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Solve Equations Using Integers; The Division Property of Equality

Determine whether an integer is a solution of an equation

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.

Example

Try it.

Determine whether each of the following is a solution of \(2x-5=-13\text{:}\)

ⓐ \(\ x=4\\)
ⓑ \(\ x=-4\\)
ⓒ \(\ x=-9.\)

Solution

ⓐ Substitute 4 for x in the equation to determine if it is true.


Multiply.
Subtract.

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

ⓑ Substitute −4 for x in the equation to determine if it is true.

Multiply.
Subtract.

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

ⓒ Substitute −9 for x in the equation to determine if it is true.

Substitute −9 for x.
Multiply.
Subtract.

Since \(x=-9\) does not result in a true equation, \(-9\) is not a solution to the equation.

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.

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

Example

Try it.

Solve: \(y+9=5.\)

Solution


Subtract 9 from each side to undo the addition.
Simplify.

Check the result by substituting \(-4\) into the original equation.

	\(y+9=5\\)
Substitute −4 for y	\(-4+9\overset{?}{=}5\\)
	\(5=5✓\)

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

Example

Try it.

Solve: \(a-6=-8\)

Solution


Add 6 to each side to undo the subtraction.
Simplify.
Check the result by substituting \(-2\) into the original equation:
Substitute \(-2\) for \(a\)

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.

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 .

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 shown in ? 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 \(2x=6.\)

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

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

Example

Try it.

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

Solution

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

Write the equation.
Divide both sides by 4.
Simplify.

There are \(\text{2 counters}\) in each envelope.

Condensed — the full section is in OpenStax Prealgebra 2e.

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.

Example

Try it.

\(\text{Solve:}\ 7x=-49.\)

Solution

To isolate \(x,\) we need to undo multiplication.


Divide each side by 7.
Simplify.

Check the solution.

	\(7x=-49\\)
Substitute −7 for x.	\(7(-7)\overset{?}{=}-49\\)
	\(-49=-49✓\)

Therefore, \(-7\) is the solution to the equation.

Example

Try it.

Solve: \(-3y=63.\)

Solution

To isolate \(y,\) we need to undo the multiplication.


Divide each side by −3.
Simplify

Check the solution.

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

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

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

Try it.

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.

\(\begin{array}{l} \\ \\ x+5=-3\ \\ -8+5\overset{?}{=}-3\ \\ -3=-3✓\end{array}\)

Example

Try it.

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.

\(\begin{array}{l} \\ n-6=-10\ \\ -4-6\overset{?}{=}-10\ \\ -10=-10✓\end{array}\)

Example

Try it.

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.	\(108=-9y\)
Divide by \(-9\).	\(\frac{108}{-9}=\frac{-9y}{-9}\)
Simplify.	\(-12=y\)

Check the answer by substituting it into the original equation.

\(\begin{array}{l}108=-9y \\ 108\overset{?}{=}-9(-12) \\ 108=108✓\end{array}\)

Key Concepts

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.

Properties of Equalities

Subtraction Property of Equality	Addition Property of Equality
\(\text{For any numbers}\ a,b,c,\) \(\text{if}\ a=b\ \text{then}\ a-c=b-c.\)	\(\text{For any numbers}\ a,b,c,\) \(\text{if}\ a=b\ \text{then}\ a+c=b+c.\)

Division Property of Equality
- For any numbers \(a,b,c,\) and \(c\ne 0\)
  If \(a=b\), then \(\frac{a}{c}=\frac{b}{c}\).

Chapter Practice Test

In the following exercises, compare the numbers, using \(<\ \text{or}\ >\ \text{or}\ \text{=.}\)

In the following exercises, find the opposite of each number.

In the following exercises, simplify.

In the following exercises, evaluate.

In the following exercises, translate each phrase into an algebraic expression and then simplify, if possible.

In the following exercises, solve.

Condensed — the full section is in OpenStax Prealgebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

\(\text{Evaluate}\ x+4\ \text{when}\ x=-4.\)
If you missed this problem, review .

Kusonyeza yankho

\(0\)
\(\text{Solve:}\ y-6=10.\)
If you missed this problem, review .

Kusonyeza yankho

\(16\)
Translate into an algebraic expression \(5\) less than \(x.\)
If you missed this problem, review .

Kusonyeza yankho

\(x-5\)
Determine whether each of the following is a solution of \(2x-5=-13\text{:}\)
1. ⓐ \(\ x=4\\)
2. ⓑ \(\ x=-4\\)
3. ⓒ \(\ x=-9.\)
Kusonyeza yankho

ⓐ Substitute 4 for x in the equation to determine if it is true.
Multiply.
Subtract.
Since \(x=4\) does not result in a true equation, \(4\) is not a solution to the equation.
ⓑ Substitute −4 for x in the equation to determine if it is true.
Multiply.
Subtract.
Since \(x=-4\) results in a true equation, \(-4\) is a solution to the equation.
ⓒ Substitute −9 for x in the equation to determine if it is true.
Substitute −9 for x.
Multiply.
Subtract.
Since \(x=-9\) does not result in a true equation, \(-9\) is not a solution to the equation.
Determine whether each of the following is a solution of \(2x-8=-14\text{:}\)
1. ⓐ \(\ x=-11\\)
2. ⓑ \(\ x=11\\)
3. ⓒ \(\ x=-3\)
Kusonyeza yankho
1. ⓐ no
2. ⓑ no
3. ⓒ yes
Determine whether each of the following is a solution of \(2y+3=-11\text{:}\)
1. ⓐ \(\ y=4\\)
2. ⓑ \(\ y=-4\\)
3. ⓒ \(\ y=-7\)
Kusonyeza yankho
1. ⓐ no
2. ⓑ no
3. ⓒ yes
Solve: \(y+9=5.\)

