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2.1: Solving Linear Equations in One Variable using Integers

Created by: CK-12

Objectives

The lesson objectives for Solving Linear Equations with Variables are:

  • Solving linear equations with variables on one side of the equation
  • Solving linear equations with variables on both sides of the equation
  • Solving linear equations that apply the distributive property

Solving Linear Equations with Variables on One Side of the Equation

Introduction

In this concept you will begin your study of mathematical equations by learning how to solve equations with only one variable. What this means is that the mathematical expression will have a letter and possibly some number on one side of the equal sign and numbers on the other side of the equal sign. It is your job to find a value for the letter that makes the expression true. Once you have mastered this, you can solve for two variables and also add more complex numbers.

It is interesting to note that the variables are known as literal coefficients. Literal means “letter”. So the literal coefficient is the unknown quantity in the mathematical expression represented by a letter (also called a variable). What do you think a numerical coefficient is?

Right! A numerical coefficient is a number in the mathematical expression. In the algebraic term 5abc, the numerical coefficient is 5.

Watch This

Khan Academy Slightly More Complicated Equations

Guidance

Erin, Jillian, Stephanie and Jacob went to the movies. The total bill for the tickets and snacks came to $72.00. How much should each teen pay to split the bill evenly.

There are four teens going to the movies (Erin, Jillian, Stephanie, and Jacob). The total bill was $72.00. Therefore our equation is 4x = 72. We divide by 4 to find our answer.

x &= \frac{72}{4}\\x &= 18

Therefore each teen will have to pay $18.00 for their movie ticket and snack.

Example A

5a + 2 = 17

The problem can be solved if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance.

In order to solve the equation, we have to get the literal coefficient (or the a) all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side.

Let’s first subtract 2 from both sides to get rid of the 2 on the left.

Since 5 is multiplied by a, we can get a by itself (or isolate it) by dividing by 5. Remember that whatever we do to one side, we have to do to the other.

If we simplify this expression, we get:

Therefore a = 3.

We can check our answer to see if we are correct.

5a + 2 &= 17\\5({\color{red}3}) + 2 &= 17\\15 + 2 &= 17\\17 &= 17 \ \ Y

Example B

7b - 7 = 42

Again, we can solve the problem if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance.

In order to solve the equation, we have to get the literal coefficient (or the b) all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side.

Let’s first add 7 from both sides to get rid of the 7 on the left.

Since 7 is multiplied by b, we can get a by itself (or isolate it) by dividing by 7. Remember that whatever we do to one side, we have to do to the other.

If we simplify this expression, we get:

Therefore b = 7.

We can check our answer to see if we are correct.

7b - 7 &= 42\\7({\color{red}7}) - 7 &= 42\\49 - 7 &= 42\\42 &= 42 \ \ Y

Example C

This same method can be extended by using algebra tiles. If we let white tiles represent the variable, green tiles represent the positive numbers and white tiles to represent the negative numbers, we can solve the equations using an alternate method.

The green algebra x-tiles represent variables (or the literal coefficients). Therefore there are 3 c blocks for the equation. The other green blocks represent the numbers or constants. There is a 2 on the left side of the equation so there are 2 green blocks. There is an 11 on the right side of the equation so there are 11 green blocks on the right side of the equation.

To solve, add two negative tiles to the right and left hand sides. The same rule applies to this problem as to all of the previous problems. Whatever we do to one side we have to do to the other.

This leaves us with the following:

We can reorganize these to look like the following:

Organizing the remaining algebra tiles allows us to realize the answer to be x = 3 or for our example b = 3.

Let’s do our check as with the previous two problems.

3c + 2 &= 11\\3({\color{red}3}) + 2 &= 11\\9 + 2 &= 11\\11 &= 11 \ \ Y

Vocabulary

Constant
A constant is also a numerical coefficient but does not contain a variable. For example in the equation 4x + 72 = 0, the 72 is a constant.
Equation
An equation is a mathematical equation with expressions separated by an equal sign.
Numerical Coefficient
In mathematical equations, the numerical coefficients are the numbers associated with the variable. For example, with the expression 4x, 4 is the numerical coefficient and x is the literal coefficient.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.

Guided Practice

  1. Use a model to solve for the variable in the problem x-5=12.
  2. Use a different model than used in question (1) to solve for the variable in the problem 3y+9=12.
  3. Using one of the models from the lesson, solve for x in the equation 3x-2x+16=-3.

