Skip to content
Math
Elementary Algebra
Solving Linear Equations and Inequalities
Solve Equations Using the Subtraction and Addition Properties of Equality

Solve Equations Using the Subtraction and Addition Properties of Equality

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, and translate and solve applications.

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!

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.

Example. Determine whether x=32x = \tfrac{3}{2} is a solution of 4x−2=2x+14x - 2 = 2x + 1.

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting 32\tfrac{3}{2} for xx.

4x−2=2x+1Substitute 32 for x.4(32)−2=?2(32)+1Multiply.6−2=?3+1Subtract.4=4  ✓ \begin{array}{lrcl} & 4x - 2 &=& 2x + 1 \\[4pt] \text{Substitute } \tfrac{3}{2} \text{ for } x. & 4\left(\tfrac{3}{2}\right) - 2 &\overset{?}{=}& 2\left(\tfrac{3}{2}\right) + 1 \\[4pt] \text{Multiply.} & 6 - 2 &\overset{?}{=}& 3 + 1 \\[4pt] \text{Subtract.} & 4 &=& 4 \; \checkmark \end{array}

Since x=32x = \tfrac{3}{2} results in a true equation (44 is in fact equal to 44), 32\tfrac{3}{2} is a solution to the equation 4x−2=2x+14x - 2 = 2x + 1.

Substitute y=43y = \tfrac{4}{3} into both sides of 9y+2=6y+39y + 2 = 6y + 3 and simplify. What value does the left side, 9y+29y + 2, come out to?

Since the left side simplifies to the same value as the right side (66), y=43y = \tfrac{4}{3} is a solution of 9y+2=6y+39y + 2 = 6y + 3.

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. Suppose both sides of a workspace have the same number of counters, but some counters are “hidden” in the envelope on the left, alongside three loose counters, while the right side shows eight loose counters. Can you tell how many counters are in the envelope?

Perhaps you are thinking: “I need to remove the three counters at the bottom left to get the envelope by itself. The three counters on the left can be matched with three on the right, so I can take them away from both sides. That leaves five on the right — so there must be five counters in the envelope.”

The algebraic equation that matches this situation is x+3=8x + 3 = 8, where each side of the workspace represents an expression and the equal sign takes the place of the center line. Let’s write algebraically the steps we took to discover how many counters were in the envelope:

x+3=8First, we took away three from each side.x+3−3=8−3Then we were left with five.x=5 \begin{array}{lrcl} & x + 3 &=& 8 \\[4pt] \text{First, we took away three from each side.} & x + 3 - 3 &=& 8 - 3 \\[4pt] \text{Then we were left with five.} & x &=& 5 \end{array}

Check: five in the envelope plus three more does equal eight!

5+3=85 + 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 aa, bb, and cc: if a=ba = b, then a−c=b−ca - c = b - c. When you subtract the same quantity from both sides of an equation, you still have 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.

Example. Solve: y+37=−13y + 37 = -13.

To get yy by itself, we will undo the addition of 3737 by using the Subtraction Property of Equality.

y+37=−13Subtract 37 from each side to ’undo’ the addition.y+37−37=−13−37Simplify.y=−50 \begin{array}{lrcl} & y + 37 &=& -13 \\[4pt] \text{Subtract 37 from each side to 'undo' the addition.} & y + 37 - 37 &=& -13 - 37 \\[4pt] \text{Simplify.} & y &=& -50 \end{array}

Check: substitute y=−50y = -50.

y+37=−13−50+37=?−13−13=−13  ✓ \begin{array}{rcl} y + 37 &=& -13 \\ -50 + 37 &\overset{?}{=}& -13 \\ -13 &=& -13 \; \checkmark \end{array}

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

Solve: x+19=−27x + 19 = -27.

Solve: x+16=−34x + 16 = -34.

What happens when an equation has a number subtracted from the variable, as in the equation x−5=8x - 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 aa, bb, and cc: if a=ba = b, then a+c=b+ca + c = b + c. When you add the same quantity to both sides of an equation, you still have equality.

In the previous example, 3737 was added to yy, and so we subtracted 3737 to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality.