Kusonyeza yankho

Subtract 9 from each side to undo the addition.
Simplify.
Check the result by substituting \(-4\) into the original equation.
\(y+9=5\\)
Substitute −4 for y \(-4+9\overset{?}{=}5\\)
\(5=5✓\)
Since \(y=-4\) makes \(y+9=5\) a true statement, we found the solution to this equation.
Solve:
\(y+11=7\)

Kusonyeza yankho

−4
Solve:
\(y+15=-4\)

Kusonyeza yankho

−19
Solve: \(a-6=-8\)

Kusonyeza yankho

Add 6 to each side to undo the subtraction.
Simplify.
Check the result by substituting \(-2\) into the original equation:
Substitute \(-2\) for \(a\)
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.
Solve:
\(a-2=-8\)

Kusonyeza yankho

−6
Solve:
\(n-4=-8\)

Kusonyeza yankho

−4
Write an equation modeled by the envelopes and counters, and then solve it.

Kusonyeza yankho

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

Write the equation.
Divide both sides by 4.
Simplify.
There are \(\text{2 counters}\) in each envelope.
Write the equation modeled by the envelopes and counters. Then solve it.

Kusonyeza yankho

4x = 12; x = 3
Write the equation modeled by the envelopes and counters. Then solve it.

Kusonyeza yankho

3x = 6; x = 2
\(\text{Solve:}\ 7x=-49.\)

Kusonyeza yankho

To isolate \(x,\) we need to undo multiplication.

Divide each side by 7.
Simplify.
Check the solution.
\(7x=-49\\)
Substitute −7 for x. \(7(-7)\overset{?}{=}-49\\)
\(-49=-49✓\)
Therefore, \(-7\) is the solution to the equation.
Solve:
\(8a=56\)

Kusonyeza yankho

7
Solve:
\(11n=121\)

Kusonyeza yankho

11
Solve: \(-3y=63.\)

Kusonyeza yankho

To isolate \(y,\) we need to undo the multiplication.

Divide each side by −3.
Simplify
Check the solution.
\(-3y=63\\)
Substitute −21 for y. \(-3(-21)\overset{?}{=}63\\)
\(63=63✓\)
Since this is a true statement, \(y=-21\) is the solution to the equation.
Solve:
\(-8p=96\)

Kusonyeza yankho

−12
Solve:
\(-12m=108\)

Kusonyeza yankho

−9
Translate and solve: five more than \(x\) is equal to \(-3.\)

Kusonyeza yankho

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.
\(\begin{array}{l} \\ \\ x+5=-3\ \\ -8+5\overset{?}{=}-3\ \\ -3=-3✓\end{array}\)
Translate and solve:
Seven more than \(x\) is equal to \(-2\).

Kusonyeza yankho

x + 7 = −2; x = −9
Translate and solve:
\(\text{Eleven more than}\ y\ \text{is equal to 2.}\)

Kusonyeza yankho

y + 11 = 2; y = −9
Translate and solve: the difference of \(n\) and \(6\) is \(-10.\)

Kusonyeza yankho

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.
\(\begin{array}{l} \\ n-6=-10\ \\ -4-6\overset{?}{=}-10\ \\ -10=-10✓\end{array}\)
Translate and solve:
The difference of \(p\) and \(2\) is \(-4\).

Kusonyeza yankho

p − 2 = −4; p = −2
Translate and solve:
The difference of \(q\) and \(7\) is \(-3\).

Kusonyeza yankho

q − 7 = −3; q = 4
Translate and solve: the number \(108\) is the product of \(-9\) and \(y.\)

Kusonyeza yankho

the number of \(108\) is the product of \(-9\) and \(y\)
Translate. \(108=-9y\)
Divide by \(-9\). \(\frac{108}{-9}=\frac{-9y}{-9}\)
Simplify. \(-12=y\)
Check the answer by substituting it into the original equation.
\(\begin{array}{l}108=-9y \\ 108\overset{?}{=}-9(-12) \\ 108=108✓\end{array}\)
Translate and solve:
The number \(132\) is the product of \(-12\) and \(y\).

Kusonyeza yankho

132 = −12y; y = −11
Translate and solve:
The number \(117\) is the product of \(-13\) and \(z\).

Kusonyeza yankho

117 = −13z; z = −9
\(x+(-2)=-18\)

Kusonyeza yankho

x = −16
\(y+(-3)=-10\)
\(r-(-5)=-9\)

Kusonyeza yankho

r = −14
\(s-(-2)=-11\)
\(-14p=-42\)

Kusonyeza yankho

p = 3
\(-120=10q\)

Kusonyeza yankho

q = −12

Symbols used here

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\approx

approximately equal
Equal to the precision shown, not exactly.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

a \bmod n

remainder
What is left after dividing a by n.

per cent
Per hundred: 15% = 15/100.

a : b,\ \frac{a}{b}

ratio, fraction
a for every b; a divided by b.

How to: Solve Equations Using Integers; The Division Property of Equality

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
Substitute the number for the variable in the equation.
Simplify the expressions on both sides of the equation.
Determine whether the resulting equation is true.

Questions people ask

Why does the order of operations matter?

Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.

How do I check an arithmetic answer?

Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.

Why are fractions harder than decimals?

They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.

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Parts of this page are adapted from OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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