Answers

1. x-5=12

Therefore, x=17. Let's do a check to make sure.

x - 5 &= 12\\({\color{red}17}) - 5 &= 12\\17 - 5 &= 12\\12 &= 12 \ \ Y

2. 3y+9=12

First you have to subtract 9 from both sides of the equation in order to start to isolate the variable.

Now, in order to get y all by itself, you have to divide both sides by 3. This will isolate the variable y.

Therefore, y=1. Let's do a check to make sure.

3y + 9 &= 12\\3({\color{red}1}) +9 &= 12\\3 + 9 &= 12\\12 &= 12 \ \ Y

3. 3x-2x+16=-3

You can use any method to solve this equation. Remember to isolate the x variable. You will notice here that there are two x values on the left. First let’s combine these terms.

3x-2x+16 &= -3\\x+16 &= -3

Now you can use any method to solve the equation. You now should just have to subtract 16 from both sides to isolate the x variable.

x+16 {\color{red}-16} &= -3 {\color{red}-16}\\x &= -19

Let's do a check to make sure.

3x-2x+16 &=-3 \quad \text{(original problem)}\\x+16 &= -3 \quad \text{(simplified problem)}\\{\color{red}-19}+16 &= -3\\-3 &= -3 \ \ Y

Summary

Solving equations with variables on one side can be done using models such as a balance (or a seesaw) or by using algebra tiles.

Remember that when solving equations with variables on one side of the equation there is one main rule to follow: whatever you do to one side of the equal sign you must do the same to the other side of the equal sign. For example, if you add a number to the left side of an equal sign, you must add the same number to the right side of the equal sign.

Problem Set

Use the model of the balance (or seesaw) to solve for each of the following variables.

  1. a+3=-5
  2. 2b-1=5
  3. 4c-3=9
  4. 2-d=3
  5. 4=3e=-2

Use algebra tiles to solve for each of the following variables.

  1. x+3=14
  2. 2y-7=5
  3. 3z+6=9
  4. 5+3x=-3
  5. 2x+2=-4

Use the models that you have learned to solve for the variables in the following problems.

  1. -4x+13=5
  2. 3x-5=22
  3. 11-2x=5
  4. 2x-4=4
  5. 5x+3=28

For each of the following models, write a problem involving an equation with a variable on one side of the equation expressed by the model and then solve for the variable.

Answers

Using the balance (seesaw) model...

  1. a+3=-5 Therefore a = - 8 \text{Check} &\\a+3 &=-5\\({\color{red}-8})+3 &= -5\\-5 &= -5 \ \ Y
  1. 4c-3=9 Therefore c = 3 \text{Check} &\\4c-3 &= 9\\4({\color{red}3})-3 &= 9\\12-3 &= 9\\9 &= 9 \ \ Y
  1. 4-3e=-2 Therefore e = 2 \text{Check} &\\4-3e &= -2\\4-3({\color{red}2}) &= -2\\4-6 &= -2\\-2 &= -2 \ \ Y

Using algebra tiles...

  1. x+3=14 Therefore x = 11 \text{Check} &\\x+3 &= 14\\{\color{red}11}+3 &= 14\\14 &= 14 \ \ Y
  1. 3z+6=9 Therefore z = 1 \text{Check} &\\3z+6 &= 9\\3({\color{red}1})+6 &= 9\\3+6 &= 9\\9 &= 9 \ \ Y
  1. 2x+2=-4 Therefore x = -3 \text{Check}&\\2x+2 &= -4\\2({\color{red}-3})+2 &= -4\\-6 + 2 &= -4\\-4 &= -4 \ \ Y

Use a model of your choice...

  1. -4x+13=5 Therefore x = 2 \text{Check} &\\-4x+13 &= 5\\-4({\color{red}2})+13 &= 5\\-8+13 &= 5\\5 &= 5 \ \ Y
  1. 11-2x=5 Therefore x = 3 \text{Check} &\\11-2x &= 5\\11-2(\color{red}3) &= 5\\11-6 &= 5\\5 &= 5 \ \ Y
  1. 5x+3=28 Therefore x = 5 \text{Check} &\\5x+3 &= 28\\5({\color{red}5})+3 &= 28\\25+3 &= 28\\28 &= 28 \ \ Y

For each of the following models...

