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Mathematics LibreTexts

3.5: Solve Equations Using Integers; The Division Property of Equality (Part 1)

  • Page ID
    4988
  • [ "article:topic", "authorname:openstax" ]

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    Skills to Develop

    • Determine whether an integer is a solution of an equation
    • Solve equations with integers using the Addition and Subtraction Properties of Equality
    • Model the Division Property of Equality
    • Solve equations using the Division Property of Equality
    • Translate to an equation and solve

    be prepared!

    Before you get started, take this readiness quiz.

    1. Evaluate x + 4 when x = −4. If you missed this problem, review Example 3.22.
    2. Solve: y − 6 = 10. If you missed this problem, review Example 2.33.
    3. Translate into an algebraic expression 5 less than x. If you missed this problem, review Table 1.34.

    Determine Whether a Number is a Solution of an Equation

    In Solve Equations with the Subtraction and Addition Properties of Equality, we saw that a solution of an equation is a value of a variable that makes a true statement when substituted into that equation. In that section, we found solutions that were whole numbers. Now that we’ve worked with integers, we’ll find integer solutions to equations.

    The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number or an integer.

    HOW TO: DETERMINE WHETHER A NUMBER IS A SOLUTION TO AN EQUATION.

    Step 1. Substitute the number for the variable in the equation.

    Step 2. Simplify the expressions on both sides of the equation.

    Step 3. Determine whether the resulting equation is true.

    • If it is true, the number is a solution.
    • If it is not true, the number is not a solution.

    Example 3.60:

    Determine whether each of the following is a solution of 2x − 5 = −13: (a) x = 4 (b) x = −4 (c) x = −9.

    Solution
    (a) Substitute 4 for x in the equation to determine if it is true. 2x − 5 = −13
    Substitute \(\textcolor{red}{4}\) for x. $$2(\textcolor{red}{4}) - 5 \stackrel{?}{=} -13$$
    Multiply. $$8 - 5 \stackrel{?}{=} -13$$
    Subtract. $$3 \neq -13$$

    Since x = 4 does not result in a true equation, 4 is not a solution to the equation.

    (b) Substitute -4 for x in the equation to determine if it is true. 2x − 5 = −13
    Substitute \(\textcolor{red}{-4}\) for x. $$2(\textcolor{red}{-4}) - 5 \stackrel{?}{=} -13$$
    Multiply. $$-8 - 5 \stackrel{?}{=} -13$$
    Subtract. $$-13 = -13 \; \checkmark$$

    Since x = −4 results in a true equation, −4 is a solution to the equation.

    (b) Substitute -9 for x in the equation to determine if it is true. 2x − 5 = −13
    Substitute \(\textcolor{red}{-9}\) for x. $$2(\textcolor{red}{-9}) - 5 \stackrel{?}{=} -13$$
    Multiply. $$-18 - 5 \stackrel{?}{=} -13$$
    Subtract. $$-23 \neq -13$$

    Since x = −9 does not result in a true equation, −9 is not a solution to the equation.

     

    Exercise 3.119:

    Determine whether each of the following is a solution of 2x − 8 = −14: (a) x = −11 (b) x = 11 (c) x = −3

    Exercise 3.120:

    Determine whether each of the following is a solution of 2y + 3 = −11: (a) y = 4 (b) y = −4 (c) y = −7

    Solve Equations with Integers Using the Addition and Subtraction Properties of Equality

    In Solve Equations with the Subtraction and Addition Properties of Equality, we solved equations similar to the two shown here using the Subtraction and Addition Properties of Equality. Now we can use them again with integers.

    $$\begin{split} x + 4 & = 12 \qquad \qquad \qquad y - 5 = 9 \\ x + 4 \textcolor{red}{-4} & = 12 \textcolor{red}{-4} \qquad \; \; y - 5 \textcolor{red}{+5} = 9 \textcolor{red}{+5} \\ x & = 8 \qquad \qquad \qquad \qquad \; y = 14 \end{split}$$

    When you add or subtract the same quantity from both sides of an equation, you still have equality.

    Definition: Properties of Equalities

    Subtraction Property of Equality Addition Property of Equality
    For any numbers a, b, c, if a = b then a − c = b − c. For any numbers a, b, c, if a = b then a + c = b + c.

    Example 3.61:

    Solve: y + 9 = 5.

    Solution
    Subtract 9 from each side to undo the addition. $$y + 9 \textcolor{red}{-9} = 5 \textcolor{red}{-9}$$
    Simplify. $$y = -4$$

    Check the result by substituting −4 into the original equation.

