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Solving Equations Using the Subtraction and Addition Properties of Equality

Determine whether a number is a solution of an equation

Determine Whether a Number is a Solution of an Equation

Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!

To find the solution to an equation means to find the value of the variable that makes the equation true. Can you recognize the solution of \(x+2=7?\) If you said \(5,\) you’re right! We say \(5\) is a solution to the equation \(x+2=7\) because when we substitute \(5\) for \(x\) the resulting statement is true.

\[\begin{array}{l} \\ x+2=7 \\ 5+2\overset{?}{=}7 \\ \\ 7=7✓\end{array}\]

Since \(5+2=7\) is a true statement, we know that \(5\) is indeed a solution to the equation.

The symbol \(\overset{?}{=}\) asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign \(\text{(=)}\) or not-equal sign \(\text{(\ne ).}\)

Example

Try it.

\(\text{Determine whether}\ x=5\ \text{is a solution of}\ 6x-17=16.\)

Solution



Multiply.
Subtract.

So \(x=5\) is not a solution to the equation \(6x-17=16.\)

Example

Try it.

\(\text{Determine whether}\ y=2\ \text{is a solution of}\ 6y-4=5y-2.\)

Solution

Here, the variable appears on both sides of the equation. We must substitute \(2\) for each \(y.\)



Multiply.
Subtract.

Since \(y=2\) results in a true equation, we know that \(2\) is a solution to the equation \(6y-4=5y-2.\)

Model the Subtraction Property of Equality

We will use a model to help you understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable – since its contents are unknown – and each counter represents one.

Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk as in . Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?

What steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking “I need to remove the \(3\) counters from the left side to get the envelope by itself. Those \(3\) counters on the left match with \(3\) on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be \(5\) counters in the envelope.” shows this process.

What algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope \(x,\) so the number of counters on the left side of the desk is \(x+3.\) On the right side of the desk are \(8\) counters. We are told that \(x+3\) is equal to \(8\) so our equation is\(x+3=8.\)

\[x+3=8\]

Let’s write algebraically the steps we took to discover how many counters were in the envelope.


First, we took away three from each side.
Then we were left with five.

Now let’s check our solution. We substitute \(5\) for \(x\) in the original equation and see if we get a true statement.

Our solution is correct. Five counters in the envelope plus three more equals eight.

Example

Try it.

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

Solution

On the left, write \(x\) for the contents of the envelope, add the \(4\) counters, so we have \(x+4\).	\(x+4\)
On the right, there are \(5\) counters.	\(5\)
The two sides are equal.	\(x+4=5\)
Solve the equation by subtracting \(4\) counters from each side.

We can see that there is one counter in the envelope. This can be shown algebraically as:

Substitute \(1\) for \(x\) in the equation to check.

Since \(x=1\) makes the statement true, we know that \(1\) is indeed a solution.

Solve Equations Using the Subtraction Property of Equality

Our puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equations. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.

Think about twin brothers Andy and Bobby. They are \(17\) years old. How old was Andy \(3\) years ago? He was \(3\) years less than \(17,\) so his age was \(17-3,\) or \(14.\) What about Bobby’s age \(3\) years ago? Of course, he was \(14\) also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages \(3\) years ago.

\[\begin{array}{l}a=b \\ a-3=b-3\end{array}\]

Example

Try it.

Solve: \(x+8=17.\)

Solution

We will use the Subtraction Property of Equality to isolate \(x.\)


Subtract 8 from both sides.
Simplify.

Since \(x=9\) makes \(x+8=17\) a true statement, we know \(9\) is the solution to the equation.

Example

Try it.

Solve: \(100=y+74.\)

Solution

To solve an equation, we must always isolate the variable—it doesn’t matter which side it is on. To isolate \(y,\) we will subtract \(74\) from both sides.


Subtract 74 from both sides.
Simplify.
Substitute \(26\) for \(y\) to check.

Since \(y=26\) makes \(100=y+74\) a true statement, we have found the solution to this equation.

Solve Equations Using the Addition Property of Equality

In all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to “undo” the addition in order to isolate the variable.

But suppose we have an equation with a number subtracted from the variable, such as \(x-5=8.\) We want to isolate the variable, so to “undo” the subtraction we will add the number to both sides.

