2. Solving One-Step Equations in One Variable
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Chapter 1
2.
Solving One-Step Equations in One Variable
This lesson delves into the topic of one-step equations, emphasizing the role of inverse operations and properties of equality in finding solutions. It provides real-life examples to illustrate how these mathematical concepts are not just academic exercises but also practical tools for everyday decision-making. For instance, the material covers scenarios like calculating distances, determining ages, and budgeting. The aim is to equip learners with the skills to solve simple equations quickly and accurately, whether they are students trying to excel in algebra or adults looking to make informed decisions in their daily lives.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving One-Step Equations in One Variable
Slide of 13
After writing an equation modeling some real-life situation, the next step is to solve it. To do so, it is necessary to apply different Properties of Equality. Along this lesson, these properties will be introduced and applied to equations that can be solved in one step.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
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Different Ages of Family Members
Zosia's brother is 10 years older than she is. Also, Zosia's brother is half their father's age.
a If Zosia is 16 years old, write equations for the ages of her brother and father.
b Solve these equations to find their ages.
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Reflexive, Symmetric, and Transitive Properties of Equality
The Properties of Equality are rules that allow manipulation of an equation in such a way that an equivalent equation is obtained. These properties will be reviewed in sets. The first set of properties is shown below.
Rule
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Reflexive Property of Equality
For any real number, the number is equal to itself.
a=a
This property is an axiom, so it does not need a proof. This property is used to solve equations. For example, consider the equation below. x + 3 = 8
The sum on the left-hand side can be interpreted as a number, so this sum has to equal 8. This implies that x must be equal to 5.Rule
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Symmetric Property of Equality
For all real numbers, the order of an equality does not matter. Let a and b be real numbers.
If a=b, then b=a.
Rule
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Transitive Property of Equality
For all real numbers, if two numbers are equal to the same number, then they are equal to each other. Let a, b, and c be real numbers.
If a=b and b=c, then a=c.
This property can be used together with other Properties of Equality to solve equations. x = 5y-30 5y-30 = 20 ⇒ x &=20
Since this property is an axiom, it does not need a proof to be accepted as true. The Transitive Property of Equality also holds true if a, b, and c are complex numbers.Pop Quiz
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Which is the Correct Property?
Select the appropriate property for each example.
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Isolating the Variable
When solving equations, certain operations usually need to be undone to isolate the variable on one side. For example, consider the following equation. x+5 = 7 To isolate x, the addition of 5 on the left-hand side has to be undone. Here is where inverse operations come into play.
Concept
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Inverse Operations
Inverse operations are two operations that undo one another. For example, adding 6 and subtracting 6 are inverse operations because they cancel each other out. This means that adding 6 to any number and then subtracting 6 results in the original number.
&10 + 6 - 6 &10 + 6 - 6 &10
Equations are solved by using inverse operations. By the Properties of Equality, any operation performed on one side of an equation must also be performed on the other side of the equation to maintain equality. Consider an example.
x-4=9
This equation can be solved by adding 4 to both sides.
In this case, the subtraction on the left-hand side of the equation can only be eliminated by adding 4 on both sides. The result of applying the Properties of Equality on an equation is an equivalent equation.
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Addition and Subtraction Properties of Equality
Some of the most commonly used inverse operations are addition and subtraction. These operations fall under the Addition Property of Equality and the Subtraction Property of Equality.
Rule
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Addition Property of Equality
Adding the same number to both sides of an equation results in an equivalent equation. Let a, b, and c be real numbers.
If a = b, then a + c = b + c.
The Addition Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider an example. x-3=5 By adding 3 to both sides of the equation, the variable x can be isolated and the solution to the equation can be found.
Rule
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Subtraction Property of Equality
Subtracting the same number from both sides of an equation results in an equivalent equation. Let a, b, and c be real numbers.
If a = b, then a - c = b - c.
The Subtraction Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider an example. x+2=7 By subtracting 2 from both sides of the equation, the variable x can be isolated and the solution to the equation can be found.
Example
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Finding How Much Money Davontay Had
One of Davontay's hobbies is playing the saxophone. He plays in the school band and wants to join a local community band as well. He goes to the music store to buy more reeds for his saxophone since he will be spending more time playing. After spending $20 on reeds, he is left with $140.
a Write an equation for this situation.
b How much money did Davontay originally have?
Answer
a x - 20 = 140
b $160
Hint
a Use a variable for the original amount.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a It is possible write this situation as an equation using variables. Let x be the original amount of money that Davontay had. It is given that Davontay spent $20. Therefore, after buying the reeds, Davontay had x - 20 dollars. This expression is the first part of the desired equation.
x - 20 Then, it is given that Davontay has $140 left after spending $20. The equation can be completed by equating the previous expression with 140.
x - 20 = 140 b To find how much money Davontay had initially, the equation set in Part A needs to be solved for x. To do so, $20 will be added to both sides of the equation. This is valid by the Addition Property of Equality.
Therefore, Davontay originally had $160.