  1. This is \text{Check} &\\x+3 &= 7\\{\color{red}4}+3 &= 7\\7 &= 7 \ \ Y x+3 &= 7\\x+3 {\color{red}-3} &= 7 {\color{red}-3}\\x &= 4
  1. This is 3x+6 &= 3\\3x+6 {\color{red}-6} &= 3 {\color{red}-6}\\3x &= -3\\\frac{3x}{3} &= \frac{-3}{3}\\x &= -1 \text{Check} &\\3x+6 &= 3\\3({\color{red}-1})+6 &= 3\\-3+6 &= 3\\3 &= 3 \ \ Y
  1. This is -3x+5 &= 8\\-3x+5 {\color{red}-5} &= 8 {\color{red}-5}\\-3x &= 3\\\frac{-3x}{-3} &= \frac{3}{-3}\\x &= -1 \text{Check} &\\-3x+5 &= 8\\-3({\color{red}-1})+5 &= 8\\3+5 &= 8\\8 &= 8 \ \ Y

Solving Linear Equations with Variables on Both Sides of the Equation

Introduction

In this concept you will learn to solve equations where there are variables on both sides of the equal sign. You will use the same models to solve these equations but the number of steps increases in these problems because now you have to combine like terms for the variables. Combining like terms means that you are putting all of the variables that match on the same side of the equation. A like term is one in which the degrees match and the variables match. So, for example, 3x and 4x are like terms, 3x and 4z are not. Three apples and four apples are like terms, three apples and four oranges are not.

Remember that your goal for solving any of these problems is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equal sign. Then you isolate the variables by multiplying or dividing. Learning the following lesson in solving equations with variables will help you when you learn the rules for solving more complex problems involving the distributive property in the next lesson.

Watch This

Khan Academy Equations with Variables on Both Sides

Guidance

Thomas has $50 and Jack has $100. Thomas is saving $10 per week for his new bike. Jack is saving $5 a week for his new bike. How long will it be before the two boys have the same amount of money?

If we let w be the number of weeks, we can write the following equation.

\underbrace{ 10x+50 }_{\text{Thomas's money:} \ \$10 \ \text{per week} + \$50}= \underbrace{ 5x+100 }_{\text{Jack's money:} \ \$5 \ \text{per week} + \$100}

We can solve the equation now by first combining like terms.

10x+50 &= 5x+100\\10x {\color{red}-5x}+50 &= 5x {\color{red}-5x}+100 && \text{-combining the} \ x \ \text{variables to left side of the equation}\\5x+50 {\color{red}-50} &= 100 {\color{red}-50} && \text{-combining the constants to left side of the equation}\\5x &= 50

We can now solve for x to find the number of weeks until the boys have the same amount of money.

5x &= 50\\\frac{5x}{5} &= \frac{50}{5}\\x &= 10

Therefore in 10 weeks Jack and Thomas will each have the same amount of money!

Example A

x+4=2x-6

We will solve this problem using the balance (or seesaw) used in the previous lesson.

You could first try to get the variables all on one side of the equation. You do this by subtracting x from both sides of the equation.

Next, isolate the x variable by adding 6 to both sides.

Therefore x = 10.

\text{Check}&\\x+4 &= 2x-6\\({\color{red}10})+4 &= 2({\color{red}10})-6\\14 &= 20-6\\14 &= 14 \ \ Y

Example B

14-3y=4y

We will solve this problem using algebra tiles used in the previous lesson.

We first have to combine our variables (x) tiles onto the same side of the equation. We do this by adding 3 x tiles to both sides of the equal sign. In this way the -3y will be eliminated from the left hand side of the equation.

By isolating the variable (y) we are left with these algebra tiles.

Rearranging we will get the following. {Note: Remember that rearranging is not necessary, it simply allows you to quickly see what the value for the variable is.}

\text{Check} &\\14 - 3y &= 4y\\14-3({\color{red}2}) &= 4({\color{red}2})\\14-6 &= 8\\8 &= 8 \ \ Y

Therefore y = 2.

Example C

We can use these same methods for any of the equations involving variables. Sometimes, however, numbers are so large that one method is more valuable than the other. Let’s look at the following problem.

53a-99=42a

To solve this problem, we may need to have a large number of algebra tiles! It might be more efficient to use the balance method to solve this problem.