We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.

Remember the \(17\text{-year-old}\) twins, Andy and Bobby? In ten years, Andy’s age will still equal Bobby’s age. They will both be \(27.\)

\[\begin{array}{l}a=b \\ a+10=b+10\end{array}\]

We can add the same number to both sides and still keep the equality.

Example

Try it.

Solve: \(x-5=8.\)

Solution

We will use the Addition Property of Equality to isolate the variable.


Add 5 to both sides.
Simplify.

Example

Try it.

Solve: \(27=a-16.\)

Solution

We will add \(16\) to each side to isolate the variable.


Add 16 to each side.
Simplify.

The solution to \(27=a-16\) is \(a=43.\)

Translate Word Phrases to Algebraic Equations

Remember, an equation has an equal sign between two algebraic expressions. So if we have a sentence that tells us that two phrases are equal, we can translate it into an equation. We look for clue words that mean equals. Some words that translate to the equal sign are:

is equal to
is the same as
is
gives
was
will be

It may be helpful to put a box around the equals word(s) in the sentence to help you focus separately on each phrase. Then translate each phrase into an expression, and write them on each side of the equal sign.

We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts with no variables to solve for. Some sentences will translate into equations with variables. The focus right now is just to translate the words into algebra.

Example

Try it.

Translate the sentence into an algebraic equation: The sum of \(6\) and \(9\) is \(15.\)

Solution

The word is tells us the equal sign goes between 9 and 15.

Locate the “equals” word(s).
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression.
Translate the words to the right of the equals word into an algebraic expression.

Example

Try it.

Translate the sentence into an algebraic equation: The product of \(8\) and \(7\) is \(56.\)

Solution

The location of the word is tells us that the equal sign goes between 7 and 56.

Locate the “equals” word(s).
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression.
Translate the words to the right of the equals word into an algebraic expression.

Example

Try it.

Translate the sentence into an algebraic equation: Twice the difference of \(x\) and \(3\) gives \(18.\)

Solution

Locate the “equals” word(s).
Recognize the key words: twice; difference of …. and ….	Twice means two times.
Translate.

Translate to an Equation and Solve

Now let’s practice translating sentences into algebraic equations and then solving them. We will solve the equations by using the Subtraction and Addition Properties of Equality.

Example

Try it.

Translate and solve: Three more than \(x\) is equal to \(47.\)

Solution

	Three more than x is equal to 47.
Translate.
Subtract 3 from both sides of the equation.
Simplify.
We can check. Let \(x=44\).

So \(x=\ 44\) is the solution.

Example

Try it.

Translate and solve: The difference of \(y\) and \(14\) is \(18.\)

Solution

	The difference of y and 14 is 18.
Translate.
Add 14 to both sides.
Simplify.
We can check. Let \(y=32\).

So \(y=32\) is the solution.

Key Concepts

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.
Subtraction Property of Equality
- For any numbers \(a\), \(b\), and \(c\),
  if \(a=b\)
  then \(a-c=b-c\)
Solve an equation using the Subtraction Property of Equality.
1. Use the Subtraction Property of Equality to isolate the variable.
2. Simplify the expressions on both sides of the equation.
3. Check the solution.
Addition Property of Equality
- For any numbers \(a\), \(b\), and \(c\),
  if \(a=b\)
  then \(a+c=b+c\)
Solve an equation using the Addition Property of Equality.
1. Use the Addition Property of Equality to isolate the variable.
2. Simplify the expressions on both sides of the equation.
3. Check the solution.

Solving Equations Using the Subtraction and Addition Properties of Equality

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each given value is a solution to the equation.

Try it.

\(x+13=21\)

ⓐ \(\ x=8\)
ⓑ \(\ x=34\)

Solution

ⓐ yes
ⓑ no

Try it.

\(y+18=25\)

ⓐ \(\ y=7\)
ⓑ \(y=43\)

Try it.

\(m-4=13\)

ⓐ \(\ m=9\)
ⓑ \(m=17\)

Solution

ⓐ no
ⓑ yes

Try it.

\(n-9=6\)

ⓐ \(\ n=3\)
ⓑ \(n=15\)

Try it.

