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# 2.2: Solving Linear Equations in One Variable using Decimals and Fractions

Difficulty Level: At Grade Created by: CK-12

Objectives

The lesson objectives for Solving Linear Equations in One Variable using Decimals and Fractions are:

• Solving linear equations with one variable involving decimals
• Solving linear equations with one variable involving fractions
• Solving linear equations with decimals and fractions that apply the distributive property

## Solving Linear Equations with One Variable Involving Decimals

Introduction

In the previous concept you learned to solve equations with one variable that involved whole numbers. In this concept you will apply the same rules to solve mathematical problems with equations that have decimals. The equations in this concept still only have one variable and it is still your mission to isolate your variable and then find a solution for it.

In the previous lesson you used two main methods to solve for equations with on variable. You used the balance (seesaw) method as well as algebra tiles. In this concept, you will use the balance (seesaw) method but will begin to use algebra statements to solve these equations. You can still use algebra tiles if you like by following the method learned in the previous concept.

What is most important when solving problems involving decimals is to remember that they look worse than they are and what's even better is that you can make them look easier. Take for example the problem below.

\begin{align*}0.1x+0.4=0.5\end{align*}

At first this looks difficult. But multiply all of the numbers by 10 and see what happens.

\begin{align*}{\color{red}(10)}0.1x+{\color{red}(10)}0.4 &= {\color{red}(10)}0.5\\ 1x+4 &= 5\\ or\\ x+4 &= 5\end{align*}

Now can you see the solution for \begin{align*}x\end{align*}? That's right! It is 1.

So you have learned that when you have decimals in the mathematical problem involving decimals, you are first going to remove the decimals by multiplying by 10 (if one decimal place), 100 (if two decimal places), or 1000(if three decimal places).

Watch This

Guidance

Karen wants to design a garden for her back yard. She knows she only has space for a rectangular garden of a perimeter equal to 60 feet. She needs to know the dimensions. The width will be half the length. What will be the dimensions of the garden?

\begin{align*}Perimeter \ (P) &= 60 \ feet\\ P &= 2l + 2w\\ w &= \frac{1}{2} l\end{align*}

Therefore: \begin{align*}P=2l+2 \left(\frac{1}{2}\right)l\end{align*}

or

\begin{align*}P &= 2l+\cancel{2} \left(\frac{1}{\cancel{2}}\right)l\\ P &= 2l+l\\ P &= 3l\end{align*}

Since you know that Karen has 60 feet for her perimeter, you can substitute 60 in for \begin{align*}P\end{align*}.

\begin{align*}60=3l\end{align*}

Now you can solve for \begin{align*}l\end{align*} (the length).

\begin{align*}\frac{60}{3} &= \frac{3l}{3}\\ l &= 20 \ feet\end{align*}

You now know that the length is 20 feet. You also know that the width is \begin{align*}\frac{1}{2}\end{align*} the length. So you can solve for the width.

\begin{align*}w &= \frac{1}{2} l\\ w &= \frac{1}{2} (20)\\ w &= 10 \ feet\end{align*}

Therefore the dimensions of Karen’s garden are \begin{align*}20 \ feet \times 10 \ feet\end{align*}.

Example A

\begin{align*}a + 2.3 = 4.7\end{align*}

Again, you can think of this problem being solved if you think about the problem in terms of a balance (or a seesaw). You 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.

Like always, in order to solve the equation, we have to get the literal coefficient (or the \begin{align*}a\end{align*}) all by itself. Always remember that you need to keep the balance horizontal. This means that whatever you do to one side of the equation, you have to do to the other side.

Let’s subtract 2.3 from both sides to get rid of the 2.3 on the left and isolate the variable.

If you simplify this expression, you get:

Therefore \begin{align*}a = 2.4\end{align*}.

You can, as always, check your answer to see if you are correct.

\begin{align*}a + 2.3 &= 4.7\\ ({\color{red}2.4}) + 2.3 &= 4.7\\ 4.7 &= 4.7 \ \ Y\end{align*}

Example B

\begin{align*}b - 1.6 = 2.4\end{align*}

This time let’s solve the problem using algebra tiles. Now you know that algebra tiles are used previously with whole numbers and they are not easily broken into pieces to make fractions. But we can make this problem more user friendly for solving with algebra tiles. Let’s multiply this problem by 10 to get rid of the decimal.

\begin{align*}{\color{red}(10)}b - {\color{red}(10)}1.6 &= {\color{red}(10)}2.4 \\ 10b - 16 &= 24\end{align*}

Now you could use algebra tiles to solve for the variable.

Your first step would be to add 16 to both sides of the equal sign.

