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Mathematics LibreTexts

2.1: Solve Equations Using the Subtraction and Addition Properties of Equality

  • Page ID
    18932
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    Learning Objectives

    By the end of this section, you will be able to:

    • Verify a solution of an equation
    • Solve equations using the Subtraction and Addition Properties of Equality
    • Solve equations that require simplification
    • Translate to an equation and solve
    • Translate and solve applications

    Quiz

    Before you get started, take this readiness quiz.

    1. Evaluate \(x+4\) when \(x=−3\).
      If you missed this problem, review Exercise 1.5.25.
    2. Evaluate \(15−y\) when \(y=−5\).
      If you missed this problem, review Exercise 1.5.31.
    3. Simplify \(4(4n+1)−15n\).
      If you missed this problem, review Exercise 1.10.49.
    4. Translate into algebra “5 is less than x.”
      If you missed this problem, review Exercise 1.3.43.

    Verify a Solution of an Equation

    Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same – so that we end up with a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

    Definition: Solution of an Equation

    A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

    TO DETERMINE WHETHER A NUMBER IS A Solution TO AN EQUATION

    1. Substitute the number in for the variable in the equation.
    2. Simplify the expressions on both sides of the equation.
    3. Determine whether the resulting equation is true (the left side is equal to the right side)
      • If it is true, the number is a solution.
      • If it is not true, the number is not a solution.

    Exercise \(\PageIndex{1}\)

    Determine whether \(x = \frac{3}{2}\) is a solution of \(4x−2=2x+1\).

    Answer

    Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

    \(4 x-2=2 x+1\)
    . \(4\left(\color{red}\frac{3}{2}\color{black}\right)-2 \stackrel{?}{=} 2\left(\color{red}\frac{3}{2}\color{black}\right)+1\)
    Multiply. \(6-2 \stackrel{?}{=} 3+1\)
    Subtract. \(4=4 \checkmark \)

    Since \(x = \frac{3}{2}\) results in a true equation (4 is in fact equal to 4), \(\frac{3}{2}\) is a solution to the equation \(4x−2=2x+1\).

    Exercise \(\PageIndex{2}\)

    Is \(y = \frac{4}{3}\) a solution of \(9y+2=6y+3\)?

    Answer

    no

    Exercise \(\PageIndex{3}\)

    Is \(y = \frac{7}{5}\) a solution of \(5y+3=10y-4\)?

    Answer

    yes

    Solve Equations Using the Subtraction and Addition Properties of Equality

    We are going to use a model to clarify the process of solving an equation. An envelope represents the variable – since its contents are unknown – and each counter represents one. We will set out one envelope and some counters on our workspace, as shown in Figure \(\PageIndex{1}\). Both sides of the workspace have the same number of counters, but some counters are “hidden” in the envelope. Can you tell how many counters are in the envelope?

    This image illustrates a workspace divided into two sides. The content of the left side is equal to the content of the right side. On the left side, there are three circular counters and an envelope containing an unknown number of counters. On the right side are eight counters.
    Figure \(\PageIndex{1}\): The illustration shows a model of an equation with one variable. On the left side of the workspace is an unknown (envelope) and three counters, while on the right side of the workspace are eight counters.

    What are you thinking? 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 at the bottom left to get the envelope by itself. The 3 counters on the left can be matched with 3 on the right and so I can take them away from both sides. That leaves five on the right—so there must be 5 counters in the envelope.” See Figure \(\PageIndex{2}\) for an illustration of this process.

    This figure contains two illustrations of workspaces, divided each into two sides. On the left side of the first workspace there are three counters circled in purple and an envelope containing an unknown number of counters. On the right side are eight counters, three of which are also circled in purple. An arrow to the right of the workspace points to the second workspace. On the left side of the second workspace, there is just an envelope. On the right side are five counters. This workspace is identical to the first workspace, except that the three counters circled in purple have been removed from both sides.
    Figure \(\PageIndex{2}\): The illustration shows a model for solving an equation with one variable. On both sides of the workspace remove three counters, leaving only the unknown (envelope) and five counters on the right side. The unknown is equal to five counters.