\text{Check} &\\53a-99 &= 42a\\53({\color{red}9}) &= 42({\color{red}9})\\477-99 &= 378\\378 &= 378 \ \ Y

Therefore a = 9.

Vocabulary

Degrees
The degree is the exponent on the variable in a term. For example, in the term 4x, the exponent is 1 so the degree is 1.
Like Terms
Like terms refer to terms in which the degrees match and the variables match. For example 3x and 4x are like terms.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.

Guided Practice

  1. Use algebra tiles to solve for the variable in the problem 6x+4=5x-5.
  2. Use the balance (seesaw) method to solve for the variable in the problem 7r-4=3+8r.
  3. Determine the most efficient method to solve for the variable in the problem 10b-22=29-7b. Explain your choice of method for solving this problem.

Answers

1. 6x+4=5x-5

Therefore x = -9.

\text{Check}&\\6x+4 &= 5x-5\\6({\color{red}-9})+4 &= 5({\color{red}-9})-5\\-54+4 &= -45-5\\-50 &= -50 \ \ Y

2. 7r-4=3+8r

You can begin by combining the r terms. Subtract 8r from both sides of the equation.

You next have to isolate the variable. To do this, add 4 to both sides of the equation.

But there is still a negative sign with the r term. You now have to divide both sides by -1 to finally isolate the variable.

Therefore r = -7.

\text{Check} &\\7r-4 &= 3+8r\\7({\color{red}-7})-4 &= 3+8({\color{red}-7})\\-49-4 &= 3-56\\-53 &= -53 \ \ Y

3. 10b-22=29-7b

You could choose either method but there are larger numbers in this equation. With larger numbers, the use of algebra tiles is not an efficient manipulative. You should solve the problem using the balance (or seesaw) method. Work through the steps to see if you can follow them.

Therefore b = 17.

\text{Check} &\\10b-22 &= 29-7b\\10({\color{red}3})-22 &= 29-7({\color{red}3})\\30-22 &= 29-21\\8 &= 8 \ \ Y

Summary

The methods used for solving equations with variables on both sides of the equation are the same as the methods used to solve equations with variables on one side of the equation. What differs in this lesson is that there is the added step of combining like terms with the variables before isolating the variable to find the solution.

You must remember in these problems, as with any math problems involving an equal sign, whatever function (addition, subtraction, multiplication, or division) you do to one side of the equal sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance.

Problem Set

Use the balance (seesaw) method to find the solution for the variable in each of the following problems.

  1. 5p+3=-3p-5
  2. 6b-13=2b+3
  3. 2x-5=x+6
  4. 3x-2x=-4x+4
  5. 4t-5t+9=5t-9

Use algebra tiles to find the solution for the variable in each of the following problems.

  1. 6-2d=15-d
  2. 8-s=s-6
  3. 5x+5=2x-7
  4. 3x-2x=-4x+4
  5. 8+t=2t+2

Use the methods that you have learned for solving equations with variables on both sides to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable.

  1. 4p-7=21-3p
  2. 75-6x=4x-15
  3. 3t+7=15-t
  4. 5+h=11-2x
  5. 9-2e=3-e

For each of the following models, write a problem to represent the model and then find the variable for the problem.

Answers

Using the balance (seesaw) method...

  1. Therefore p = -1. \text{Check} &\\5p+3 &= -3p-5\\5({\color{red}-1})+3 &= -3({\color{red}-1})-5\\-5+3 &= 3-5\\-2 &= -2 \ \ Y
  1. Therefore x = 11. \text{Check}&\\2x-5 &= x+6\\2({\color{red}11})-5 &= ({\color{red}11})+6\\22-5 &= 11+66\\17 &= 17 \ \ Y
  1. Therefore t = 3. \text{Check} &\\4t-5t+9 &= 5t-9\\4({\color{red}3})-5({\color{red}3})+9 &= 5{\color{red}(3)}-9\\12-15+9 &= 15-9\\6 &= 6 \ \ Y

Using algebra tiles...

  1. Therefore x = -9 \text{Check}&\\6-2d &= 15-d\\6-2({\color{red}-9}) &= 15-({\color{red}-9})\\6+18 &= 15+9\\24 &= 24 \ \ Y
  1. Therefore x = -4. \text{Check} &\\5x+5 &= 2x-7\\5({\color{red}-4})+5 &= 2({\color{red}-4})-7\\-20+5 &= -8-7\\-15 &= -15 \ \ Y
  1. Therefore t = 6. \text{Check} &\\8+t &= 2t+2\\8+ {\color{red}6} &= 2({\color{red}6})+2\\14 &= 12+2\\14 &= 14 \ \ Y

Choose a method...

