**Basics of Expressions / Equations**

**Click here for definitions of terms**

**Contents**

**Sample Expression/Equation Problems**

*Expressions* are groups of mathematical symbols that can be written and manipulated according to definite rules, for simplifying or for identifying relationships. For example, (a + 2b)(a – 2b) is an expression. *Equations* consist of expressions on the two sides of an equals sign, which can be manipulated in order to find the value of one or more variables or “unknowns.” Word problems are usually solved through the formulation and solution of one or more equations.

**Expressions**

**Polynomials**

Most expressions that occur in algebra involve polynomials. A *polynomial *is an algebraic expression consisting of multiple *terms* connected by plus (+) or minus (-) signs. For example, x³ + 3x – 4 is a polynomial with 3 terms. The first term, x³ has a variable, x, and x has the exponent 3. This exponent indicates that x is “raised to the third power.” The second term, 3x, consists of the variable x and its *coefficient*, 3. The third term is -4, and is a constant term.

**Factoring Polynomials**

A polynomial can be broken into simpler pieces, called *factors,* which when multiplied together result in the original polynomial. This *factorization* is an expression which shows the polynomial as a product of two or more smaller polynomials. For example, the following is a factorization of a polynomial:

\(\begin{align} x^2 – 6x + 8 = (x – 2)(x – 4)\end{align}\)

The expression on the right of the equals sign is a factorization of the expression on the left. It consists of two factors: x – 2, and x -4.

Factoring polynomials is very useful in solving equations of which polynomials are a part.

**Expanding Factored Polynomials**

Suppose we start with the expression (x – 2)(x – 4), and we want to expand it to a single polynomial. Then we use the FOIL method: Multiply the First parts, then the Outside parts, then the Inside parts, and finally the Last parts.

(x – 2)(x – 4) = x·x – 4·x – 2·x + (-2)(-4) = x² – 6x + 8

**Simplifying Expressions**

Expressions involving polynomials can often be simplified. For example, suppose we have an algebraic expression involving two polynomials shown in a division relationship:

\(\begin{align}\frac{ x^2 – 6x + 8}{x^2 – 3x + 2}\end{align}\)

These two polynomials can be factored as:

\(\begin{align}\frac{ (x – 2)(x – 4)}{(x – 2)(x – 1)}\end{align}\)

Since both numerator and denominator contain the polynomial factor (x – 2), these can be canceled, and we’re left with:

\(\begin{align}\frac{(x – 4)}{(x – 1)}\end{align}\)

**Simplifying Expressions Containing Radicals**

Expressions containing square roots, or radicals, occur often and should be simplified where possible. As an example:

\(\begin{align}9\sqrt{50} + \sqrt{18}\end{align}\)

This expression contains the square roots of two numbers that don’t look related and therefore can’t be combined as they stand. But if we take the prime factorization of each of these two numbers, 50 and 18, we get:

\(\begin{align}9\sqrt{2·5²} + \sqrt{2·3²}\end{align}\)

which shows that each of these equals 2 times a number that is a square. Taking the square root of these two squares, we get

\(\begin{align}9·5\sqrt{2} + 3\sqrt{2}\end{align}\)

which finally results in

\(\begin{align}48\sqrt{2}\end{align}\)

And this is much simpler than the original.

**Sample Expression/Equation Problems**

**Equations**

An *equation* is an algebraic statement that shows two algebraic expressions separated by an equals sign. This signifies that the value on the left of the equals sign is the same as the value on the right.

Example: 3x – 5 = 2x + 9.

This equations states that the expression on the left, 3x – 5, is equal to the expression on the right, 2x + 9. When we solve this equation, we determine that these two expressions can only be equal when x = 14.

We will only deal with two types of equations: Linear Equations and Quadratic Equations.

**Equivalent Equations**

Two equations are *equivalent* if one can be transformed to the other by a simple operation, such as multiplying both sides of the equation by the same value. Example: 2x + 7 = y is equivalent to 4x + 14 = 2y, because if we multiply both sides of the first equation by 2, then we obtain the second equation. If one equation is equivalent to another, then the second equation gives us no new information that we didn’t already have in the first equation. Every solution to one equation is also a solution to the other. If two equations are not equivalent, then they are said to be *independent.*

**Sample Expression/Equation Problems**

**Linear Equations**

A *linear equation* is an equation with x, or x and y, as variables, but with every *variable* having an exponent of 1. Examples:

\(\begin{align}4x + 2 = 6\end{align}\)

\(\begin{align}5x – 3y = 17\end{align}\)

They are called “linear” because the graph of the second of these equations on the coordinate plane is a straight line.

