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1. Equations & Inequalities
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Algebra 1

1.03 Properties of equality

Lesson
Practice

Properties of equality

An equation is a mathematical relation statement where two equivalent expressions and values are separated by an equal sign. The solutions to an equation are the values of the variable(s) that make the equation true. Equivalent equations are equations that have the same solutions.

Equations, particularly in real-world contexts, are sometimes referred to as constraints as they describe restrictions or limitations of the given situation. One familiar type of equation is a linear equation.

Linear equation

An equation that contains a variable term with an exponent of 1, and no variable terms with exponents other than 1.

Example:

2x + 3 = 5

Equations are often used to solve mathematical and real-world problems. To solve equations, we use a variety of inverse operations to "undo" what was done to a variable. For instance, if a variable was multiplied by a number, we would use division to get the variable by itself.

The distributive property is an important property that we use frequently to simplify and solve equations:

\displaystyle a\left(b+c\right)=ab + ac
\bm{a,\,b,\,c}
are real numbers

Exploration

Consider the undeniably true statement 5=5. Perform each of the following operations:

  • Add 3 to both sides
  • Subtract 1 from both sides
  • Multiply -2 to both sides
  • Divide both sides by 3
  1. What do you notice about the equation that results from performing each operation?

Properties of equality are facts about equations. They describe different operations that can be performed on an equation that would maintain the truth of the equation statement. The following are the properties of equality and identity:

Symmetric property of equality\text{If } a=b, \text{then } b=a
Transitive property of equality\text{If } a=b \text{ and } b=c, \text{then } a=c
Addition property of equality\text{If } a=b, \text{then } a+c=b+c
Subtraction property of equality\text{If } a=b, \text{then } a-c=b-c
Multiplication property of equality\text{If } a=b, \text{then } ac=bc
Division property of equality\text{If } a=b \text{ and } c \neq0, \text{then } \dfrac{a}{c}=\dfrac{b}{c}
Substitution property of equality\text{If } {a=b}, \text{ then } b \text{ may be substituted for } a \text{ in any expression}
Additive identity\text{If } {a=b}, \text{ then } a+0=b \text{ and } a=b+0
Multiplicative identity\text{If } {a=b}, \text{ then } a\cdot 1=b \text{ and } a=b \cdot 1

Using algebra tiles, we can represent these properties to show the balance of the two sides. The balance below represents -2=3x+2(-x+1).

A balance scale with algebra tiles. On the left side, there are two -1 tiles. On the right side, there are three +x tiles, two -x tiles, and two +1 tiles.

Keeping the two sides balanced, we want to add or remove tiles to work towards a single x tile. We see from the balance that 2(-x+1) represents -2x+2.

A balance scale with algebra tiles. On the left side, there are two -1 tiles. On the right side, there are one +x tile, and two +1 tiles.

By eliminating the zero pairs on the right side, we are left with -2=x+2.

By using the subtraction property of equality, we will make zero pairs with the +2 on the right side of the equation to isolate the variable.

A zero pair on the right side leaves us with x+0 and using the additive identity we are left with just x.

A balance scale with algebra tiles. On the left side, there are four -1 tiles. On the right side, there are one +x tile, two +1 tiles, and two -1 tiles.
-2 - 2 = x + 0
A balance scale with algebra tiles. On the left side, there are four -1 tiles. On the right side, there is one +x tile.
-4=x

Solutions can be verified a variety of ways, including visually using algebra tiles. After solving -2=3x+2(-x+1), we want to verify that x=-4.

Each +x tile will be replaced with four -1 tiles. A -x tile will change the four -1 tiles to +1 tiles.

A balance scale with algebra tiles. On the left side, there are two -1 tiles. On the right side, there are twelve -x tiles, and ten +1 tiles.
Substituting x=-4 into -2=3x +2(-x+1)
A balance scale with algebra tiles. On the left side, there are two -1 tiles. On the right side, there are two -1 tiles.
Zero pairs keep the equation balanced at -2=-2

Examples

Example 1

Viola and Akim were asked to solve the equation -10\left(3x+7\right)+4=54.