Example. Solve: a−28=−37a - 28 = -37.

a−28=−37Add 28 to each side to ’undo’ the subtraction.a−28+28=−37+28Simplify.a=−9 \begin{array}{lrcl} & a - 28 &=& -37 \\[4pt] \text{Add 28 to each side to 'undo' the subtraction.} & a - 28 + 28 &=& -37 + 28 \\[4pt] \text{Simplify.} & a &=& -9 \end{array}

Check: substitute a=−9a = -9.

a−28=−37−9−28=?−37−37=−37  ✓ \begin{array}{rcl} a - 28 &=& -37 \\ -9 - 28 &\overset{?}{=}& -37 \\ -37 &=& -37 \; \checkmark \end{array}

The solution to a−28=−37a - 28 = -37 is a=−9a = -9.

Solve: n−61=−75n - 61 = -75.

Solve: p−41=−73p - 41 = -73.

Example. Solve: x−58=34x - \tfrac{5}{8} = \tfrac{3}{4}.

x−58=34Use the Addition Property of Equality.x−58+58=34+58Find the LCD to add the fractions on the right.x−58+58=68+58Simplify.x=118 \begin{array}{lrcl} & x - \tfrac{5}{8} &=& \tfrac{3}{4} \\[4pt] \text{Use the Addition Property of Equality.} & x - \tfrac{5}{8} + \tfrac{5}{8} &=& \tfrac{3}{4} + \tfrac{5}{8} \\[4pt] \text{Find the LCD to add the fractions on the right.} & x - \tfrac{5}{8} + \tfrac{5}{8} &=& \tfrac{6}{8} + \tfrac{5}{8} \\[4pt] \text{Simplify.} & x &=& \tfrac{11}{8} \end{array}

Check: substitute x=118x = \tfrac{11}{8}.

x−58=34118−58=?3468=?3434=34  ✓ \begin{array}{rcl} x - \tfrac{5}{8} &=& \tfrac{3}{4} \\[2pt] \tfrac{11}{8} - \tfrac{5}{8} &\overset{?}{=}& \tfrac{3}{4} \\[2pt] \tfrac{6}{8} &\overset{?}{=}& \tfrac{3}{4} \\[2pt] \tfrac{3}{4} &=& \tfrac{3}{4} \; \checkmark \end{array}

The solution to x−58=34x - \tfrac{5}{8} = \tfrac{3}{4} is x=118x = \tfrac{11}{8}.

Solve: p−23=56p - \tfrac{2}{3} = \tfrac{5}{6}.

Solve: q−12=56q - \tfrac{1}{2} = \tfrac{5}{6}.

The next example is an equation with decimals.

Example. Solve: n−0.63=−4.2n - 0.63 = -4.2.

n−0.63=−4.2Use the Addition Property of Equality.n−0.63+0.63=−4.2+0.63Add.n=−3.57 \begin{array}{lrcl} & n - 0.63 &=& -4.2 \\[4pt] \text{Use the Addition Property of Equality.} & n - 0.63 + 0.63 &=& -4.2 + 0.63 \\[4pt] \text{Add.} & n &=& -3.57 \end{array}

Check: let n=−3.57n = -3.57.

n−0.63=−4.2−3.57−0.63=?−4.2−4.2=−4.2  ✓ \begin{array}{rcl} n - 0.63 &=& -4.2 \\ -3.57 - 0.63 &\overset{?}{=}& -4.2 \\ -4.2 &=& -4.2 \; \checkmark \end{array}

Solve: b−0.47=−2.1b - 0.47 = -2.1.

Solve: c−0.93=−4.6c - 0.93 = -4.6.

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.

Example. Solve 9x−5−8x−6=79x - 5 - 8x - 6 = 7.

Step 1. Simplify the expressions on each side as much as possible. Rearrange the terms, using the Commutative Property of Addition, and combine like terms. Notice that each side is now simplified as much as possible.

9x−5−8x−6=7  ⇒  9x−8x−5−6=7  ⇒  x−11=79x - 5 - 8x - 6 = 7 \;\Rightarrow\; 9x - 8x - 5 - 6 = 7 \;\Rightarrow\; x - 11 = 7

Step 2. Isolate the variable. Now isolate xx. Undo subtraction by adding 1111 to both sides.

x−11+11=7+11x - 11 + 11 = 7 + 11

Step 3. Simplify the expressions on both sides of the equation.

x=18x = 18

Step 4. Check the solution. Substitute x=18x = 18.