If there is only a single equation, and it has only one variable or *unknown *it can be solved – that is, the value of the unknown can be determined – if the equation is consistent and is not an identity. For example, consider the equations:

x + 2 = x +3

2(3x + 5) = 6x + 10

The first “equation” is inconsistent, because if we subtract x from both sides we get 2 = 3, which is false. So the first is not a legitimate equation and has no solution. The second equation is actually an identity, since expanding the left side by multiply 2 by (3x + 5) just gives us 6x + 10, which is the right side. This equation is true for all values of x, so it has an infinite number of solutions. In fact, that’s an easy way to recognize an identity: If you always obtain a constant on the left of the equation and the same constant on the right no matter what value you plug in for x, then the equation is an identity.

If the equation is consistent and not an identity, then it has a solution, so, for example, we solve this equation as shown:

4x + 2 = 6 +x

4x + 2 – 2 = 6 + x – 2 Subtract 2 from both sides

4x = x + 4

4x – x = x – x + 4 Subtract x from both sides

3x = 4

3x/3 = 4/3 Divide both sides by 3

x = 4/3

Linear equations with one unknown can normally be solved by the repeated application of this principle: Add, Subtract, Multiply, or Divide both sides by the same number. This is done until the equation shows x on one side and a real number on the other.

**Simultaneous Linear Equations**

If there are two or more equations with multiple unknowns, then we have *simultaneous linear equations.* These are multiple linear equations in several variables, each of which has the same value in each equation. If the number of *independent* linear equations is the same as the number of variables or unknowns in the equations, then the values of the unknowns can usually be determined. Example:

3x – 7 = y + 4

2x + 6 = 4y

In this system there are 2 equations in 2 unknowns. The solution to the system of equations is x = 5 and y = 4.

The exception to this rule is illustrated by the following pair of equations:

3x + y = 7

9x + 3y = 10

In this case, if we multiply both sides of the first equation by 3, the two equations together look like this:

9x + 3y = 21

9x + 3y = 10

Obviously, 9x + 3y cannot be equal to both 21 and 10, so there is no solution for x and y that will satisfy both these equations. An interpretation of these equations in coordinate geometry illustrates why there is no solution. The graph of these two equations on the coordinate plane shows two parallel lines. They don’t have a common solution because the two lines never intersect.

**Solving Simultaneous Linear Equations**

There are several ways of solving simultaneous linear equations. Let’s look again at the two equations given above.

3x – 7 = y + 4

2x + 6 = 4y

We show two standard approaches:

**Method 1 – Substitution:** In this approach we first solve one of the equations for one of the variables in terms of the other. We’ll use the first equation and solve for y in terms of x:

3x – 7 = y + 4

3x – 7 -4 = y + 4 – 4

3x – 11 = y

Now we’ll substitute 3x – 11 for y in the *other* equation, and then we’ll solve for x.

2x + 6 = 4y

2x + 6 = 4(3x – 11)

2x + 6 = 12x – 44

2x + 6 + 44 = 12x – 44 + 44

2x + 50 = 12x

2x – 2x + 50 = 12x – 2x

50 = 10x

5 = x

We have the value for x. To obtain the value for y, we substitute 5 for x in the equation:

3x – 11 = y

3·5 – 11 = y

4 = y

This solution is easily validated by plugging 5 for x and 4 for y in the original equations.

**Method 2 – Adding or Subtracting Equations:** In this approach we quickly eliminate one of the variables by subtracting one equation from the other. We start with the 2 equations:

3x – 7 = y + 4

2x + 6 = 4y

To make the comparable, we multiply both sides of the first equation by 4.

12x – 28 = 4y + 16

2x + 6 = 4y

Now we subtract the left side of equation 2 from the left side of equation 1, and the right side of equation 2 from the right side of equation 1. This eliminates y. We then solve the resulting equation for x:

12x – 28 = 4y + 16

2x + 6 = 4y

———————-

10x – 34 = 16

10x = 16 + 34 = 50

x = 5

This is the same value we got for x when we used the substitution method. If we now plug that value into either equation and solve for y, we get y = 4.