    What algebraic equation would match this situation? In Figure \(\PageIndex{3}\) each side of the workspace represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope x.

    This image illustrates a workspace divided into two sides. The content of the left side is equal to the content of the right side. On the left side, there are three circular counters and an envelope containing an unknown number of counters. On the right side are eight counters. Underneath the image is the equation modeled by the counters: x plus 3 equals 8.
    Figure \(\PageIndex{3}\): The illustration shows a model for the equation \(x+3=8\).

    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. .

    Table \(\PageIndex{1}\)

    Check:

    Five in the envelope plus three more does equal eight!

    \[5+3=8\]

    Our model has given us an idea of what we need to do to solve one kind of equation. The goal is to isolate the variable by itself on one side of the equation. To solve equations such as these mathematically, we use the Subtraction Property of Equality.

    SUBTRACTION PROPERTY OF EQUALITY

    For any numbers a, b, and c,

    \[\begin{array} {ll} {\text{If}} &{a = b} \\ {\text{then}} &{a - c = b - c} \end{array}\]

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

    Note

    Doing the Manipulative Mathematics activity “Subtraction Property of Equality” will help you develop a better understanding of how to solve equations by using the Subtraction Property of Equality.

    Let’s see how to use this property to solve an equation. Remember, the goal is to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement.

    Exercise \(\PageIndex{4}\)

    Solve: \(y+37=−13\).

    Answer

    To get y by itself, we will undo the addition of 37 by using the Subtraction Property of Equality.

    .
    Subtract 37 from each side to ‘undo’ the addition. .
    Simplify. .
    Check: .
    Substitute \(y=−50\) .
    .

    Since y=−50 makes y+37=−13 a true statement, we have the solution to this equation.

    Exercise \(\PageIndex{5}\)

    Solve: \(x+19=−27\).

    Answer

    \(x=−46\)

    Exercise \(\PageIndex{6}\)

    Solve: \(x+16=−34\).

    Answer

    \(x=−50\)

    What happens when an equation has a number subtracted from the variable, as in the equation \(x−5=8\)? We use another property of equations to solve equations where a number is subtracted from the variable. 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.

    ADDITION PROPERTY OF EQUALITY

    For any numbers a, b, and c,

    \[\begin{array} {ll} {\text{If}} &{a = b} \\ {\text{then}} &{a + c = b + c} \end{array}\]

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

    In Exercise \(\PageIndex{4}\), 37 was added to the y and so we subtracted 37 to ‘undo’ the addition. In Exercise \(\PageIndex{7}\), we will need to ‘undo’ subtraction by using the Addition Property of Equality.

    Exercise \(\PageIndex{7}\)

    Solve: \(a−28=−37\).

    Answer
    .
    Add 28 to each side to ‘undo’ the subtraction. .
    Simplify. .
    Check: .
    Substitute \(a=−9\) .
    .
    The solution to \(a−28=−37\) is \(a=−9\).

    Exercise \(\PageIndex{8}\)

    Solve: \(n−61=−75\).

    Answer

    \(n=−14\)

    Exercise \(\PageIndex{9}\)

    Solve: \(p−41=−73\).

    Answer

    \(p=−32\)

    Exercise \(\PageIndex{10}\)

    Solve: \(x - \frac{5}{8} = \frac{3}{4}\)

    Answer
    .
    Use the Addition Property of Equality. .
    Find the LCD to add the fractions on the right. \(x-\frac{5}{8}+\frac{5}{8}=\frac{6}{8}+\frac{5}{8}\)
    Simplify. \(x=\frac{11}{8}\)
    Check: .
    Substitute \(x= \frac{11}{8}\) .
    Subtract. .
    Simplify. .
    The solution to \(x - \frac{5}{8} = \frac{3}{4}\) is \(x= \frac{11}{8}\).

    Exercise \(\PageIndex{11}\)

    Solve: \(p−\frac{2}{3}=\frac{5}{6}\).

    Answer

    \(p = \frac{9}{6} p =\frac{3}{2}\)

    Exercise \(\PageIndex{12}\)

    Solve: \(q−\frac{1}{2}=\frac{5}{6}\).

    Answer

    \(q =\frac{4}{3}\)

    The next example will be an equation with decimals.