  1. Therefore p = 4. \text{Check} &\\4p-7 &= 21-3p\\4({\color{red}4})-7 &= 21-3({\color{red}4})\\16-7 &= 21-12\\9 &= 9
  1. Therefore t = 2. \text{Check} &\\3t+7 &= 15-t\\3({\color{red}2})+7 &= 15-({\color{red}2})\\6+7 &= 13\\13 &= 13 \ \ Y
  1. Therefore e = 6. \text{Check} &\\9-2e &= 3-e\\9-2({\color{red}6}) &= 3-({\color{red}6})\\9-12 &= -3\\-3 &= -3 \ \ Y

For each of the following models...

  1. -3x+17 &= 5-x\\-3x {\color{red}+x}+17 &= 5-x {\color{red}+x}\\-2x+17 &= 5\\-2x+17 {\color{red}-17} &= 5 {\color{red}-17}\\-2x &= -12\\\frac{-2x}{-2} &= \frac{-12}{-2}\\x &= 6 \text{Check} &\\-3x+17 &= 5-x\\-3({\color{red}6})+17 &= 5-({\color{red}6})\\-18+17 &= 5-6\\-1 &= -1 \ \ Y
  1. \text{Check} &\\5x+2 &= 3x-6\\5 ({\color{red}-4})+2 &= 3({\color{red}-4})-6\\-20+2 &= -12-6\\-18 &= -18 \ \ Y 5x+2 &= 3x-6\\5x {\color{red}-3x}+2 &= 3x {\color{red}-3x}-6\\2x+2 &= -6\\2x+2 {\color{red}-2} &= -6{\color{red}-2}\\2x &= -8\\\frac{2x}{2} &= \frac{-8}{2}\\x &= -4
  1. 7x+8 &= 4x-4\\7x {\color{red}-4x}+8 &= 4x {\color{red}-4x}-4\\3x+8 &= -4\\3x+8 {\color{red}-8} &= -4 {\color{red}-8}\\3x &= -12\\\frac{3x}{3} &= \frac{-12}{3}\\x &= -4 \text{Check} &\\7x+8 &= 4x-4\\7({\color{red}-4})+8 &= 4({\color{red}-4})-4\\-28+8 &= -16-4\\-20 &= -20 \ \ Y

Solving Linear Equations that Apply to the Distributive Property

Introduction

In this concept you will learn to solve equations with variables in which the equations involve the distributive property. The distributive property is a mathematical way of grouping terms. The distributive property states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. Wow! Let’s see what this looks like in math terms. Say you had {\color{red}3}({\color{blue}x + 5}). The distributive property states that the product of a number ({\color{red}3}) and a sum ({\color{blue}x + 5}) is equal to the sum of the individual products of the number ({\color{red}3}) and the addends ({\color{blue}x} and {\color{blue}5}).

When solving equations using the distributive property, you simply have one more step to follow than you have experienced in the previous lessons. You still have the same goal for solving problems with variables. That is to get the variables on one side and the constants on the other side. You do this by adding and subtracting terms from both sides of the equal sign. Then you isolate the variables by multiplying or dividing. In this lesson, your first step will be to use the distributive property to remove any brackets.

Watch This

Khan Academy Solving Equations with the Distributive Property

Guidance

Morgan, Connor, and Jake are going on a class field trip with 12 of their class mates. Each student is required to have a survival kit with a flashlight, a first aid kit, and enough food rations for the trip. A flashlight costs $10. A first aid kit costs $9. Each days food ration costs $7. If the class has only $500 for the survival kits, how many days can they go on their trip?

If First let’s write down what we know:

\# Students going on the trip = 15 (Morgan, Connor, Jake + 12 others)

Total money available = $500

Each flashlight costs $10

Each first aid kit costs $9

Each days food ration costs $7

One survival kit contains 1 flashlight + 1 first aid kit + x day’s rations

Therefore cost of each survival kit: \$10 + \$9 + \$7x = \$19 + \$7x

For the 15 students, the total cost of the survival kits would be:

15 (\$19+\$7x)=\$285+\$105x

Since the class has $1000 to spend, you can calculate how many days they can go on their trip.