Viola solved it like this:

\displaystyle -10\left(3x+7\right)+4\displaystyle =\displaystyle 54Given equation
\displaystyle -30x-70+4\displaystyle =\displaystyle 54
\displaystyle -30x-66\displaystyle =\displaystyle 54
\displaystyle -30x-66+66\displaystyle =\displaystyle 54+66
\displaystyle -30x+0\displaystyle =\displaystyle 120Evaluate the addition
\displaystyle -30x\displaystyle =\displaystyle 120
\displaystyle -30x\div\left(-30\right)\displaystyle =\displaystyle 120\div\left(-30\right)
\displaystyle 1 \cdot x\displaystyle =\displaystyle -4Evaluate the division
\displaystyle x\displaystyle =\displaystyle -4

Akim solved the equation like this:

\displaystyle -10\left(3x+7\right)+4\displaystyle =\displaystyle 54Given equation
\displaystyle -10\left(3x+7\right)+4-4\displaystyle =\displaystyle 54-4
\displaystyle -10\left(3x+7\right)+0\displaystyle =\displaystyle 50Evaluate the addition
\displaystyle -10\left(3x+7\right)\displaystyle =\displaystyle 50
\displaystyle \dfrac{-10\left(3x+7\right)}{-10}\displaystyle =\displaystyle \dfrac{50}{-10}
\displaystyle 1\left(3x+7\right)\displaystyle =\displaystyle -5Evaluate the division
\displaystyle 3x+7\displaystyle =\displaystyle -5
\displaystyle 3x+7-7\displaystyle =\displaystyle -5-7
\displaystyle 3x+0\displaystyle =\displaystyle -12Evaluate the addition
\displaystyle 3x\displaystyle =\displaystyle -12
\displaystyle \dfrac{3x}{3}\displaystyle =\displaystyle \dfrac{-12}{3}
\displaystyle 1x\displaystyle =\displaystyle -4Evaluate the division
\displaystyle x\displaystyle =\displaystyle -4
a

Use properties of equality and identities to justify each missing step of their work.

Worked Solution
Create a strategy

From line to line, identify what changed and what operation was used, then find the corresponding property or identity.

Apply the idea

Viola's work:

\displaystyle -10\left(3x+7\right)+4\displaystyle =\displaystyle 54Given equation
\displaystyle -30x-70+4\displaystyle =\displaystyle 54Distributive property
\displaystyle -30x-66\displaystyle =\displaystyle 54Evaluate the addition
\displaystyle -30x-66+66\displaystyle =\displaystyle 54+66Addition property of equality
\displaystyle -30x+0\displaystyle =\displaystyle 120Evaluate the addition
\displaystyle -30x\displaystyle =\displaystyle 120Additive identity
\displaystyle -30x\div\left(-30\right)\displaystyle =\displaystyle 120\div\left(-30\right)Division property of equality
\displaystyle 1 \cdot x\displaystyle =\displaystyle -4Evaluate the division
\displaystyle x\displaystyle =\displaystyle -4Multiplicative identity

Akim's work:

\displaystyle -10\left(3x+7\right)+4\displaystyle =\displaystyle 54Given equation
\displaystyle -10\left(3x+7\right)+4-4\displaystyle =\displaystyle 54-4Subtraction property of equality
\displaystyle -10\left(3x+7\right)+0\displaystyle =\displaystyle 50Evaluate the addition
\displaystyle -10\left(3x+7\right)\displaystyle =\displaystyle 50Additive identity
\displaystyle \dfrac{-10\left(3x+7\right)}{-10}\displaystyle =\displaystyle \dfrac{50}{-10}Division property of equality
\displaystyle 1\left(3x+7\right)\displaystyle =\displaystyle -5Evaluate the division
\displaystyle 3x+7\displaystyle =\displaystyle -5Multiplicative identity
\displaystyle 3x+7-7\displaystyle =\displaystyle -5-7Subtraction property of equality
\displaystyle 3x+0\displaystyle =\displaystyle -12Evaluate the addition
\displaystyle 3x\displaystyle =\displaystyle -12Additive identity
\displaystyle \dfrac{3x}{3}\displaystyle =\displaystyle \dfrac{-12}{3}Division property of equality
\displaystyle 1x\displaystyle =\displaystyle -4Evaluate the division
\displaystyle x\displaystyle =\displaystyle -4Multiplicative identity
Reflect and check

Although Viola and Akim solved the problem in different ways, they both used inverse operations to arrive at the same answer.

b

Compare their strategies.

Worked Solution
Create a strategy

We want to consider how their strategies are similar and how they are different. We can look at how many steps were taken and which properties were used.

Apply the idea

Viola decided to expand and simplify the left side first, then use inverse operations and apply properties of equality. Akim solved the equation by directly using inverse operations from the start. They both arrived at the same answer, but Viola used less steps.