    Exercise \(\PageIndex{13}\)

    Solve: \(n−0.63=−4.2\).

    Answer
    \(n-0.63=-4.2\)
    Use the Addition Property of Equality. .
    Add. \(n=-3.57\)
    Check: \(n=-3.57\)
    Let \(n=−3.57\). .
    .

    Exercise \(\PageIndex{14}\)

    Solve: \(b−0.47=−2.1\).

    Answer

    \(b=−1.63\)

    Exercise \(\PageIndex{15}\)

    Solve: \(c−0.93=−4.6\).

    Answer

    \(c=−3.67\)

    Solve Equations That Require Simplification

    In the previous examples, we were able to isolate the variable with just one operation. Most of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality.

    You should always simplify as much as possible before you try to isolate the variable. Remember that to simplify an expression means to do all the operations in the expression. Simplify one side of the equation at a time. Note that simplification is different from the process used to solve an equation in which we apply an operation to both sides.

    Exercise \(\PageIndex{16}\): How to Solve Equations That Require Simplification

    Solve: \(9x−5−8x−6=7\).

    Answer

    This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Simplify the expressions on each side as much as possible.” The text in the second cell reads: “Rearrange the terms, using the Commutative Property of Addition. Combine like terms. Notice that each side is now simplified as much as possible.” The third cell contains the equation 9 x minus 5 minus 8 x minus 6 equals 7. Below this is the same equation, with the terms rearranged: 9 x minus 8 x minus 5 minus 6 equals 7. Below this is the equation with like terms combined: x minus 11 equals 7.In the second row of the table, the first cell says “Step 2. Isolate the variable.” In the second cell, the instructions say “Now isolate x. Undo subtraction by adding 11 to both sides.” The third cell contains the equation x minus 11 plus 11 equals 7 plus 11, with “plus 11” written in red on both sides.In the third row of the table, the first cell says: “Step 3. Simplify the equation on both sides of the equation.” The second cell is left blank. The third cell contains x equals 18. Step 4.  We check the solution to be sure 18 makes both sides of the equation equal.

    Exercise \(\PageIndex{17}\)

    Solve: \(8y−4−7y−7=4\).

    Answer

    \(y=15\)

    Exercise \(\PageIndex{18}\)

    Solve: \(6z+5−5z−4=3\).

    Answer

    \(z=2\)

    Exercise \(\PageIndex{19}\)

    Solve: 5(n−4)−4n=−8.

    Answer

    We simplify both sides of the equation as much as possible before we try to isolate the variable.

    \(5(n-4)-4 n=-8\)

    Distribute on the left. \(5 n-20-4 n=-8\)
    Use the Commutative Property to rearrange terms. \(5 n-4 n-20=-8\)
    Combine like terms. \(n-20=-8\)
    Each side is as simplified as possible. Next, isolate n.
    Undo subtraction by using the Addition Property of Equality. \(n-20 + \color{red}+ 20 \color{black}=-8 \color{red}+20}\)
    Add. \(n=12\)

    Check. Substitute n=12.

    .

    The solution to \(5(n−4)−4n=−8\) is \(n=12\).

    Exercise \(\PageIndex{20}\)

    Solve: \(5(p−3)−4p=−10\).

    Answer

    \(p=5\)

    Exercise \(\PageIndex{21}\)

    Solve: \(4(q+2)−3q=−8\).

    Answer

    \(q=−16\)

    Exercise \(\PageIndex{22}\)

    Solve: \(3(2y−1)−5y=2(y+1)−2(y+3)\).

    Answer

    We simplify both sides of the equation before we isolate the variable.

    \(3(2 y-1)-5 y=2(y+1)-2(y+3)\)
    Distribute on both sides. \(6 y-3-5 y=2 y+2-2 y-6\)
    Use the Commutative Property of Addition. \(6 y-5 y-3=2 y-2 y+2-6\)
    Combine like terms. \(y-3=-4\)
    Each side is as simplified as possible. Next, isolate y.
    Undo subtraction by using the Addition Property of Equality. \(y-3 \color{red} + 3 \color{black} = -4 \color{red} +3\)
    Add. \(y=-1\)
    Check. Let y=−1.
    .