Example
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Finding the Number of Pairs of Pants
Heichi like collecting a particular brand of clothing. He is inspecting his wardrobe before going to the mall. He notices that he has 7 shirts and 4 more shirts than pairs of pants.
a If p is the number of pairs of pants that Heichi has, which equation correctly describes the given situation?
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b How many pairs of pants does Heichi have?
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Hint
a Use variables to describe the situations.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a To write an equation that describes the given situation, variables need to be assigned to each item.
Number of Pairs of Pants:& p Number of Shirts:& s Since Heichi has 4 more shirts than pants, the sum of p and 4 is equal to s. p + 4 = s Also, it is given that Heichi has 7 shirts. Therefore, by the Transitive Property of Equality, the equation can be rewritten using this information. p+4 = s s = 7 ⇒ p + 4 = 7
b To find out how many pairs of pants Heichi has, the equation set in Part A needs to be solved for p. To do so, 4 will be subtracted from both sides of the equation. This is valid by the Subtraction Property of Equality.
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Multiplication and Division Properties of Equality
The other most common type of inverse operations are the multiplication and division operations. These are valid by the following properties of equality.
Rule
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Multiplication Property of Equality
Given an equation, multiplying each side of the equation by the same number yields an equivalent equation. Let a, b, and c be real numbers.
If a = b, then a * c = b * c.
The Multiplication Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider the following example. x÷4&=2 x÷4 * 4&=2 * 4 x&=8
Here, by multiplying both sides of the equation by 4, the variable x was isolated and the solution of the equation was found. Note that the Multiplication Property of Equality also holds true if a, b, and c are complex numbers.Rule
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Division Property of Equality
Dividing each side of an equation by the same nonzero number yields an equivalent equation. Let a, b, and c be real numbers.
If a = b and c≠ 0, then a ÷ c = b ÷ c.
The Division Property of Equality is an axiom, so it does not need a proof to be accepted as true. This property is one of the Properties of Equality that can be used when solving equations. 5x&=10 5x ÷ 5&=10 ÷ 5 x&=2
As can be observed, by dividing both sides of the equation by 5, the variable x was isolated and the solution of the equation was found. Note that the Division Property of Equality also holds true if a, b, and c are complex numbers.Example
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Distances From Home
LaShay's hobbies include playing golf with her father. She wants to go to the golf course to practice. She knows that her school is one quarter of the way from her house to the golf course.
Consider that the school is about 2.5 miles from LaShay's house.
a If g is the distance from LaShay's house to the golf course, which equation correctly describes the situation?
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b How far is the golf course from LaShay's house?
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Hint
a Use variables for each distance.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a To write an equation that describes the given situation, variables need to be assigned to each distance.
Distance to School:&s Distance to the Golf Course:&g Since the school is one quarter of the way to the golf course, the value of the division of g by 4 is equal to the value of s. g/4 = s Also, it is given that the distance to school is 2.5 miles. s = 2.5 Using the Transitive Property of Equality, it is possible to rewrite the first equation. g/4 = s s = 2.5 ⇒ g/4 = 2.5
b To find out how far is the golf course from LaShay's house, the equation set in Part A needs to be solved. To do so, both sides of the equation will be multiplied by 4. This is valid by the Multiplication Property of Equality.
Therefore, the golf course is 10 miles away from LaShay's house.
g/4 = 2.5
MultEqn
LHS * 4=RHS* 4
4* g/4 = 4* 2.5
DenomMultFracToNumber
4 * a/4= a
g = 4 * 2.5
Multiply
Multiply
g = 10
Example
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The Number of Chess Pieces
Zain is in the chess club at their school. During one game, their opponent has double the pieces that Zain has. Zain's opponent has 6 pieces left on the board.
a If z is the number of pieces that Zain has, which equation correctly describes the situation?
{"type":"choice","form":{"alts":["2z = 6","6z = 2","z\/6=2","z\/2=6"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
b How many pieces does Zain have left?
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Hint
a Use variables for the number of pieces of each player.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a An equation can be written if variables are used for the number of pieces each player has left.
Number of Zain's Pieces: & z Number of Opponent's Pieces: & o Since Zain's opponent has twice the pieces that Zain has, multiplying z by 2 is equal to o. 2z = o It is given that the opponent has 6 pieces left. Therefore, o is equal to 6. Using the Transitive Property of Equality, it is possible to rewrite the equation. 2z = o o = 6 ⇒ 2z = 6
b To find the number of pieces that Zain has left, both sides of the equation will be divided by 2. This is valid by the Division Property of Equality.
2z = 6
DivEqn
.LHS / 2.=.RHS / 2.
2z/2 = 6/2
CancelCommonFac
Cancel out common factors
2z/2 = 6/2
SimpQuot
Simplify quotient
z = 6/2
CalcQuot
Calculate quotient
z = 3
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Solving Equations Using the Properties of Equality
Find the value of the variable on each equation using the Properties of Equality.
Closure
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Finding the Ages of Family Members
The challenge at the beginning of the lesson gave some information about Zosia's family.
- Zosia is 16 years old.
