$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 1.6: Solving Equations by Addition and Subtraction

• • David Arnold
• Retired Professor (Mathematics) at College of the Redwoods
$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

Variable

A variable is a symbol (usually a letter) that stands for a value that may vary

Next we follow with the definition of an equation.

Equation

An equation is a mathematical statement that equates two mathematical expressions.

The key difference between a mathematical expression and an equation is the presence of an an equals sign. So, for example,

2 + 3[5 − 4 · 2], x2 + 2x − 3, and x + 2y + 3

are mathematical expressions (two of which contain variables), while

3 + 2(7 − 3) = 11, x +3=4, and 3x = 9

are equations. Note that each of the equations contain an equals sign, but the expressions do not.

Next we have the definition of a solution of an equation.

What it Means to be a Solution

A solution of an equation is a numerical value that satisfies the equation. That is, when the variable in the equation is replaced by the solution, a true statement results.

Example 1

Show that 3 is a solution of the equation x + 8 = 11.

Solution

Substitute 3 for x in the given equation and simplify.

$\begin{array}{rlrl}{x+8} & {=11} & {} & { \textcolor{red}{ \text { The given equation. }}} \\ {3+8} & {=11} & {} & {\textcolor{red}{ \text { Substitute } 3 \text { for } x .}} \\ {11} & {=11} & {} & {\textcolor{red}{ \text { Simplify both sides. }}}\end{array}\nonumber$

Since the left- and right-hand sides of the last line are equal, this shows that when 3 is substituted for x in the equation a true statement results. Therefore, 3 is a solution of the equation.

Exercise

Show that 27 is a solution of the equation x − 12 = 15

Example 2

Is 23 a solution of the equation 4 = y − 11?

Solution

Substitute 23 for y in the given equation and simplify.

$\begin{array}{ll}{4=y-11} & {\textcolor{red}{ \text { The given equation. }}} \\ {4=23-11} & {\textcolor{red}{ \text { Substitute } 23 \text { for } y}} \\ {4=12} & {\textcolor{red} {\text { Simplify both sides. }}}\end{array}\nonumber$

Since the left- and right-hand sides of the last line are not equal, this shows that when 23 is substituted for y in the equation a false statement results. Therefore, 23 is not a solution of the equation.

Exercise

Is 8 a solution of 5 = 12 − y?

No.

## Equivalent Equations

Equivalent Equations

Two equations are equivalent if they have the same solution set.

Example 3

Are the equations x + 2 = 9 and x = 7 equivalent?

Solution

The number 7 is the only solution of the equation x + 2 = 9. Similarly, 7 is the only solution of the equation x = 7. Therefore x + 2 = 9 and Answer: No. x = 7 have the same solution sets and are equivalent.

Exercise

Are the equations x = 4 and x + 8 = 3 equivalent?

No.

Example 4

Are the equations x2 = x and x = 1 equivalent?

Solution

By inspection, the equation x2 = x has two solutions, 0 and 1. On the other hand, the equation x = 1 has a single solution, namely 1. Hence, the equations x2 = x and x = 1 do not have the same solution sets and are not equivalent.

Exercise

Are the equations x = 2 and x2 = 2x equivalent?

No.

## Operations that Produce Equivalent Equations

There are many operations that will produce equivalent operations. In this section we look at two: addition and subtraction.

Adding the Same Quantity to Both Sides of an Equation

Adding the same quantity to both sides of an equation does not change the solution set. That is, if

$a = b,\nonumber$

then adding c to both sides of the equation produces the equivalent equation

$a + c = b + c.\nonumber$

Let’s see if this works as advertised. Consider the equation x − 4=3. By inspection, 7 is the only solution of the equation. Now, let’s add 4 to both sides of the equation to see if the resulting equation is equivalent to x − 4 = 3.

$\begin{array}{rlrl}{x-4} & {=3} & {} & {\textcolor{red}{ \text{The given equation. }}} \\ {x-4+4} & {=3+4} & {} & {\textcolor{red}{ \text{Add 4 to both sides of the equation.} }} \\ {x} & {=7} & {} & {\textcolor{red}{ \text{Simplify both sides of the equation. }}}\end{array}\nonumber$

The number 7 is the only solution of the equation x = 7. Thus, the equation x = 7 is equivalent to the original equation x − 4 = 3 (they have the same solutions).

Important Point

Adding the same amount to both sides of an equation does not change its solutions.

It is also a fact that subtracting the same quantity from both sides of an equation produces an equivalent equation.