\$1000 &= \$285+\$105x\\\$1000 {\color{red}- \$285} &= \$285 {\color{red}-\$285}+\$105x\\\$715 &= \$105x\\\frac{\$715}{\$105} &= \frac{\$105x}{\$105}\\6.81 &= x

Since the class does not have enough money to buy 7 days of food rations for each student, they will buy six days of food rations and the class will go on a class trip for six days.

Example A

2(3x+5)=-2

We will solve this problem using the balance (or seesaw) used in the previous lesson.

Your first step is to remove the brackets. To do this multiply the 2 by the numbers inside the brackets. Therefore multiply 2 by 3x and 2 by 5.

Next, isolate the x variable by subtracting 10 from both sides.

Simplifying you get:

Now divide by 6 to solve for the x variable.

To finally solve for the variable, simplify each side.

Therefore x = -2.

\text{Check}&\\2(3x+5) &= -2\\2(3({\color{red}-2})+5) &= -2\\2(-6+5) &= -2\\2(-1) &= -2\\-2 &= -2 \ \ Y

Example B

3(4+3y)=-6

We will solve this problem using algebra tiles used in previous lessons.

Since all of our variables (x tiles) are on the same side, you only need to subtract the 3 groups of 4 from both sides of the equal sign to isolate the variable.

Simplifying the right side of the equation and rearranging the tiles leave the following:

Therefore y = -2.

\text{Check} &\\3(4+3y) &= -6\\3(4+3({\color{red}-2})) &= -6\\3(4-6) &= -6\\3(-2) &= -6\\-6 &= -6 \ \ Y

Example C

6(4x+1)=-18

Therefore x = -1.

\text{Check} &\\6(4x+1) &= -18\\6(4({\color{red}-1})+1) &= -18\\6(-4+1) &= -18\\6(-3) &= -18\\-18 &= -18 \ \ Y

Vocabulary

Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: {\color{red}3}({\color{blue}x + 5}), the distributive property states that the product of a number ({\color{red}3}) and a sum ({\color{blue}x + 5}) is equal to the sum of the individual products of the number ({\color{red}3}) and the addends ({\color{blue}x} and {\color{blue}5}).
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.

Guided Practice

  1. Use balance (seesaw) method to solve for the variable in the problem 6(-3x+2)=-6.
  2. Use the algebra tiles to solve for the variable in the problem 2(r-2)=-6.
  3. Determine the most efficient method to solve for the variable in the problem 2(j+1)+3=3(j-1). Explain your choice of method for solving this problem.

Answers

1. 6(-3x+2)=-6.

You can begin by using the distributive property on the left side of the equation.

Next subtract 12 from both sides of the equation to get the variable term alone.

Simplifying, you get:

You next have to isolate the variable. To do this, divide both sides of the equation by -18.

Finally, simplifying leaves you with:

Therefore x = 1.

\text{Check} &\\6(-3x+2) & =-6\\6(-3({\color{red}1})+2) &= -6\\6(-3+2) &= -6\\6(-1) &= -6\\-6 &= -6 \ \ Y

2. 2(r-2)=-6

First, you need to add 4 to both sides to get the variables alone on the one side of the equal sign.

Simplifying leaves you with:

Therefore r = -1.

\text{Check} &\\2(r-2) &= -6\\2(({\color{red}-1})-2) &= -6\\2(-3) &= -6\\-6 &= -6 \ \ Y

3. 2(j+1)+3=3(j-1).

You could choose either method for this problem. Below you will see the balance method used to solve the problem. Work through the steps to see if you can follow them. Remember since there are brackets, you must start with the distributive property and remove the brackets.

Therefore j = 5.

\text{Check} &\\2(j+1) &= 3(j-1)\\2(({\color{red}5})+1) &= 3(({\color{red}5})-1)\\2(6) &= 3(4)\\12 &= 12 \ \ Y

Summary

The previous three lessons began your study of solving equations with variables. In this final lesson of the section, you were working with the distributive property. Equations with one variable that involve the distributive property are solved in the exact same way as any other equations with one variable. The only difference is that the very first step with distributive property problems is to remove the brackets.

To remove brackets, remember that you must multiply all the numbers inside the brackets by the number outside the brackets. After you remove brackets, you can then solve the equation by combining like terms, moving constants to one side of the equal sign, variables to the other side of the equal sign, then isolating the variable to find the solution.