Reflect and check

We can notice that Viola and Akim used different notation for simple operations. For example, Viola represented division using the \div symbol, however Akim used fractions. Also, Viola used explicit multiplication in their second last step of their working, where as Akim used implicit multiplication. In both cases, the difference between notation has no impact on the properties or the results of the identities they use.

Example 2

Verify that the x=3 is a solution to the equation 3 − 6x + 2x = -9.

Worked Solution
Create a strategy

A value is a solution if it can be substituted into the equation and make the equation true.

Apply the idea

We will substitute x=3 into the equation and see if both sides of the equation are the same after evaluating.

\displaystyle 3-6\cdot(3)+2\cdot(3)\displaystyle =\displaystyle -9Substitute x=3
\displaystyle 3-18+6\displaystyle =\displaystyle -9Evaluate the multiplication
\displaystyle -15+6\displaystyle =\displaystyle -9Evaluate the subtraction
\displaystyle -9\displaystyle =\displaystyle -9Evaluate the addition

Since the two sides of the equation are equal, x=3 is a solution.

Reflect and check

Algebra tiles can also be used to verify that a value is a solution to an equation. The original equation 3-6x+2x=-9 is shown.

The image shows algebra tiles. On the left side of the equals sign there are three +1 tiles, six -x tiles, and two +x tiles. On the right side there are nine -1 tiles.

Each +x tile will be replaced by three +1 tiles and -x tiles will change the +1 to -1.

The image shows algebra tiles. On the left side of the equals sign there are three +1 tiles, 2 groups of nine -1 tiles, and six +1 tiles. On the right side there are nine -1 tiles.

A solution will show the same value on both sides of the equals sign after zero pairs cancel out.

The image shows algebra tiles. On the left side of the equals sign there are three +1 tiles, 2 groups of nine -1 tiles, and six +1 tiles. On the right side there are nine -1 tiles. Ask your teacher for more information.
The image shows algebra tiles. On the left side of the equals sign there are nine -1 tiles. On the right side there are nine -1 tiles.

Since the same number of -1 tiles are shown on each side, x=3 is a solution to 3-6x+2x=-9.

Example 3

Solve the following equations and justify each step.

a

2\left(-\dfrac{a}{3}\right) + 7 = 15

Worked Solution
Create a strategy

Consider how the expression was constructed, starting from the variable. Once this is identified, we can solve the equation by applying the inverse operations in the reverse order.

Apply the idea
\displaystyle 2\left(-\dfrac{a}{3}\right) + 7\displaystyle =\displaystyle 15Given equation
\displaystyle 2\left(-\dfrac{a}{3}\right)+7-7\displaystyle =\displaystyle 15-7Subtraction property of equality
\displaystyle 2\left(-\dfrac{a}{3}\right)+0\displaystyle =\displaystyle 8Evaluate the subtraction
\displaystyle 2\left(-\dfrac{a}{3}\right)\displaystyle =\displaystyle 8Additive identity
\displaystyle 2\left(-\dfrac{a}{3}\right)\div 2\displaystyle =\displaystyle 8\div 2Division property of equality
\displaystyle 1 \cdot \left(-\dfrac{a}{3}\right)\displaystyle =\displaystyle 4Evaluate the division
\displaystyle -\dfrac{a}{3}\displaystyle =\displaystyle 4Multiplicative identity
\displaystyle -\dfrac{a}{3} \cdot \left(-3\right)\displaystyle =\displaystyle 4 \cdot \left(-3\right)Multiplication property of equality
\displaystyle 1 \cdot a\displaystyle =\displaystyle -12Evaluate the multiplication
\displaystyle a\displaystyle =\displaystyle -12Multiplicative identity
Reflect and check

Sometimes, it is easier to simplify the problem before solving for a. In this case, we could have multiplied by 2 first, then solved the equation. Either way, we get the same answer.

\displaystyle 2\left(-\dfrac{a}{3}\right)+7\displaystyle =\displaystyle 15Given equation
\displaystyle -\dfrac{2a}{3}+7\displaystyle =\displaystyle 15Evaluate the multiplication
\displaystyle -\dfrac{2a}{3}+7-7\displaystyle =\displaystyle 15-7Subtraction property of equality
\displaystyle -\dfrac{2a}{3}+0\displaystyle =\displaystyle 8Evaluate the subtraction
\displaystyle -\dfrac{2a}{3}\displaystyle =\displaystyle 8Additive identity
\displaystyle -\dfrac{2a}{3}\div -\dfrac{2}{3}\displaystyle =\displaystyle 8\div -\dfrac{2}{3}Division property of equality
\displaystyle 1 \cdot a\displaystyle =\displaystyle -12Evaluate the division
\displaystyle a\displaystyle =\displaystyle -12Multiplicative identity
b

\dfrac{5b}{3}-2b=-1

Worked Solution
Create a strategy

Since each term on the left hand side of the equation has the variable b in it, we can start by removing the fraction so we can collect like terms.

