# Steps for Solving Equations

 Steps for Solving Equations

To perform these steps you will need to use a number of mathematical properties of addition, subtraction, multiplication, and division. The use of these properties, both in combining like terms and isolating terms and variables, will be reviewed in this unit. In the next unit we will solve equations containing two variables, called multivariate equations.

### Step 1: Combine Like Terms

As we learned in the last unit, like terms are terms that contain the same variable or group of variables raised to the same exponent, regardless of their numerical coefficient. Keeping in mind that an equation is a mathematical statement that two expressions are equal, in this step we will focus on combining like terms for the two expressions contained in an equation. Since this unit deals only with equations containing a single variable, there are not many like terms to deal with. Let's look at an example. If we are given the equation 3z + 5 +2z = 12 + 3z, we need to first combine like terms in each expression of this equation.

 The two expressions in this equation are3z + 5 +2z and 12 + 4z. 3z + 5 +2z = 12 + 4z There are three terms that contain the variable z: 3z, 2z, and 4z. We combine 3z, and 2z on the left side of the equation, then subtract 4z from both sides. (3z +2z) + 5 = 12 + 4z 5z + 5 – 4z = 12 + 4z – 4z z + 5 = 12

Notice we chose to subtract 4z from both sides rather than 5z. We chose to do this because consolidating in this manner left z positive. However, subtracting 5z from both sides would also be correct.

 See how it works when we subtract 5z from both sides. 5z + 5 – 5z = 12 + 4z – 5z 5 = 12 – z

### Step 2: Isolate the Terms that Contain the Variable

The main idea in solving equations is to isolate the variable you want to solve for. This means we want to get terms containing that variable on one side of the equation, with all other variables and constants "moved" to the opposite side of the equation. This section will address how we "move" terms from one side of an equation to another, in order to isolate a variable, using addition and its inverse property of subtraction.

 The Addition Property of Equality and Its Inverse Property of Subtraction If a = b, then a + c = b + c If a = b, then a – d = b – d In other words, adding the same quantity to both sides of an equation produces an equivalent equation. Since subtraction is simply adding a negative number, this rule applies when subtracting the same quantity from both sides.

Let's try this with the example: x + 3 = 15.

The key to solving this equation is to isolate x. On the left side of the equation, x is added to 3. To undo this addition we must subtract 3 from both sides of the equation. It is important that we subtract 3 from both sides of the equation, otherwise we will lose equality. We now have:

 Subtract 3 from both sides. (x + 3) – 3 = 15 – 3 x = 2

Here we solved for x by isolating it through the use of inverse operations or in other words, by using opposite operations. Addition and subtraction are examples of inverses, and so are multiplication and division.

We can easily check the result we found above by substituting 12 for x in to the original equation\

 The 12 works in this equation, so the answer is correct. 12+ 3 = 15

The example we started in step one, 3z + 5 +2z = 12 + 4z, is an example of an equation that contains more than one term with a variable. In step one we combined all terms containing the variable z. In this step we want to isolate the variable z. To isolate z, choose one term containing the variable z (usually the smaller one) to subtract from both sides.

 Subtract 5 from both sides to isolate the z. z + 5 – 5 = 12 – 5 This gives us our final result. z = 7

To be sure our answer is correct, we can check it by substituting the solution back to the original equation, 3z + 5 +2z = 12 + 4z:

 The left side becomes: 3 (7) + 5 + 2 (7)= 21 + 5 + 14 = 40 The right side becomes: 12 + 4 (7) = 12 + 28 = 40

Notice that the right and left sides are equal, therefore we have the correct solution. Now let’s look at some examples of using addition and subtraction to solve equations.

### Example

Solve the following equations for the variable in the equation.

1. 38 = z + 15
2. 9x + 3 = 8x + 19

Solve the following equations for the variable in the equation.

 1.   38 = z + 15 23 = z 2.   9x + 3 = 8x + 19 x = 16

Solve the following equations for the variable in the equation.

1. 38 = z + 15 23 = z

For this problem, we wish to isolate z on one side of the equation.

 Notice that in the original equation, 15 is added to z, therefore we must perform the operation that is the inverse of addition. This is subtraction. We must subtract 15 from both sides of the equation. 38 = z + 15 38 – 15 = z + 15 – 15 23 = z

Again, to check your answer, replace the variable with the solution in the original equation

 The left side becomes: 38 The right side becomes: (23) + 15 = 38

Both sides are equal, so our solution is correct.