9x−5−8x−6=79(18)−5−8(18)−6=?7162−5−144−6=?7157−144−6=?713−6=?77=7  ✓ \begin{array}{rcl} 9x - 5 - 8x - 6 &=& 7 \\ 9(18) - 5 - 8(18) - 6 &\overset{?}{=}& 7 \\ 162 - 5 - 144 - 6 &\overset{?}{=}& 7 \\ 157 - 144 - 6 &\overset{?}{=}& 7 \\ 13 - 6 &\overset{?}{=}& 7 \\ 7 &=& 7 \; \checkmark \end{array}

The solution to 9x−5−8x−6=79x - 5 - 8x - 6 = 7 is x=18x = 18.

Solve: 8y−4−7y−7=48y - 4 - 7y - 7 = 4.

Solve: 6z+5−5z−4=36z + 5 - 5z - 4 = 3.

Example. Solve 5(n−4)−4n=−85(n - 4) - 4n = -8.

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

5(n−4)−4n=−8Distribute on the left.5n−20−4n=−8Use the Commutative Property to rearrange terms.5n−4n−20=−8Combine like terms.n−20=−8Undo subtraction by using the Addition Property of Equality.n−20+20=−8+20Add.n=12 \begin{array}{lrcl} & 5(n - 4) - 4n &=& -8 \\[4pt] \text{Distribute on the left.} & 5n - 20 - 4n &=& -8 \\[4pt] \text{Use the Commutative Property to rearrange terms.} & 5n - 4n - 20 &=& -8 \\[4pt] \text{Combine like terms.} & n - 20 &=& -8 \\[4pt] \text{Undo subtraction by using the Addition Property of Equality.} & n - 20 + 20 &=& -8 + 20 \\[4pt] \text{Add.} & n &=& 12 \end{array}

Check: substitute n=12n = 12.

5(n−4)−4n=−85(12−4)−4(12)=?−85(8)−48=?−840−48=?−8−8=−8  ✓ \begin{array}{rcl} 5(n - 4) - 4n &=& -8 \\ 5(12 - 4) - 4(12) &\overset{?}{=}& -8 \\ 5(8) - 48 &\overset{?}{=}& -8 \\ 40 - 48 &\overset{?}{=}& -8 \\ -8 &=& -8 \; \checkmark \end{array}

The solution to 5(n−4)−4n=−85(n - 4) - 4n = -8 is n=12n = 12.

Solve: 5(p−3)−4p=−105(p - 3) - 4p = -10.

Solve: 4(q+2)−3q=−84(q + 2) - 3q = -8.

Example. Solve 3(2y−1)−5y=2(y+1)−2(y+3)3(2y - 1) - 5y = 2(y + 1) - 2(y + 3).

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

3(2y−1)−5y=2(y+1)−2(y+3)Distribute on both sides.6y−3−5y=2y+2−2y−6Use the Commutative Property of Addition.6y−5y−3=2y−2y+2−6Combine like terms.y−3=−4Undo subtraction by using the Addition Property of Equality.y−3+3=−4+3Add.y=−1 \begin{array}{lrcl} & 3(2y - 1) - 5y &=& 2(y + 1) - 2(y + 3) \\[4pt] \text{Distribute on both sides.} & 6y - 3 - 5y &=& 2y + 2 - 2y - 6 \\[4pt] \text{Use the Commutative Property of Addition.} & 6y - 5y - 3 &=& 2y - 2y + 2 - 6 \\[4pt] \text{Combine like terms.} & y - 3 &=& -4 \\[4pt] \text{Undo subtraction by using the Addition Property of Equality.} & y - 3 + 3 &=& -4 + 3 \\[4pt] \text{Add.} & y &=& -1 \end{array}

Check: let y=−1y = -1.

3(2y−1)−5y=2(y+1)−2(y+3)3(2(−1)−1)−5(−1)=?2(−1+1)−2(−1+3)3(−2−1)+5=?2(0)−2(2)3(−3)+5=?0−4−9+5=?−4−4=−4  ✓ \begin{array}{rcl} 3(2y - 1) - 5y &=& 2(y + 1) - 2(y + 3) \\ 3(2(-1) - 1) - 5(-1) &\overset{?}{=}& 2(-1 + 1) - 2(-1 + 3) \\ 3(-2 - 1) + 5 &\overset{?}{=}& 2(0) - 2(2) \\ 3(-3) + 5 &\overset{?}{=}& 0 - 4 \\ -9 + 5 &\overset{?}{=}& -4 \\ -4 &=& -4 \; \checkmark \end{array}

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=−1y = -1.