    Substitute −4 for y $$-4 + 9 \stackrel{?}{=} 5$$
      $$5 = 5 \; \checkmark$$

    Since y = −4 makes y + 9 = 5 a true statement, we found the solution to this equation

    Exercise 3.121:

    Solve: y + 11 = 7

    Exercise 3.122:

    Solve: y + 15 = −4

    Example 3.62:

    Solve: a − 6 = −8

    Solution
    Add 6 to each side to undo the subtraction. $$a - 6 \textcolor{red}{+6} = -8 \textcolor{red}{+6}$$
    Simplify. $$a = -2$$
    Check the result by substituting −2 into the original equation. $$a - 6 = -8$$
    Substitute −2 for a. $$-2 - 6 \stackrel{?}{=} -8$$
      $$-8 = -8 \; \checkmark$$

    The solution to a − 6 = −8 is −2. Since a = −2 makes a − 6 = −8 a true statement, we found the solution to this equation.

    Exercise 3.123:

    Solve: a − 2 = −8

    Exercise 3.124:

    Solve: n − 4 = −8

    Model the Division Property of Equality

    All of the equations we have solved so far have been of the form x + a = b or x − a = b. We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division. We will model an equation with envelopes and counters in Figure 3.21.

    Figure 3.21

    Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

    To determine the number, separate the counters on the right side into 2 groups of the same size. So 6 counters divided into 2 groups means there must be 3 counters in each group (since 6 ÷ 2 = 3).

    What equation models the situation shown in Figure 3.22? There are two envelopes, and each contains x counters. Together, the two envelopes must contain a total of 6 counters. So the equation that models the situation is 2x = 6.

    Figure 3.22

    We can divide both sides of the equation by 2 as we did with the envelopes and counters.

    $$\begin{split} \frac{2x}{\textcolor{red}{2}} & = \frac{6}{\textcolor{red}{2}} \\ x & = 3 \end{split}$$

    We found that each envelope contains 3 counters. Does this check? We know 2 • 3 = 6, so it works. Three counters in each of two envelopes does equal six. Figure 3.23 shows another example.

    Figure 3.23

    Now we have 3 identical envelopes and 12 counters. How many counters are in each envelope? We have to separate the 12 counters into 3 groups. Since 12 ÷ 3 = 4, there must be 4 counters in each envelope. See Figure 3.24.

    Figure 3.24

    The equation that models the situation is 3x = 12. We can divide both sides of the equation by 3.

    $$\begin{split} \frac{3x}{\textcolor{red}{3}} & = \frac{12}{\textcolor{red}{3}} \\ x & = 4 \end{split}$$

    Does this check? It does because 3 • 4 = 12.

    Example 3.63:

    Write an equation modeled by the envelopes and counters, and then solve it.

    Solution

    There are 4 envelopes, or 4 unknown values, on the left that match the 8 counters on the right. Let’s call the unknown quantity in the envelopes x.

    Write the equation. $$4x = 8$$
    Divide both sides by 4. $$\frac{4x}{\textcolor{red}{4}} = \frac{8}{\textcolor{red}{4}}$$
    Simplify. $$x = 2$$

    There are 2 counters in each envelope.

    Exercise 3.125:

    Write the equation modeled by the envelopes and counters. Then solve it.

    Exercise 3.126:

    Write the equation modeled by the envelopes and counters. Then solve it.

    Solve Equations Using the Division Property of Equality

    The previous examples lead to the Division Property of Equality. When you divide both sides of an equation by any nonzero number, you still have equality.

    Definition: Division Property of Equality

    For any numbers a, b, c, and c ≠ 0,

    $$If\; a = b\; then\; \frac{a}{c} = \frac{b}{c} \ldotp$$

    Example 3.64:

    Solve: 7x = −49.

    Solution

    To isolate x, we need to undo multiplication.

    Divide each side by 7. $$\frac{7x}{\textcolor{red}{7}} = \frac{-49}{\textcolor{red}{7}}$$
    Simplify $$x = -7$$

    Check the solution.

    Substitute −7 for x. $$7(-7) \stackrel{?}{=} -49$$
      $$-49 = -49 \; \checkmark$$

    Therefore, −7 is the solution to the equation.

    Exercise 3.127:

    Solve: 8a = 56

    Exercise 3.128:

    Solve: 11n = 121

    Example 3.65:

    Solve: −3y = 63.

    Solution

    To isolate y, we need to undo the multiplication.

    Divide each side by −3. $$\frac{-3y}{\textcolor{red}{-3}} = \frac{63}{\textcolor{red}{-3}}$$
    Simplify. $$y = -21$$

    Check the solution.

    Substitute −21 for y. $$-3(-21) \stackrel{?}{=} 63$$
      $$63 = 63 \; \checkmark$$

    Since this is a true statement, y = −21 is the solution to the equation.

    Exercise 3.129:

    Solve: −8p = 96

    Exercise 3.130:

    Solve: −12m = 108

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