- Zosia's brother is 10 years older than she is.
- Zosia's brother is half the age of their father.
Then, the following exercises were presented.
a Write equations for the ages of Zosia's brother and father.
b Solve these equations to find their ages.
Answer
a b - 10 = 16 f2 = b
b Brother's Age: 26 years
Father's Age: 52 years
Father's Age: 52 years
Hint
a Use variables for the ages of each person.
b Use the Properties of Equality to solve each equation set in Part A.
Solution
a To write equations with the given information, it is convenient to use variables to represent each person's age.
Zosia's Age:& z Zosia's Brother's Age:& b Zosia's Father's Age:& f Zosia's brother is 10 years older than Zosia. This means that subtracting 10 from b is equal to z. b - 10 = z Also, it is given that Zosia is 16 years old. Using the Transitive Property of Equality, it is possible to rewrite this equation. b - 10 = z z = 16 ⇒ b - 10 = 16 Finally, since Zosia's brother's age is half of their father's age, the dividing f by 2 is equal to b. f/2 = b Therefore, these are the two equations for the ages of Zosia's brother and father. b - 10 = 16 f2 = b
b To find the age of Zosia's brother, 10 will be added to both sides of the first equation. This is valid by the Addition Property of Equality.
f/2 = 26
MultEqn
LHS * 2=RHS* 2
2* f/2 = 2* 26
DenomMultFracToNumber
2 * a/2= a
f = 2 * 26
Multiply
Multiply
f = 52
Solving One-Step Equations in One Variable
Exercises
The price of candy is $0.40 per gram. How much candy can Tiffaniqua buy if she has $2 to spend?
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Let x be the amount of candy in grams that we can buy for $2. If we multiply the price of a gram of candy $0.40 by the amount of candy that we can buy x, the resulting product is $2. This can be written as an equation. 0.40x = 2 To find the value of x, we can divide both sides of the equation by 0.40.
Therefore, Tiffaniqua can buy 5 grams of candy for $2.
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Solve the following equations.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"z=","formTextAfter":null,"answer":{"text":["21"]}}
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We can solve the equation by adding 10 to both sides. Doing this isolates x on the left-hand side of the equation.
Therefore, x=37 is the solution to the equation!
On the left-hand side we have 5 times the value of y. Since we want to find the value of y, we need to divide both sides of the equation by 5.
The value of y is 4.
In this equation, we have one third of z. We will multiply both sides of the equation by 3 to isolate z and find its value.
The value of z is 21.
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Tearrik worked 2.5 hours and earned $9. How much would he earn if he worked 4.5 hours at the same salary?
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To find out how much money Tearrik would earn working 4.5 hours, we need to find out how much he earns per hour. Let x be Tearrik's salary. The result of multiplying 2.5 by x is equal to 9. This can be written as an equation. 2.5x = 9 To find Tearrik's salary, we will divide both sides of the equation by 2.5.
We found that Tearrik earns $ 3.60 each hour he works. Now we can multiply his salary by 4.5 to find how much he would earn if he worked for 4.5 hours. 4.5 * 3.60 = 16.20 Tearrik would earn $16.20 if he worked 4.5 hours.
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Solve the following equations. Check your answer.
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To solve the equation we need to isolate x on one side of the equation. We will do this by subtracting 3 from both sides of the equation.
We found that x=4 is the solution to the equation. To check the answer, we will substitute 4 for x into the original equation.
Since both sides of the equation are equal, the solution is correct.
We will isolate a on one side of the equation to find the solution. To do so, we will add 5 to both sides of the equation.
The value of a is 24. To check if this value is correct, we will substitute 24 for a into the original equation.
Since both sides of the equation are equal, the solution is correct.
To solve the equation, we have to isolate k on one side of the equation. We will do this by subtracting 6 from both sides of the equation.
We found that k=-23. To verify if this is a correct solution, we will substitute -23 for k into the original equation.
Both sides of the equation are equal. Now we can be sure that we found the correct value.
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Solve the following equations. Check your answer.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"m=","formTextAfter":null,"answer":{"text":["7"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"t=","formTextAfter":null,"answer":{"text":["6"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"q=","formTextAfter":null,"answer":{"text":["-72"]}}
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To solve the equation, we need to isolate m on one side of the equation. We will do this by dividing both sides of the equation by -8.
We found that m=7 is the solution to the equation. To check the answer, we will substitute 7 for m into the original equation.
Since both sides of the equation are equal, the solution is correct.
To find the solution, we will isolate t on one side of the equation. We will do this by dividing both sides of the equation by 2.5.
The value of t is 6. To check if this solution is correct, we will substitute 6 for t into the original equation.
Since both sides of the equation are equal, the solution is correct.
To solve the equation, we have to isolate q on one side of the equation. We will do this by multiplying both sides of the equation by 6.
We found the value of q is -72. To verify if this is the correct solution, we will substitute -72 for q in the equation.
Both sides of the equation are equal, so we can be sure that we found the correct value.
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Solving One-Step Equations in One Variable
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