Simplifying and rearranging you would get:

Therefore \begin{align*}b = 4\end{align*}.

Let’s do your check as with the previous problems.

\begin{align*}3b - 1.6 &= 2.4\\ ({\color{red}4}) - 1.6 &= 2.4\\ 2.4 &= 2.4 \ \ Y\end{align*}

It could be noted that since the numbers were so large with this problem, the balance (or seesaw) method is more efficient. This is true with most problems involving decimals. In fact, algebra tiles are rarely more efficient when solving problems with variables involving decimals. Having said this, you may find it a valuable solution method for you in your problem solving.

Example C

\begin{align*}6.4c - 2.1 = 7.5\end{align*}

You can again use the balance (seesaw) method to solve this problem.

Let’s first add 2.1 to both sides to get rid of the 2.1 on the left.

Simplifying you get:

Since 6.4 is multiplied by \begin{align*}c\end{align*}, you can get a by itself (or isolate it) by dividing by 6.4. Remember that whatever you do to one side, you have to do to the other.

If we simplify this expression, we get:

Therefore \begin{align*}c = 1.5\end{align*}.

\begin{align*}6.4c - 2.1 &= 7.5\\ 6.4({\color{red}1.5}) - 2.1 &= 7.5\\ 9.6 - 2.1 &= 7.5\\ 7.5 &= 7.5 \ \ Y\end{align*}

Vocabulary

Equation
An equation is a mathematical equation with expressions separated by an equal sign.
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 \begin{align*}t-1.4=0.6\end{align*}.
2. Use a model than used to solve for the variable in the problem \begin{align*}1.2s+3.9=4.1\end{align*}.
3. Using the rules from the lesson, solve for x in the equation \begin{align*}7-0.03x=1.72x+1.75\end{align*}.

1. \begin{align*}t-1.4=0.6\end{align*}

You can use either method to solve this problem but let’s use algebra tiles for this one. Remember since there are decimals (to one decimal place to be exact), you first must multiply the equation by 10 to get rid of the decimals.

\begin{align*}{\color{red}(10)}t-{\color{red}(10)}1.4 &= {\color{red}(10)}0.6\\ 10t-14 &= 6\end{align*}

You now can add 14 to each side to isolate the variable.

Simplifying and rearranging leaves you with:

Therefore, \begin{align*}t=2\end{align*}. Let's do a check to make sure.

\begin{align*}t - 1.4 &= 0.7\\ ({\color{red}2}) - 1.4 &= 0.7\\ 0.7 &= 0.7 \ \ Y\end{align*}

2. \begin{align*}1.2s+3.9=4.1\end{align*}

First you have to subtract 3.9 from both sides of the equation in order to start to isolate the variable.

Now, in order to get \begin{align*}s\end{align*} all by itself, you have to divide both sides by 1.2. This will isolate the variable \begin{align*}s\end{align*}

Therefore, \begin{align*}s=0.17\end{align*}. Let's do a check to make sure.

\begin{align*}1.2s + 3.9 &= 4.1\\ 1.2({\color{red}0.17}) +3.9 &= 4.1\\ 0.20 + 3.9 &= 4.1\\ 4.1 &= 4.1 \ \ Y\end{align*}

3. \begin{align*}7-0.03x=1.72x+1.75\end{align*}

You can use any method to solve this equation. Remember to isolate the \begin{align*}x\end{align*} variable. You will notice here that there are two \begin{align*}x\end{align*} values, one on each side of the equation. First let’s first combine these terms by adding \begin{align*}0.03x\end{align*} to both sides of the equation.

\begin{align*}7-0.03x {\color{red}+0.03x}=1.72x {\color{red}+0.03x}+1.75\end{align*}

Simplifying you get:

\begin{align*}7=1.75x+1.75\end{align*}

Now you can use any method to solve the equation. You now should just have to subtract 1.75 from both sides to isolate the \begin{align*}x\end{align*} variable.

\begin{align*}7{\color{red}-1.75}=1.75x+1.75 {\color{red}-1.75}\end{align*}

Simplifying you get:

\begin{align*}5.25=1.75x\end{align*}

Now to solve for the variable, you need to divide both sides by 1.75.

\begin{align*}\frac{5.25}{1.75}=\frac{1.75x}{1.75}\end{align*}

You can now solve for \begin{align*}x\end{align*}.

\begin{align*}x = 3\end{align*}

Let's do a check to make sure.

\begin{align*}7-0.03x &= 1.72x+1.75 \quad (\text{original problem})\\ 7 &= 1.75x+1.75 \quad (\text{simplified problem})\\ 7 &= 1.75({\color{red}3})+1.75\\ 7 &= 7 \ \ Y\end{align*}

Summary

Solving equations with one variable with decimals involves the same rules when there are whole numbers. The goal is still to isolate the variable by combining like terms and putting all constant terms on one side of the equal sign with variables on the opposite side of the equal sign. The same two methods for solving equations with one variable can be used for these problems. These two methods are the balance (seesaw) model and using algebra tiles.