    The solution to \(3(2y−1)−5y=2(y+1)−2(y+3)3(2y−1)−5y=2(y+1)−2(y+3)\) is \(y=−1\).

    Exercise \(\PageIndex{23}\)

    Solve: \(4(2h−3)−7h=6(h−2)−6(h−1)\).

    Answer

    \(h = 6\)

    Exercise \(\PageIndex{24}\)

    Solve: \(2(5x+2)−9x=3(x−2)−3(x−4)\).

    Answer

    \(x=2\)

    Translate to an Equation and Solve

    To solve applications algebraically, we will begin by translating from English sentences into equations. Our first step is to look for the word (or words) that would translate to the equals sign. Here are some of the words that are commonly used.

    Equals =

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

    The steps we use to translate a sentence into an equation are listed below.

    TRANSLATE AN ENGLISH SENTENCE TO AN ALGEBRAIC EQUATION

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

    Exercise \(\PageIndex{25}\)

    Translate and solve: Eleven more than x is equal to 54.

    Answer
    Translate. .
    Subtract 11 from both sides. .
    Simplify. .
    Check: Is 54 eleven more than 43?
    \[\begin{array} {rrr} {43 + 11} &{\stackrel{?}{=}} &{54}\\ {54} &{=} &{54\checkmark} \end{array}\]

    Exercise \(\PageIndex{26}\)

    Translate and solve: Ten more than x is equal to 41.

    Answer

    \(x+10=41;x=31\)

    Exercise \(\PageIndex{27}\)

    Translate and solve: Twelve less than x is equal to 51.

    Answer

    y−12=51;y=63

    Exercise \(\PageIndex{28}\)

    Translate and solve: The difference of 12t and 11t is −14.

    Answer
    Translate. .
    Simplify. .
    Check:
    \[\begin{array} {rrl} {12(-14) - 11(-14)} &{\stackrel{?}{=}} &{-14}\\{-168 + 154} &{\stackrel{?}{=}} &{-14} \\ {-14} &{=} &{-14\checkmark}\end{array}\]

    Exercise \(\PageIndex{29}\)

    Translate and solve: The difference of 4x and 3x is 14.

    Answer

    \(4x−3x=14;x=14\)

    Exercise \(\PageIndex{30}\)

    Translate and solve: The difference of 7a and 6a is −8.

    Answer

    \(7a−6a=−8;a=−8\)

    Translate and Solve Applications

    Most of the time a question that requires an algebraic solution comes out of a real life question. To begin with that question is asked in English (or the language of the person asking) and not in math symbols. Because of this, it is an important skill to be able to translate an everyday situation into algebraic language.

    We will start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for. For example, you might use q for the number of quarters if you were solving a problem about coins.

    Exercise \(\PageIndex{31}\): How to Solve Translate and Solve Applications

    The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?

    Answer

    This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains text and algebra. In the top row, the first cell says “Step 1. Read the problem. Make sure all the words and ideas are understood.” The text in the second cell says “The problem is about the weight of newspapers.” The third cell is blank.In the second row, the first cell says “Step 2. Identify what we are asked to find.” The second cell says “What are we asked to find?” The third cell says: “How much did the newspapers weigh the 2nd month?”In the third row, the first cell says “Step 3. Name what we are looking for. Choose a variable to represent that quantity.” The second cell says “Choose a variable.” The third cell says “Let w equal weight of the newspapers the 1st month.”In the fourth row, the first cell says “Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.” The second cell says “Restate the problem. We know that the weight of the newspapers the second month is 28 pounds.” The third cell says “Weight of newspapers the 1st month plus the weight of the newspapers the 2nd month equals 57 pounds. Weight from 1st month plus 28 equals 57.” One line down, the second cell says “Translate into an equation using the variable w.” The third cell contains the equation w plus 28 equals 57.In the fifth row, the first cell says “Step 5. Solve the equation using good algebra techniques.” The second cell says “Solve.” The third cell contains the equation with 28 being subtracted from both sides: w plus 28 minus 28 equals 57 minus 28, with minus 28 written in red. Below this is w equals 29.In the sixth row, the first cell says “Step 6. Check the answer and make sure it makes sense.” The second cell says “Does 1st month’s weight plus 2nd month’s weight equal 57 pounds?” The third cell contains the equation 29 plus 28 might equal 57. Below this is 57 equals 57 with a check mark next to it.In the seventh and final row, the first cell says ‘Step 7. Answer the question with a complete sentence.” The second cell says “Write a sentence to answer ‘How much did the newspapers weigh the 2nd month?’” The third cell contains the sentence “The 2nd month the newspapers weighed 29 pounds.”