Apply the idea
\displaystyle \dfrac{5b}{3}-2b\displaystyle =\displaystyle -1Given equation
\displaystyle \dfrac{5b}{3}\cdot 3-2b\cdot 3\displaystyle =\displaystyle -1\cdot 3Multiplication property of equality
\displaystyle 1 \cdot 5b-6b\displaystyle =\displaystyle -3Evaluate the multiplication
\displaystyle 5b-6b\displaystyle =\displaystyle -3Multiplicative identity
\displaystyle -b\displaystyle =\displaystyle -3Collect like terms
\displaystyle -b\cdot \left(-1\right)\displaystyle =\displaystyle -3\cdot \left(-1\right)Multiplication property of equality
\displaystyle b\displaystyle =\displaystyle 3Evaluate the multiplication
Reflect and check

You may find that the working could have been condensed to something like this:

\displaystyle \dfrac{5b}{3}-2b\displaystyle =\displaystyle -1Given equation
\displaystyle 5b-6b\displaystyle =\displaystyle -3Multiplication property of equality
\displaystyle -b\displaystyle =\displaystyle -3Collect like terms
\displaystyle b\displaystyle =\displaystyle 3Multiplication property of equality

Often, steps such as evaluating sums or products, adding 0, or multiplying by 1, are combined into one step.

c

x+\dfrac{4x+7}{3}=1

Worked Solution
Create a strategy

It is easiest to work with whole numbers, so we can multiply everything by 3 to eliminate the fraction.

Apply the idea
\displaystyle x+\dfrac{4x+7}{3}\displaystyle =\displaystyle 1Given equation
\displaystyle 3x+4x+7\displaystyle =\displaystyle 3Multiplication property of equality
\displaystyle 7x+7\displaystyle =\displaystyle 3Combine like terms
\displaystyle 7x\displaystyle =\displaystyle -4Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle -\dfrac{4}{7}Division property of equality
Reflect and check

We can check our answer by plugging it back into the original equation.

\displaystyle \left(-\dfrac{4}{7}\right)+\dfrac{4\left(-\frac{4}{7}\right)+7}{3}\displaystyle =\displaystyle 1Substitue x=-\dfrac{4}{7}
\displaystyle -\dfrac{4}{7}+\dfrac{-\frac{16}{7}+7}{3}\displaystyle =\displaystyle 1Evaluate the multiplication
\displaystyle -\dfrac{4}{7}+\dfrac{\frac{33}{7}}{\frac{3}{1}}\displaystyle =\displaystyle 1Evaluate the addition in the numerator
\displaystyle -\dfrac{4}{7}+\dfrac{11}{7}\displaystyle =\displaystyle 1Evaluate the division
\displaystyle \dfrac{7}{7}\displaystyle =\displaystyle 1Evaluate the addition
\displaystyle 1\displaystyle =\displaystyle 1Evaluate the division
d

0.5x+2\left(1.2x+3\right)=11.8

Worked Solution
Create a strategy

Before solving for x, we need to simplify the left side of the equation.

Apply the idea
\displaystyle 0.5x+2\left(1.2x+3\right)\displaystyle =\displaystyle 11.8Given equation
\displaystyle 0.5x+2.4x+6\displaystyle =\displaystyle 11.8Distributive property
\displaystyle 2.9x+6\displaystyle =\displaystyle 11.8Combine like terms
\displaystyle 2.9x\displaystyle =\displaystyle 5.8Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle 2Division property of equality
Reflect and check

If you prefer working with whole numbers instead of decimals, you could have multiplied everything by 10 after the second step.

\displaystyle 0.5x+2\left(1.2x+3\right)\displaystyle =\displaystyle 11.8Given equation
\displaystyle 0.5x+2.4x+6\displaystyle =\displaystyle 11.8Distributive property
\displaystyle 5x+24x+60\displaystyle =\displaystyle 118Multiplication property of equality
\displaystyle 29x+60\displaystyle =\displaystyle 118Combine like terms
\displaystyle 29x\displaystyle =\displaystyle 58Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle 2Division property of equality

Example 4

Yolanda works at a restaurant 5 nights a week and receives tips. On the first three nights, the total tips she received was \$32,\,\$27,\, and \$26. She earned twice as much in tips on the fourth night compared to the fifth night. The average amount of tips received per night for the week was \$29.