When using algebra tiles, the first step is always to multiply by 10, 100, or 1000 in order to remove the decimal. In other words, a tenths decimal place value (0.5) would be multiplied by 10. A hundredths place value decimal would be multiplied by 100 (0.55). A thousandths place value decimal would be multiplied by 1000 (0.555). Having said this, with equations with one variable involving the decimal, the use of algebra tiles may become more cumbersome as the numbers are larger.

Problem Set

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

1. \begin{align*}a+0.3=-0.5\end{align*}
2. \begin{align*}2b-1.5=6.3\end{align*}
3. \begin{align*}1.4c-3.2=4.9\end{align*}
4. \begin{align*}2.1-1.5d=3.2\end{align*}
5. \begin{align*}4.5-3.1e=-2.2\end{align*}

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

1. \begin{align*}0.5x+0.2=0.7\end{align*}
2. \begin{align*}0.2y-0.9=0.5\end{align*}
3. \begin{align*}0.2z-0.3=0.7\end{align*}
4. \begin{align*}0.07+0.05x=-0.03\end{align*}
5. \begin{align*}0.05x+0.16=-0.04\end{align*}

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

1. \begin{align*}0.87+0.15x=-0.03\end{align*}
2. \begin{align*}0.52x+0.12=-0.4\end{align*}
3. \begin{align*}0.25z-3.3=0.7\end{align*}
4. \begin{align*}0.6x-1.25=0.4x+0.35\end{align*}
5. \begin{align*}x-0.3+0.05x=2-1.4x\end{align*}

Using the balance (seesaw) model...

1. \begin{align*}a+0.3=-0.5\end{align*} Therefore \begin{align*}a = - 0.8\end{align*} \begin{align*}\text{Check} &\\ a+0.3 &= -0.5\\ ({\color{red}-0.8})+0.3 &= -0.5\\ -0.5 &= -0.5 \ \ Y\end{align*}
1. \begin{align*}1.4c-3.2=4.9\end{align*} Therefore \begin{align*}c = 5.78\end{align*} \begin{align*}\text{Check}&\\ 1.4c-3.2 &= 4.9\\ 1.4({\color{red}5.78})-3.2 &= 4.9\\ 8.1-3.2 &= 4.9\\ 4.9 &= 4.9 \ \ Y\end{align*}
1. \begin{align*}4.5-3.1e=-2.2\end{align*} Therefore \begin{align*}e = 2.16\end{align*} \begin{align*}\text{Check} &\\ 4.5-3.1e &= -2.2\\ 4.5-3.1({\color{red}2.16}) &= -2.2\\ 4.5-6.7 &= -2.2\\ -2.2 &= -2.2 \ \ Y\end{align*}

Using algebra tiles...

1. \begin{align*}0.5x+0.2=0.7\end{align*} First step: multiply by 10. \begin{align*}{\color{red}(10)}0.5x + {\color{red}(10)}0.2 &= {\color{red}(10)}0.7\\ 5x + 2 &= 7\end{align*} Therefore \begin{align*}x = 1\end{align*} \begin{align*}\text{Check}&\\ 0.5x+0.2 &= 0.7\\ 0.5({\color{red}1})+0.2 &= 0.7\\ 0.7 &= 0.7 \ \ Y\end{align*}
1. \begin{align*}0.2z-0.3=0.7\end{align*} First step: multiply by 10. \begin{align*}{\color{red}(10)}0.2z - {\color{red}(10)}0.3 &= {\color{red}(10)}0.7\\ 2z - 3 &= 7\end{align*} Therefore \begin{align*}z = 5\end{align*} \begin{align*}\text{Check}&\\ 0.2z-0.3 &= 0.7\\ 0.2({\color{red}5})-0.3 &= 0.7\\ 1.0-0.3 &= 0.7\\ 0.7 &= 0.7 \ \ Y\end{align*}
1. \begin{align*}0.05x+0.16=-0.04\end{align*} First multiply by 100. \begin{align*}{\color{red}(100)}0.05x+{\color{red}(100)}0.16={\color{red}(100)}-0.04\end{align*} \begin{align*}5x+16=-4\end{align*} Therefore \begin{align*}x = -4\end{align*}. \begin{align*}\text{Check} &\\ 0.05x+0.16 &= -0.04\\ 0.05({\color{red}-4})+0.16 &= -0.04\\ -0.20+0.16 &= -0.04\\ -0.04 &= -0.04 \ \ Y\end{align*}

Using rules from the lesson...