    Exercise \(\PageIndex{32}\)

    Translate into an algebraic equation and solve:

    The Pappas family has two cats, Zeus and Athena. Together, they weigh 23 pounds. Zeus weighs 16 pounds. How much does Athena weigh?

    Answer

    7 pounds

    Exercise \(\PageIndex{33}\)

    Translate into an algebraic equation and solve:

    Sam and Henry are roommates. Together, they have 68 books. Sam has 26 books. How many books does Henry have?

    Answer

    42 books

    SOLVE AN APPLICATION.

    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

    Exercise \(\PageIndex{34}\)

    Randell paid $28,675 for his new car. This was $875 less than the sticker price. What was the sticker price of the car?

    Answer

    \(\begin{array} {ll} {\textbf {Step 1. Read}\text{ the problem. }} &{}\\\\ {\textbf {Step 2. Identify}\text{ what we are looking for.}} &{\text{"What was the sticker price of the car?"}} \\\\ {\textbf{Step 3. Name}\text{ what we are looking for.}} &{} \\ {\text{Choose a variable to represent that quantity.}} &{\text{Let s = the sticker price of the car.}} \\\\{\textbf {Step 4. Translate}\text{ into an equation. Restate }} &{} \\ {\text{the problem in one sentence.}} &{$\text{28675 is } $\text{875 less than the sticker price}} \\ \\ {} &{$\text{28675 is } $\text{875 less than s}}\\ {}&{28675 = s - 875} \\ {\textbf {Step 5. Solve}\text{ the equation. }} &{28675 + 875 = s - 875 + 875}\\ {} &{29550 = s} \\ \\ {\textbf {Step 6. Check}\text{ the answer. }} &{} \\ {\text{Is }$875\text{ less than }$29550\text{ equal to } $28675?} &{} \\ {29550 - 875 \stackrel{?}{=} 28675} &{} \\ {28675 = 28675\checkmark} &{} \\ \\ {\textbf {Step 7. Answer}\text{ the question with }} &{\text{The sticker price of the car was }$29550.} \\ {\text{a complete sentence.}} &{} \end{array}\)

    Exercise \(\PageIndex{35}\)

    Translate into an algebraic equation and solve:

    Eddie paid $19875 for his new car. This was $1025 less than the sticker price. What was the sticker price of the car?

    Answer

    $20900

    Exercise \(\PageIndex{36}\)

    Translate into an algebraic equation and solve:

    The admission price for the movies during the day is $7.75. This is $3.25 less the price at night. How much does the movie cost at night?

    Answer

    $11.00

    Key Concepts

    • To Determine Whether a Number is a Solution to an Equation
      1. Substitute the number in for the variable in the equation.
      2. Simplify the expressions on both sides of the equation.
      3. Determine whether the resulting statement is true.
        • If it is true, the number is a solution.
        • If it is not true, the number is not a solution.
    • Addition Property of Equality
      • For any numbers a, b, and c, if a=b, then a+c=b+c.
    • Subtraction Property of Equality
      • For any numbers a, b, and c, if a=b, then a−c=b−c.
    • To Translate a Sentence to an Equation
      1. Locate the “equals” word(s). Translate to an equal sign (=).
      2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
      3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
    • To Solve an Application
      1. Read the problem. Make sure all the words and ideas are understood.
      2. Identify what we are looking for.
      3. Name what we are looking for. Choose a variable to represent that quantity.
      4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
      5. Solve the equation using good algebra techniques.
      6. Check the answer in the problem and make sure it makes sense.
      7. Answer the question with a complete sentence.

    Glossary

    solution of an equation
    A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.