Solve Equations Using Integers; The Division Property of Equality

3.5 Solve Equations Using Integers; The Division Property of Equality

The topics covered in this section are:

  1. Determine whether an integer is a solution of an equation
  2. Solve equations with integers using the Addition and Subtraction Properties of Equality
  3. Model the Division Property of Equality
  4. Solve equations using the Division Property of Equality
  5. Translate to an equation and solve

3.5.1 Determine Whether a Number is a Solution of an Equation

In Solving Equations Using 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: How to determine whether a number is a solution to an equation.

  1. Substitute the number for the variable in the equation.
  2. Simplify the expressions on both sides of the equation.
  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 1

Determine whether each of the following is a solution of $2x-5=-13$:

  1. $x=4$
  2. $x=-4$
  3. $x=-9$
Solution

Part 1. Substitute $4$ for $x$ in the equation to determine if it is true.$2x-5=-13$
Substitute $4$ for $x$.$2(4)-5 \stackrel{?}{=} -13$
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.

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

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

Part 3. Substitute $-9$ for $x$ in the equation to determine if it is true.$2x-5=-13$
Substitute $-9$ for $x$.$2(-9)-5 \stackrel{?}{=} -13$
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.

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

In Solving Equations Using 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.

This figure has two columns. The first column has the equation x plus 4 equals 12. Underneath there is x plus 4 minus 4 equals 12 minus 4. Under this there is x equals 8. The second column has the equation y minus 5 equals 9. Underneath there is the equation y minus 5 plus 5 equals 9 plus 5. Under this there is y equals 14.

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

PROPERTIES OF EQUALITIES

Subtraction Property of EqualityAddition Property of Equality
For any number $a,b,c$,
if $a=b$ then $a-c=b-c$
For any numbers $a,b,c$,
if $a=b$ then $a+c=b+c$

Example 2

Solve: $y+9=5$.

Solution

$y+9=5$
Subtract $9$ from each side to udo the addition.$y+9-9=5-9$
Simplify.$y=-4$

Check the result by substituting $-4$ into the original equation.

$y+9=5$
Substitute $-4$ for $y$.$-4+9 \stackrel{?}{=} 5$
$5=5$✓

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

Example 3

Solve: $a-6=-8$

Solution

$a-6=-8$
Add $6$ to each side to undo the subtraction.$a-6+6=-8+6$
Simplify.$a=-2$
Check the result by substituting $-2$ into the original equation:$a-6=-8$
Substitute $-2$ for $a$$-2-6 \stackrel{?}{=} -8$
$-8=-8$✓

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.

3.5.3 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 the figure below.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.

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 $6 \div 2=3$)

What equation models the situation show in the figure below? 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 $2x-6$.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.

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

This figure has two rows. The first row has the equation 2x divided by 2 equals 6 divided by 2. The second row has the equation x equals 3.

We found that each envelope contains $3$ counters. Does this check? We know $2 \cdot 3 = 6$, so it works. Three counters in each of two envelopes does equal six.

The figure below shows another example.

This image has two columns. In the first column are three envelopes. In the second column there are four rows of  three blue circles. Underneath the image is the equation 3x equals 12.

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 $12 \div 3 = 4$, there must be $4$ counters in each envelope. See the figure below.

This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.

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

This image shows the equation 3x divided by 3 equals 12 divided by 3. Below this equation is the equation x equals 4.

Does this check? It does because $3 \cdot 4 = 12$.

Example 4

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

This image has two columns. In the first column are four envelopes. In the second column there are 8 blue circles.
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 $4$.$\frac{4x}{4} = \frac{8}{4}$
Simplify.$x=2$

There are $2$ counters in each envelope.

3.5.4 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.

DIVISION PROPERTY OF EQUALITY

For any nuymbers $a,b,c$, and $c \neq 0$,

If $a=b$ then $\frac{a}{c} = \frac{b}{c}$.

Example 5

Solve: $7x=-49$.

Solution

To isolate $x$, we need to undo multiplication.

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

Check the solution.

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

Therefore, $-7$ is the solution to the equation.

Example 6

Solve: $-3y=63$.

Solution

To isolate $7$, we need to undo multiplication.

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

Check the solution.

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

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

3.5.5 Translate to an Equation and Solve

In the past several examples, we were given an equation containing a variable. In the next few examples, we’ll have to first translate word sentences into equations with variables and then we will solve the equations.

Example 7

Translate and solve: five more than $x$ is equal to $-3$.

Solution

five more than $x$ is equal to $-3$
Translate$x+5=-3$
Subtract $5$ from both sides.$x+5-5=-3-5$
Simplify.$x=-8$

Check the answer by substituting it into the original equation.

$x+5=-3$

$-8+5 \stackrel{?}{=} -3$

$-3=-3$✓

Example 8

Translate and solve: the difference of $n$ and $6$ is $-10$.

Solution

the difference of $n$ and $6$ is $-10$
Translate.$n-6=-10$
Add $6$ to each side.$n-6+6=-10+6$
Simplify.$n=-4$

Check the answer by substituting it into the original equation.

$n-6=-10$

$-4-6 \stackrel{?}{=} -10$

$-10=-10$✓

Example 9

Translate and solve: the number $108$ is the product of $-9$ and $y$.

Solution

the number of $108$ is the product of $-9$ and $y$
Translate.$108=-9y$
Add $6$ to each side.$\frac{108}{-9} = \frac{-9y}{-9}$
Simplify.$-12=y$

Check the answer by substituting it into the original equation.

$108=-9y$

$108 \stackrel{?}{=} -9(-12)$

$108=108$✓

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