**Sample Expression/Equation Problems**

**Quadratic Equations**

Quadratic equations are the simplest of the non-linear polynomial equations with one variable. Here are two examples of non-linear polynomial equations:

x² – 3x + 2 = 0

2x³ + 3x² – x + 15 = 0

Please note two things about these equations as we’ve written them. First, they both consist of a polynomial on the left and 0 on the right of the equals sign. Any equation consisting of polynomial terms and constants can be put into this form by simply moving everything from the right to the left, changing signs from + to -, or vice versa, as needed.

Second, we’ve arranged the individual terms of each polynomial in descending order according to the power of the variable in the term, with the constant term last. Thus, the order of the powers of the variable in the first equation is 2, 1, 0, and the order in the second equation is 3, 2, 1, 0. Again, this ordering of terms by their powers can be done with every polynomial. We just have to make sure we preserve the + and – signs as they apply to each individual term.

Quadratic equations are those polynomial equations in which the highest power of the variable is 2. Therefore, the first equation listed above is a quadratic equation, while the second is not. We will look at two ways of solving these types of equations – by factoring and by using the quadratic formula.

**Solving Equations by Factoring**

Consider the first equation listed above.

x² – 3x + 2 = 0

The polynomial in this equation can be factored as follows.

x² – 3x + 2 = (x – 1)(x – 2)

Thus, we replace the original polynomial with its factored version.

x² – 3x + 2 = 0

(x – 1)(x – 2) = 0

Now, the product of any two real numbers will be zero if and only if one or both of them is zero. Therefore, we can set each of the factors to 0, solve for x, and obtain the two *solutions* to – or *roots* of – this equation.

x – 1 = 0

x = 1

x – 2 = 0

x = 2

As a second example, look at this equation:

x³ + 4x² – 5x = 0

Note that this is not a quadratic, since the highest power of x is 3. But notice also that the variable x appears in every term, so we can factor it out of each term and list it separately. Therefore, in this case, the factorization process occurs in two steps.

x³ + 4x² – 5x = 0

x(x² + 4x – 5) = 0

x(x + 5)(x – 1) = 0

Now we set all 3 factors equal to 0 to determine the 3 roots of the original equation.

x = 0

x + 5 = 0

x = -5

x – 1 = 0

x = 1

**Solving Equations with the Quadratic Formula**

Any quadratic equation can be solved by using the *quadratic formula*. This formula assumes the quadratic equation has been put in this format:

\(\begin{align}ax^2 + bx + c = 0\end{align}\)

where a is the coefficient of the x² term, b is the coefficient of the x term, and c is the constant term. Of course, we assume a is not zero, but either b or c or both can be zero. Here is an example of such a quadratic equation.

\(\begin{align}2x^2 + 8x – 10 = 0\end{align}\)

The general quadratic formula for solving any quadratic equation is

\(\begin{align} x = \frac { – b \pm \sqrt {b^2 – 4ac} } {{2a}}\end{align}\)

Notice the ± in the middle of the numerator. This means that there are actually two solutions given here: one when the sign is positive (+) and the other when the sign is negative (-). Notice also that the entire solution is given in terms of the constant terms a, b, and c. Let’s plug these values in for the equation we gave above. If

\(\begin{align}2x^2 + 8x – 10 = 0\end{align}\)

then the two values for x are given by

\(\begin{align} x = \frac { – 8 \pm \sqrt {8^2 – 4·2·(-10)} } {{2·2}}\end{align}\)

\(\begin{align} x = \frac { – 8 \pm \sqrt {64 + 80} } {{4}}\end{align}\)

\(\begin{align} x = \frac { – 8 \pm \sqrt {144} } {{4}}\end{align}\)

\(\begin{align} x = \frac { – 8 \pm 12 } {{4}}\end{align}\)

\(\begin{align} x = – 2 \pm 3\end{align}\)

So the two solutions are

\(\begin{align} x = – 2 + 3 = 1\end{align}\)

\(\begin{align} x = – 2 – 3 = -5\end{align}\)

Here’s a second example:

\(\begin{align}x^2 + 6x – 10 = 0\end{align}\)

\(\begin{align} x = \frac { – 6 \pm \sqrt {6^2 – 4·1·(-10)} } {{2·1}}\end{align}\)

\(\begin{align} x = \frac { – 6 \pm \sqrt {36 + 40} } {{2}}\end{align}\)

\(\begin{align} x = \frac { – 6 \pm \sqrt {76} } {{2}}\end{align}\)

\(\begin{align} x = \frac { – 6 \pm 2\sqrt{19} } {{2}}\end{align}\)

\(\begin{align} x = – 3 \pm \sqrt{19} \end{align}\)