Subtracting the Same Quantity from Both Sides of an Equation

Subtracting the same quantity from both sides of an equation does not change the solution set. That is, if

$a = b,\nonumber$

then subtracting c from both sides of the equation produces the equivalent equation

$a − c = b − c.\nonumber$

Let’s also see if this works as advertised. Consider the equation

$x + 4 = 9.\nonumber$

By inspection, 5 is the only solution of the equation. Now, let’s subtract 4 from both sides of the equation to see if the resulting equation is equivalent to x + 4 = 9.

$\begin{array}{rlrl}{x+4} & {=9} & {} & {\textcolor{red}{ \text { The given equation. }}} \\ {x+4-4} & {=9-4} & {} & {\textcolor{red}{ \text { Subtract 4 from both sides of the equation. }}} \\ {x} & {=5} & {} & {\textcolor{red}{ \text { Simplify both sides of the equation. }}}\end{array}\nonumber$

The number 5 is the only solution of the equation $$x = 5$$. Thus, the equation $$x = 5$$ is equivalent to the original equation $$x + 4 = 9$$ (they have the same solutions).

Important Point

Subtracting the same amount from both sides of an equation does not change its solutions.

Writing Mathematics

When solving equations, observe the following rules to neatly arrange your work:

1. One equation per line. This means that you should not arrange your work like this:

$x+3=7 \quad x+3-3=7-3 \quad x=4\nonumber$

That’s three equations on a line. Rather, arrange your work one equation per line like this:

\begin{aligned} x+3 &=7 \\ x+3-3 &=7-3 \\ x &=4 \end{aligned}\nonumber

2. Add and subtract inline. Don’t do this:

$\begin{array}{r} x -7 & = & 12 \\ +7 & & + 7 \\ \hline x & = & 19 \end{array}\nonumber$

\begin{aligned} x-7 &=12 \\ x-7+7 &=12+7 \\ x &=19 \end{aligned}\nonumber

## Wrap and Unwrap

Suppose that you are wrapping a gift for your cousin. You perform the following steps in order.

1. Put the gift paper on.
2. Put the tape on.
3. Put the decorative bow on.

When we give the wrapped gift to our cousin, he politely unwraps the present, “undoing” each of our three steps in inverse order.

1. Take off the decorative bow.
2. Take off the tape.
3. Take off the gift paper.

This seemingly frivolous wrapping and unwrapping of a gift contains some deeply powerful mathematical ideas. Consider the mathematical expression $$x+ 4$$. To evaluate this expression at a particular value of x, we would start with the given value of x, then

Suppose we started with the number 7. If we add 4, we arrive at the following result: 11.

1. Subtract 4.

That is, we would take our result from above, 11, then subtract 4, which returns us to our original number, namely 7.

Addition and Subtraction as Inverse Operations

Two extremely important observations:

The inverse of addition is subtraction. If we start with a number x and add a number a, then subtracting a from the result will return us to the original number x. In symbols,

$x + a − a = x.\nonumber$

The inverse of subtraction is addition. If we start with a number x and subtract a number a, then adding a to the result will return us to the original number x. In symbols,

$x − a + a = x.\nonumber$

Example 5

Solve $$x − 8 = 10$$ for x.

Solution

To undo the effects of subtracting 8, we add 8 to both sides of the equation.

\begin{aligned} x-8=10 & \textcolor{red}{\text { Original equation. }} \\ x-8+8=10+8 & \textcolor{red}{ \text { Add 8 to both sides of the equation. }} \\ x=18 & \textcolor{red}{ \text { On the left, adding "undoes" the effect }} \\ & \textcolor{red}{ \text { of subtracting 8 and returns } x . \text { On the right, }} \\ & \textcolor{red}{10+8=18.} \end{aligned}\nonumber

Therefore, the solution of the equation is 18.

Check

To check, substitute the solution 18 into the original equation.

\begin{aligned} x - 8 = 10 & \textcolor{red}{ \text{ Original equation. }} \\ 18 - 8 = 10 & \textcolor{red}{ \text{ Substitute 18 for} x. } \\ 10 = 10 & \textcolor{red}{ \text{ Simplify both sides. }} \end{aligned}\nonumber

The fact that the last line of our check is a true statement guarantees that 18 is a solution of x − 8 = 10.

Exercise

Solve $$x + 5 = 12$$ for x.

7.

Example 6

Solve $$11 = y + 5$$ for y.

Solution

To undo the effects of adding 5, we subtract 5 from both sides of the equation.

\begin{aligned} 11 = y + 5 & \textcolor{red}{ \text{ Original equation. }} \\ 1 - 5 = y + 5 - 5 & \textcolor{red}{ \text{ Subtract 5 from both sides of the equation. }} \\ 6 = y & \textcolor{red}{ \text{ On the right, subtracting "undoes" the effect }} \\ & \textcolor{red}{ \text{ of adding 5 and returns } y. \text{ On the left, }} \\ & \textcolor{red}{ 11 - 5 = 6. } \end{aligned}\nonumber

Therefore, the solution of the equation is 6.