If the amount she received on the fifth night was \$k, determine how much she received that night.

Worked Solution
Create a strategy

The average is equal to the sum of values divided by the number of values. We can use this to build an equation to represent the constraint in terms of k about the average tips Yolanda received. Then, we can solve the equation for k.

Apply the idea

We can represent her earnings on the fourth night as 2k, as it is twice her earnings on the fifth night, k.

Since \$29 is equal to the average of her tips over five nights, we can write an equation representing the average of her tips in terms of k:29=\dfrac{32+27+26+2k+k}{5}where, 2k and k represent her earnings on the fourth and fifth night respectively.

Now, we can solve the equation for k.

\displaystyle 29\displaystyle =\displaystyle \dfrac{32+27+26+2k+k}{5}Start with equation in terms of k
\displaystyle 29\displaystyle =\displaystyle \dfrac{85+3k}{5}Combine like terms in numerator
\displaystyle 29\displaystyle =\displaystyle \dfrac{3k+85}{5}Commutative property of addition
\displaystyle 29 \cdot 5\displaystyle =\displaystyle \dfrac{3k+85}{5} \cdot 5Multiplication property of equality
\displaystyle 145\displaystyle =\displaystyle \left(3k+85\right) \cdot 1Evaluate product on both sides of equation
\displaystyle 145\displaystyle =\displaystyle 3k+85 Multiplicative identity
\displaystyle 145-85\displaystyle =\displaystyle 3k+85-85Subtraction property of equality
\displaystyle 60\displaystyle =\displaystyle 3k+0Combine like terms on both sides of equation
\displaystyle 60\displaystyle =\displaystyle 3kAdditive identity
\displaystyle \dfrac{60}{3}\displaystyle =\displaystyle \dfrac{3k}{3}Division property of equality
\displaystyle 20\displaystyle =\displaystyle 1kEvaluate the division on both sides of the equation
\displaystyle 20\displaystyle =\displaystyle kMultiplicative identity
\displaystyle k\displaystyle =\displaystyle 20Symmetric property of equality

Yolanda received \$20 on the fifth night.

Reflect and check

We can check that the solution, k=20, is correct and makes sense in this context.

Firstly, since Yolanda's earnings from tips on the fifth night is equal to \$20, her earnings on the fourth night is equal to \$40. If we use these numbers to calculate the average over the five nights, we get:

\displaystyle \text{Average tips}\displaystyle =\displaystyle \dfrac{32+27+26+40+20}{5}Calculate average over five nights
\displaystyle =\displaystyle \dfrac{145}{5}Collect like terms on numerator
\displaystyle =\displaystyle 29Evaluate the division

We can see that this gives us the same value for the average tips over the five nights as specified in the question. Notice that both of the values for the tips earned on the fourth and fifth night are positive numbers, which makes sense within the context of the question. We may have questioned our solution if we returned a negative value for tips or the average.

Idea summary

To identify the properties of equalities needed to solve an expression, consider how an expression was constructed, starting from the variable. Then, solve the equation by applying the inverse operations in the reverse order and matching the operation to the correct property.

  • Symmetric property of equality: if a=b, then b=a.
  • Transitive property of equality: if a=b and b=c, then a=c.
  • Addition property of equality: if a=b, then a+c=b+c.
  • Subtraction property of equality: if a=b, then a-c=b-c.
  • Multiplication property of equality: if a=b, then ac=bc.
  • Division property of equality: if a=b and c \neq 0, then a \div c=b\div c.
  • Substitution property of equality: if a=b, then b may be substituted for a in any expression containing a.

Outcomes

A.EI.1

The student will represent, solve, explain, and interpret the solution to multistep linear equations and inequalities in one variable and literal equations for a specified variable.

A.EI.1a

Write a linear equation or inequality in one variable to represent a contextual situation.

A.EI.1b

Solve multistep linear equations in one variable including those in contextual situations, by applying the properties of real numbers and/or properties of equality.

A.EI.1f

Verify possible solution(s) to multistep linear equations and inequalities in one variable algebraically, graphically, and with technology to justify the reasonableness of the answer(s). Explain the solution method and interpret solutions for problems given in context.

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