + 1)(4 x + 1) = 0.
But note that you can’t divide by any of the other factors, say by x, because that would amount to dividing by zero, which is an undefined operation. (Also it would cause you to lose that corresponding solution.)

• In a previous example we looked at the equation
20 x 3 + 25 x 2 + 5 x = 0.
We didn’t solve it but we did verify that it has the same three solutions as the present example, namely x = 0, x = −1 and x = −¼, by substituting them back in and checking. In fact the only difference between that previous example and our present example is that left-hand-side of the equation in this present example is in factored form and in the previous example it is not. In fact, manipulating equations into the form a · b = 0 is one of the most important applications of factoring.

 Algebra Coach Exercises

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