Again, as with previous lessons, you must remember in these problems, as with any math problems involving an equal sign, whatever function (addition, subtraction, multiplication, or division) you do to one side of the equal sign, you must do to the other side. This is a big rule to remember in order for equations to remain equal or to remain in balance.

Problem Set

Use the balance (seesaw) method to find the solution for the variable in each of the following problems.

  1. 5(4x+3)=75
  2. 3(s-4)=15
  3. 5(k-4)=10
  4. 43=4(t+6)-1
  5. 6(x+4)=3(5x+2)

Use algebra tiles to find the solution for the variable in each of the following problems.

  1. 2(d-3)=4
  2. 5+2(x+7)=20
  3. 2(3x-4)=22
  4. 2(3x+2)-x=-6
  5. 2(x+4)-x=9

Use the methods that you have learned for solving equations with variables to solve for the variables in each of the following problems. Remember to choose an efficient method to solve for the variable.

  1. -6=-6(3x-8)
  2. -2(x-2)=11
  3. 2+3(-2x+1)=20
  4. 3(x+2)-x=12
  5. 5(2-3x)=-8-6x

Answers

Using the balance (seesaw) method...

  1. Therefore x = 3. \text{Check} &\\5(4x+3) &= 75\\5(4({\color{red}3})+3) &= 75\\5(12+3) &= 75\\5(15) &= 75\\75 &= 75 \ \ Y
  1. Therefore k = 6. \text{Check} &\\5(k-4) &= 10\\5(({\color{red}6})-4) &= 10\\5(2) &= 10\\10 &= 10 \ \ Y
  1. Therefore x = 2. \text{Check} &\\6(x+4) &= 3(5x+2)\\6(({\color{red}2})+4) &= 3(5({\color{red}5})+2)\\6(6) &= 3(10+2)\\36 &= 3(12)\\36 &= 36 \ \ Y

Using algebra tiles...

  1. Therefore x = 5 \text{Check} &\\2(d-3) &= 4\\2(({\color{red}5})-3) &= 4\\2(2) &= 4\\4 &= 4 \ \ Y
  1. Therefore x = 5. \text{Check}&\\2(3x-4) &= 22\\2(3({\color{red}5})-4) &= 22\\2(15-4) &= 22\\2(11) &= 22\\22 &= 22 \ \ Y
  1. Therefore x = 1. \text{Check} &\\2(x+4)-x &= 9\\2(({\color{red}1})+4)-({\color{red}1}) &= 9\\2(5)-1 &= 9\\10-1 &= 9\\9 &= 9 \ \ Y

Choose a method...

  1. Therefore x = 3. & \text{Check} \\-6 &= -6(3x-8)\\-6 &= -6(3({\color{red}3})-8)\\-6 &= -6(9-8)\\-6 &= -6(1)\\-6 &= -6 \ \ Y
  1. Therefore x = -3. \text{Check} &\\2+3(-2x+1) &= 23\\2+3(-2(-3)+1) &= 23\\2+3(6+1) &= 23\\2+3(7) &= 23\\23 &= 23 \ \ Y
  1. Therefore x = 2. \text{Check} &\\5(2-3x) &= -8-6x\\5(2-3({\color{red}2})) &= -8-6({\color{red}2})\\5(2-6) &= -8-12\\5(-4) &= -20\\-20 &= -20 \ \ Y

Summary

In this lesson, you have worked with linear equations in one variable. You have learned how to solve problems where variables were on one side of the equation, where variables were on both sides of the equation, and also when there are brackets involved in the equations. The key to solving these equations is to make sure that whatever you do to one side of the equal sign, you do the other. In other words, if you add a number to the left hand side of the equal sign, add it to the right hand side. In this way the equation stays in balance.

When solving equations, you want to get variables on one side of the equal sign by adding or subtracting constant terms and then solving for the variable by multiplying or dividing by whatever term is attached to your variable. When you have brackets, your very first step is to use the distributive property and remove these brackets.

As well, you used two main methods to solve these equations as you begin learning to solve linear equations. You learned to solve using a balance (or seesaw) where the middle of the seesaw represents the equal sign. The other method is using algebra tiles. Although algebra tiles are great for solving these types of equations, when numbers get large, they become very cumbersome and therefore not very efficient. It may be more helpful to use another method.

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Jan 16, 2013

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Mar 10, 2014
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