Check

\begin{aligned} 11 = y + 5 & \textcolor{red}{ \text{ Original equation. }} \\ 11 = 6 + 5 & \textcolor{red}{ \text{ Substitute 6 for } y.} \\ 11 = 11 & \textcolor{red}{ \text{ Simplify both sides. }} \end{aligned}\nonumber

The fact that the last line of our check is a true statement guarantees that 6 is a solution of 11 = y + 5.

Exercise

Solve $$y - 8 = 11$$ for y.

$$y = 19.$$

## Word Problems

The solution of a word problem must incorporate each of the following steps.

Requirements for Word Problem Solutions

1. Set up a Variable Dictionary. You must let your readers know what each variable in your problem represents. This can be accomplished in a number of ways:
1. Statements such as “Let P represent the perimeter of the rectangle.”
2. Labeling unknown values with variables in a table.
3. Labeling unknown quantities in a sketch or diagram.
2. Set up an Equation. Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement.
3. Solve the Equation. You must always solve the equation set up in the previous step.
4. Answer the Question. This step is easily overlooked. For example, the problem might ask for Jane’s age, but your equation’s solution gives the age of Jane’s sister Liz. Make sure you answer the original question asked in the problem. Your solution should be written in a sentence with appropriate units.
5. Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it’s possible that your equation incorrectly models the problem’s situation, so you could have a valid solution to an incorrect equation. The important question is: “Does your answer make sense based on the words in the original problem statement.”

Let's give these requirements a test drive.

Example 7

Four more than a certain number is 12. Find the number.

Solution

In our solution, we will carefully address each step of the Requirements for Word Problem Solutions.

1. Set up a Variable Dictionary. We can satisfy this requirement by simply stating “Let x represent a certain number.”

2. Set up an Equation. “Four more than a certain number is 12” becomes

\begin{aligned} \colorbox{cyan}{4} & \text{ more than } & \colorbox{cyan}{a certain number} & \text{ is } & \colorbox{cyan}{12} \\ 4 & + & x & = & 12 \end{aligned}\nonumber

3. Solve the Equation. To “undo” the addition, subtract 4 from both sides of the equation.

\begin{aligned} 4 + x = 12 & \textcolor{red}{ \text{ Original equation.}} \\ 4 + x - 4 = 12 - 4 & \textcolor{red}{ \text{Subtract 4 from both sides of the equation. }} \\ x = 8 & \textcolor{red}{ \text{ On the left, subtracting 4 "undoes" the effect}} \\ & \textcolor{red}{ \text{ of adding 4 and returns } x. \text{ On the right, }} \\ & \textcolor{red}{12 - 4 = 8.} \end{aligned}\nonumber

4. Answer the Question. The number is 8.

5. Look Back. Does the solution 8 satisfy the words in the original problem? We were told that “four more than a certain number is 12.” Well, four more than 8 is 12, so our solution is correct.

Exercise

12 more than a certain number is 19. Find the number.

7

Example 8

Amelie withdraws $125 from her savings account. Because of the withdrawal, the current balance in her account is now$1,200. What was the original balance in the account before the withdrawal?

Solution

In our solution, we will carefully address each step of the Requirements for Word Problem Solutions.

1. Set up a Variable Dictionary. We can satisfy this requirement by simply stating “Let B represent the original balance in Amelie’s account.”

2. Set up an Equation. We can describe the situation in words and symbols.

\begin{aligned} \colorbox{cyan}{Original Balance} & \text{ minus } & \colorbox{ Amelie's Withdrawal } & \text{ is } & \colorbox{cyan}{ Current Balance } \\ B & - & 125 & = & 1200 \end{aligned}\nonumber

3. Solve the Equation. To “undo” the subtraction, add 125 to both sides of the equation.

\begin{aligned} B - 125 = 1200 & \textcolor{red}{ \text{ Original equation. }} \\ B - 125 + 125 = 1200 + 125 & \textcolor{red}{ \text{Add 125 to both sides of the equation. }} \\ B = 1325 & \textcolor{red}{ \text{ On the left, adding 125 "undoes" the effect}} \\ & \textcolor{red}{ \text{ of subtracting 125 and returns } B. \text{ On the right, }} \\ & \textcolor{red}{ 1200 + 125 = 1325.} \end{aligned}\nonumber

4. Answer the Question. The original balance was $1,325. 5. Look Back. Does the solution$1,325 satisfy the words in the original problem? Note that if Amelie withdraws $125 from this balance, the new balance will be$1,200. Hence, the solution is correct.

Exercise

Fred withdraws $230 from his account, lowering his balance to$3,500. What was his original balance?