\(3p+6=15\)

ⓐ \(\ p=3\)
ⓑ \(\ p=7\)

Solution

ⓐ yes
ⓑ no

Try it.

\(8q+4=20\)

ⓐ \(\ q=2\)
ⓑ \(\ q=3\)

Try it.

\(18d-9=27\)

ⓐ \(\ d=1\)
ⓑ \(\ d=2\)

Solution

ⓐ no
ⓑ yes

Try it.

\(24f-12=60\)

ⓐ \(\ f=2\)
ⓑ \(\ f=3\)

Try it.

\(8u-4=4u+40\)

ⓐ \(\ u=3\)
ⓑ \(u=11\)

Solution

ⓐ no
ⓑ yes

Try it.

\(7v-3=4v+36\)

ⓐ \(\ v=3\)
ⓑ \(v=11\)

Try it.

\(20h-5=15h+35\)

ⓐ \(\ h=6\)
ⓑ \(h=8\)

Solution

ⓐ no
ⓑ yes

Try it.

\(18k-3=12k+33\)

ⓐ \(k=1\)
ⓑ \(k=6\)

Model the Subtraction Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve using the subtraction property of equality.

Try it.

Solution

x + 2 = 5; x = 3

Try it.

Solution

x + 3 = 6; x = 3

Try it.

Solve Equations using the Subtraction Property of Equality

In the following exercises, solve each equation using the subtraction property of equality.

Try it.

\(a+2=18\)

Solution

a = 16

Try it.

\(b+5=13\)

Try it.

\(p+18=23\)

Solution

p = 5

Try it.

\(q+14=31\)

Try it.

\(r+76=100\)

Solution

r = 24

Try it.

\(s+62=95\)

Try it.

\(16=x+9\)

Solution

x = 7

Try it.

\(17=y+6\)

Try it.

\(93=p+24\)

Solution

p = 69

Try it.

\(116=q+79\)

Try it.

\(465=d+398\)

Solution

d = 67

Try it.

\(932=c+641\)

Solve Equations using the Addition Property of Equality

In the following exercises, solve each equation using the addition property of equality.

Try it.

\(y-3=19\)

Solution

y = 22

Try it.

\(x-4=12\)

Try it.

\(u-6=24\)

Solution

u = 30

Try it.

\(v-7=35\)

Try it.

\(f-55=123\)

Solution

f = 178

Try it.

\(g-39=117\)

Try it.

\(19=n-13\)

Solution

n = 32

Try it.

\(18=m-15\)

Try it.

\(10=p-38\)

Solution

p = 48

Try it.

\(18=q-72\)

Try it.

\(268=y-199\)

Solution

y = 467

Try it.

\(204=z-149\)

Translate Word Phrase to Algebraic Equations

In the following exercises, translate the given sentence into an algebraic equation.

Try it.

The sum of \(8\) and \(9\) is equal to \(17.\)

Solution

8 + 9 = 17

Try it.

The sum of \(7\) and \(9\) is equal to \(16.\)

Try it.

The difference of \(23\) and \(19\) is equal to \(4.\)

Solution

23 − 19 = 4

Try it.

The difference of \(29\) and \(12\) is equal to \(17.\)

Try it.

The product of \(3\) and \(9\) is equal to \(27.\)

Solution

3 ⋅ 9 = 27

Try it.

The product of \(6\) and \(8\) is equal to \(48.\)

Try it.

The quotient of \(54\) and \(6\) is equal to \(9.\)

Solution

54 ÷ 6 = 9

Try it.

The quotient of \(42\) and \(7\) is equal to \(6.\)

Try it.

Twice the difference of \(n\) and \(10\) gives \(52.\)

Solution

2(n − 10) = 52

Try it.

Twice the difference of \(m\) and \(14\) gives \(64.\)

Try it.

The sum of three times \(y\) and \(10\) is \(100.\)

Solution

3y + 10 = 100

Try it.

The sum of eight times \(x\) and \(4\) is \(68.\)

Translate to an Equation and Solve

In the following exercises, translate the given sentence into an algebraic equation and then solve it.

Try it.

Five more than \(p\) is equal to \(21.\)

Solution

p + 5 = 21; p = 16

Try it.

Nine more than \(q\) is equal to \(40.\)

Try it.