2. 9x + 3 = 8x + 19 x = 16

In this equation we want to isolate the variable x on one side of the equation. Note that x occurs on both sides of the equation.

 First apply the subtraction rule to the variables. We can subtract 8x from both sides. (You could have chosen to subtract 9x from both sides, but 8x was chosen because it’s often useful to subtract the one with the smaller coefficient since the result will be positive.) 9x + 3 = 8x + 19 9x + 3 – 8x = 8x + 19 – 8x 9x – 8x + 3 = 8x – 8x + 19 x + 3 = 19 Once we isolate the x variable on one side of the equation, we apply the subtraction property to the constants. x + 3 – 3 = 19 – 3 x = 16

Again, to check your answer, replace the variable with the solution in the original equation.

 The left side becomes: 9 (16) + 3 = 144 + 3 = 147 The right side becomes: 8 (16) + 19 = 128 + 19 = 147

Both sides are equal, so our solution is correct.

### Step 3: Isolate The Variable You Wish To Solve For

In the examples in the previous section, by isolating the terms containing the variable we wished to solve for, we were left with a term that had a numerical coefficient of one, so the variable was automatically isolated. However, if the variable does not have a coefficient of one, we will need to isolate the variable itself. When the variable we wish to isolate is either multiplied or divided by a numerical coefficient or other variables that are not equal to one, we need to use either multiplication or division to isolate the variable.

 The Multiplication Property of Equality and The Inverse Operation of Division If a = b, then ac = bc where c ¹ 0 If c = d, then c/e = d/e where e ¹ 0 Multiplying both sides of an equation by the same nonzero number produces an equivalent equation. We may adapt this property to state that if we divide both sides of an equation by the same nonzero number, we obtain an equivalent equation. This fact follows from knowing that multiplying by the reciprocal of a number is the same thing as dividing by that number. c • 1/e is equivalent to c ÷ e

For example, suppose you want to solve the following equation for x:

3x = 72

To solve this problem we want to isolate the x variable. Since the numerical is multiplied by a numerical coefficient, we can’t use addition or subtraction to do this.

 Since x is multiplied by 3, we should use the inverse operation of multiplication to isolate x. That means we want to divide by 3. To isolate x, we simply divide both sides of the equation by 3.

Again, what we do to one side of the equation, we must also do to the other side of the equation. In this example we divided both sides of the equation by 3.

Using the inverse operations procedure, work through the examples below.

### Example

Solve each of the equations for x.

1. 6 = x/3
2. (2/5) x = 8

Solve each of the equations for x.

 1.   6 = x/3 18 = x 2.   (2/5) x = 8 x = 20

Solve each of the equations for x.

 1.   6 = x/3 18 = x

 In this problem, the variable x is divided by 3. To isolate x, you need to perform the inverse operation of division, which is multiplication. This means we must multiply both sides of the equation by 3. To check, replace the variable with the solution in the original equation.

 2.   (2/5) x = 8 x = 20

 In this problem, the variable x is multiplied by 2 and divided by 5. This means we must both divide the variable by 2 and multiply by 5 to isolate x. This is the same as multiplying by the fraction 5/2. Again we should check our result by substituting the value we got for x back into the equation.

Every answer should be checked to be sure it is correct. Substitution is a process of replacing variables with numbers or expressions. By substituting, we switch or exchange values, often replacing a variable with a numerical value. We often substitute to minimize the number of variables in an expression, or to actually evaluate the expression or equation. For example, we are given the following:

a + 12 = b,   and a = 9,   find the value for b.

We can determine the value for the expression a + 12 by using substitution. Since we are given a = 9, we can substitute 9 for a in the original expression:

9 + 12 = b
21 = b

After finding the solution for a variable, substitute the answer into the original equation to be sure the equality holds true. For example, in an earlier section we solved the following equation for x:

3x = 72

We found the solution was x = 24. To check and be sure this solution is correct we substitute 24 for x in the original equation.

3x = 72   and   x = 24.
3(24) = 72
72 = 72

In each of the examples shown in this unit we have been doing this step so you should be familiar with seeing this.

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