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

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

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 — words such as “is,” “is equal to,” “is the same as,” “the result is,” “gives,” “was,” and “will be” all translate to ==.

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.

Example. Translate and solve: Eleven more than xx is equal to 5454.

Translate.x+11=54Subtract 11 from both sides.x+11−11=54−11Simplify.x=43 \begin{array}{lrcl} \text{Translate.} & x + 11 &=& 54 \\[4pt] \text{Subtract 11 from both sides.} & x + 11 - 11 &=& 54 - 11 \\[4pt] \text{Simplify.} & x &=& 43 \end{array}

Check: is 5454 eleven more than 4343?

43+11=?5454=54  ✓43 + 11 \overset{?}{=} 54 \qquad 54 = 54 \; \checkmark

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

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

Example. Translate and solve: The difference of 12t12t and 11t11t is −14-14.

Translate.12t−11t=−14Simplify.t=−14 \begin{array}{lrcl} \text{Translate.} & 12t - 11t &=& -14 \\[4pt] \text{Simplify.} & t &=& -14 \end{array}

Check: substitute t=−14t = -14.

12(−14)−11(−14)=?−14−168+154=?−14−14=−14  ✓12(-14) - 11(-14) \overset{?}{=} -14 \qquad -168 + 154 \overset{?}{=} -14 \qquad -14 = -14 \; \checkmark

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

Translate and solve: The difference of 7a and 6a is -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 qq for the number of quarters if you were solving a problem about coins.

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.

Example. The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 5757 pounds. The second month, the newspapers weighed 2828 pounds. How much did the newspapers weigh the first month?

Read the problem: it is about the weight of newspapers. We are asked to find how much the newspapers weighed the first month. Let w=w = weight of the newspapers the first month.

Restate the problem in one sentence with the important information: weight of newspapers the first month plus the weight of the newspapers the second month equals 5757 pounds. Translate into an equation, using the variable ww:

w+28=57w + 28 = 57

Solve the equation:

w+28−28=57−28w=29w + 28 - 28 = 57 - 28 \qquad w = 29

Check: does the first month’s weight plus the second month’s weight equal 5757 pounds?

29+28=?5757=57  ✓29 + 28 \overset{?}{=} 57 \qquad 57 = 57 \; \checkmark

Answer the question with a complete sentence: the first month the newspapers weighed 2929 pounds.

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, in pounds?

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?

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

Read the problem. We are asked to find the sticker price of the car. Let s=s = the sticker price of the car.

Restate the problem in one sentence: $28,675 is $875 less than the sticker price. Translate into an equation:

28,675=s−87528{,}675 = s - 875

Solve the equation:

28,675+875=s−875+87529,550=s28{,}675 + 875 = s - 875 + 875 \qquad 29{,}550 = s

Check: is $875 less than $29,550 equal to $28,675?

29,550−875=?28,67528,675=28,675  ✓29{,}550 - 875 \overset{?}{=} 28{,}675 \qquad 28{,}675 = 28{,}675 \; \checkmark

The sticker price of the car was $29,550.

Translate into an algebraic equation and solve: Eddie paid $19,875 for his new car. This was $1,025 less than the sticker price. What was the sticker price of the car, in dollars?

Translate into an algebraic equation and solve: The admission price for the movies during the day is $7.75. This is $3.25 less than the price at night. How much does the movie cost at night, in dollars?

Key terms

solution of an equation — a value of a variable that makes a true statement when substituted into the equation. Subtraction Property of Equality — if a=ba = b, then a−c=b−ca - c = b - c; subtracting the same quantity from both sides of an equation preserves equality. Addition Property of Equality — if a=ba = b, then a+c=b+ca + c = b + c; adding the same quantity to both sides of an equation preserves equality.

This section is adapted from Elementary Algebra 2e, Section 2.1: Solve Equations Using the Subtraction and Addition Properties of Equality by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the envelope-and-counters figures as prose descriptions and the “Equals” word list and Step tables as prose/Callouts; omitted the Manipulative Mathematics callout, Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Solving Linear Equations and InequalitiesSolve Equations Using the Division and Multiplication Properties of Equality