The sum of \(r\) and \(18\) is \(73.\)

Solution

r + 18 = 73; r = 55

Try it.

The sum of \(s\) and \(13\) is \(68.\)

Try it.

The difference of \(d\) and \(30\) is equal to \(52.\)

Solution

d − 30 = 52; d = 82

Try it.

The difference of \(c\) and \(25\) is equal to \(75.\)

Try it.

\(12\) less than \(u\) is \(89.\)

Solution

u − 12 = 89; u = 101

Try it.

\(19\) less than \(w\) is \(56.\)

Try it.

\(325\) less than \(c\) gives \(799.\)

Solution

c − 325 = 799; c = 1124

Try it.

\(299\) less than \(d\) gives \(850.\)

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.

\(\text{Evaluate}\ x+8\ \text{when}\ x=11.\)
If you missed this problem, review .

Odkrij odgovor

\(19\)
\(\text{Evaluate}\ 5x-3\ \text{when}\ x=9.\)
If you missed this problem, review .

Odkrij odgovor

\(42\)
Translate into algebra: the difference of \(x\) and \(8.\)
If you missed this problem, review .

Odkrij odgovor

\(x-8\)
\(\text{Determine whether}\ x=5\ \text{is a solution of}\ 6x-17=16.\)

Odkrij odgovor

Multiply.
Subtract.
So \(x=5\) is not a solution to the equation \(6x-17=16.\)
\(\text{Is}\ x=3\ \text{a solution of}\ 4x-7=16\text{?}\)

Odkrij odgovor

no
\(\text{Is}\ x=2\ \text{a solution of}\ 6x-2=10\text{?}\)

Odkrij odgovor

yes
\(\text{Determine whether}\ y=2\ \text{is a solution of}\ 6y-4=5y-2.\)

Odkrij odgovor

Here, the variable appears on both sides of the equation. We must substitute \(2\) for each \(y.\)

Multiply.
Subtract.
Since \(y=2\) results in a true equation, we know that \(2\) is a solution to the equation \(6y-4=5y-2.\)
\(\text{Is}\ y=3\ \text{a solution of}\ 9y-2=8y+1\text{?}\)

Odkrij odgovor

yes
\(\text{Is}\ y=4\ \text{a solution of}\ 5y-3=3y+5\text{?}\)

Odkrij odgovor

yes

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

Odkrij odgovor

On the left, write \(x\) for the contents of the envelope, add the \(4\) counters, so we have \(x+4\).	\(x+4\)
On the right, there are \(5\) counters.	\(5\)
The two sides are equal.	\(x+4=5\)
Solve the equation by subtracting \(4\) counters from each side.

We can see that there is one counter in the envelope. This can be shown algebraically as:

Substitute \(1\) for \(x\) in the equation to check.

Since \(x=1\) makes the statement true, we know that \(1\) is indeed a solution.

Write the equation modeled by the envelopes and counters, and then solve the equation:

Odkrij odgovor

x + 1 = 7; x = 6
Write the equation modeled by the envelopes and counters, and then solve the equation:

Odkrij odgovor

x + 3 = 4; x = 1
Solve: \(x+8=17.\)

Odkrij odgovor

We will use the Subtraction Property of Equality to isolate \(x.\)

Subtract 8 from both sides.
Simplify.
Since \(x=9\) makes \(x+8=17\) a true statement, we know \(9\) is the solution to the equation.
Solve:
\(x+6=19\)

Odkrij odgovor

x = 13
Solve:
\(x+9=14\)

Odkrij odgovor

x = 5
Solve: \(100=y+74.\)

Odkrij odgovor

To solve an equation, we must always isolate the variable—it doesn’t matter which side it is on. To isolate \(y,\) we will subtract \(74\) from both sides.

Subtract 74 from both sides.
Simplify.
Substitute \(26\) for \(y\) to check.

Since \(y=26\) makes \(100=y+74\) a true statement, we have found the solution to this equation.
Solve:
\(95=y+67\)

Odkrij odgovor

y = 28
Solve:
\(91=y+45\)

Odkrij odgovor

y = 46
Solve: \(x-5=8.\)

Odkrij odgovor

We will use the Addition Property of Equality to isolate the variable.

Add 5 to both sides.
Simplify.
Solve:
\(x-9=13\)

Odkrij odgovor

x = 22
Solve:
\(y-1=3\)

Odkrij odgovor

y = 4
Solve: \(27=a-16.\)

Odkrij odgovor

We will add \(16\) to each side to isolate the variable.

Add 16 to each side.
Simplify.

The solution to \(27=a-16\) is \(a=43.\)
Solve:
\(19=a-18\)

Odkrij odgovor

a = 37
Solve:
\(27=n-14\)

Odkrij odgovor

n = 41

Translate the sentence into an algebraic equation: The sum of \(6\) and \(9\) is \(15.\)

Odkrij odgovor

The word is tells us the equal sign goes between 9 and 15.

Locate the “equals” word(s).
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression.
Translate the words to the right of the equals word into an algebraic expression.

Translate the sentence into an algebraic equation:
The sum of \(7\) and \(6\) gives \(13.\)

Odkrij odgovor

7 + 6 = 13
Translate the sentence into an algebraic equation:
The sum of \(8\) and \(6\) is \(14.\)

Odkrij odgovor

8 + 6 = 14

Translate the sentence into an algebraic equation: The product of \(8\) and \(7\) is \(56.\)

Odkrij odgovor

The location of the word is tells us that the equal sign goes between 7 and 56.

Locate the “equals” word(s).
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression.
Translate the words to the right of the equals word into an algebraic expression.

Translate the sentence into an algebraic equation:
The product of \(6\) and \(9\) is \(54.\)

Odkrij odgovor

6 ⋅ 9 = 54
Translate the sentence into an algebraic equation:
The product of \(21\) and \(3\) gives \(63.\)

Odkrij odgovor

21 ⋅ 3 = 63
Translate the sentence into an algebraic equation: Twice the difference of \(x\) and \(3\) gives \(18.\)

Odkrij odgovor

Locate the “equals” word(s).
Recognize the key words: twice; difference of …. and …. Twice means two times.
Translate.
Translate the given sentence into an algebraic equation:
Twice the difference of \(x\) and \(5\) gives \(30.\)

Odkrij odgovor

2(x − 5) = 30
Translate the given sentence into an algebraic equation:
Twice the difference of \(y\) and \(4\) gives \(16.\)

Odkrij odgovor

2(y − 4) = 16
Translate and solve: Three more than \(x\) is equal to \(47.\)

Odkrij odgovor

Three more than x is equal to 47.
Translate.
Subtract 3 from both sides of the equation.
Simplify.
We can check. Let \(x=44\).
So \(x=\ 44\) is the solution.
Translate and solve:
Seven more than \(x\) is equal to \(37.\)

Odkrij odgovor

x + 7 = 37; x = 30
Translate and solve:
Eleven more than \(y\) is equal to \(28.\)

Odkrij odgovor

y + 11 = 28; y = 17
Translate and solve: The difference of \(y\) and \(14\) is \(18.\)

Odkrij odgovor

The difference of y and 14 is 18.
Translate.
Add 14 to both sides.
Simplify.
We can check. Let \(y=32\).
So \(y=32\) is the solution.
Translate and solve:
The difference of \(z\) and \(17\) is equal to \(37.\)

Odkrij odgovor

z − 17 = 37; z = 54
Translate and solve:
The difference of \(x\) and \(19\) is equal to \(45.\)

Odkrij odgovor

x − 19 = 45; x = 64
\(x+13=21\)
1. ⓐ \(\ x=8\)
2. ⓑ \(\ x=34\)
Odkrij odgovor
1. ⓐ yes
2. ⓑ no

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: Solving Equations Using the Subtraction and Addition Properties of Equality

Determine whether a number is a solution of an equation
Model the Subtraction Property of Equality
Solve equations using the Subtraction Property of Equality
Solve equations using the Addition Property of Equality
Translate word phrases to algebraic equations
Translate to an equation and solve
Substitute the number for the variable in the equation.
Simplify the expressions on both sides of the equation.

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.

Poskusi sam.

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

Več v Arithmetic

Order of operations Fractions Percentages Powers and roots

if	\(a=b\)
then	\(a-c=b-c\)

if	\(a=b\)
then	\(a+c=b+c\)