1. \begin{align*}0.87+0.15x=-0.03\end{align*} Therefore \begin{align*}x = -6\end{align*} \begin{align*}\text{Check} &\\ 0.87+0.15x &= -0.03\\ 0.87+0.15({\color{red}-6}) &= -0.03\\ 0.87-0.90 &= -0.03\\ -0.03 &= -0.03 \ \ Y\end{align*}
1. \begin{align*}0.25z-3.3=0.7\end{align*} Therefore \begin{align*}z = 16\end{align*} \begin{align*}\text{Check} &\\ 0.25z-3.3 &= 0.7\\ 0.25({\color{red}16})-3.3 &= 0.7\\ 4-3.3 &= 0.7\\ 0.7 &= 0.7 \ \ Y\end{align*}
1. \begin{align*}x-0.3+0.05x=2-1.4x\end{align*} For this last problem you can solve the problem using the balance method but start without the “balance”. You will use this method in the next lesson with fractions. Follow along... \begin{align*}x-0.3+0.05x &= 2-1.4x\\ -0.3+1.05x &= 2-1.4x && \boxed{\approx \ \text{Start by combining the} \ x \ \text{variables on the left side.}}\end{align*} \begin{align*}-0.3+1.05x{\color{red}+1.4x} &= 2-1.4x {\color{red}+1.4x}\\ -0.3+2.45x &= 2 && \boxed{\approx \ \text{Next and} \ 1.4x \ \text{to each side to have the variables on the same side of the equal sign. Then simplify.}}\end{align*} \begin{align*}-0.3 {\color{red}+0.3}+2.45x &= 2 {\color{red}+0.3}\\ 2.45x &= 2.3 && \boxed{\approx \ \text{Next add} \ 0.3 \ \text{to each side to isolate the variable on the left side of the equal sign. Then simplify.}}\end{align*} \begin{align*}\frac{2.45x}{2.45} &= \frac{2.3}{2.45}\\ x &= 0.94 && \boxed{\approx \ \text{Finally, divide each side of the equal sign by} \ 2.45 \ \text{to solve for the variable. Then simplify.}}\end{align*} Therefore \begin{align*}x = 0.94\end{align*} \begin{align*}\text{Check} &\\ x-0.3+0.05x &= 2-1.4x\\ {\color{red}(0.94)}-0.3+0.05 {\color{red}(0.94)} &= 2-1.4 {\color{red}(0.94)}\\ 0.94-0.3+0.047 &= 2-1.32\\ 0.68 &= 0.68 \ \ Y\end{align*}

## Solving Linear Equations with One Variable Involving Fractions

Introduction

This concept is an extension of the previous four concepts in that you will be solving equations with one variable. When introducing fractions into these equations, the same rules apply. You need to keep the equations in balance by adding, subtracting, multiplying, or dividing on both sides of the equal sign in order to isolate the variable. The goal still remains to get your variable alone on one side of the equal sign with your constant terms on the other in order to solve for this variable.

When solving equations with fractions, you can use the balance method but you can begin to solve these problems without the aid of the “balance”. This was illustrated in the final problem of the problem set in the previous lesson. For the problems in this concept, you will be guided through algebraic problem solving of linear equations with one variable involving fractions.

Watch This

Guidance

In this year’s student election for president, there were two candidates. The winner received \begin{align*}\frac{1}{3}\end{align*} more votes. If there were 584 votes cast for president, how many votes did each of the two candidates receive?

Let \begin{align*}x =\end{align*} votes for candidate 1 (the winner)

Let \begin{align*}y =\end{align*} votes for candidate 2

\begin{align*}x + y = 584\end{align*}

You need to get only one variable into the equation in order to solve it. Let’s look at another relationship from the problem.

\begin{align*}x &= y + \frac{1}{3} y && (\text{candidate} \ 1 \ \text{received} \ \frac{1}{3} \ \text{more votes than candidate} \ 2)\\ x &= \frac{3}{3} y + \frac{1}{3} y && (\approx \text{Make denominator common for both} \ y \ \text{variables})\\ x &= \frac{4}{3} y && (\approx \text{Simplify})\end{align*}

Now substitute into the original problem.

\begin{align*}\frac{4}{3} y + y &= 584 && (\approx \text{Substitute for} \ x \ \text{into the equation})\\ \frac{4}{3} y + \frac{3}{3} y &= 584 && (\approx \text{Make denominator common for both} \ y \ \text{variables})\\ \frac{7}{3} y &= 584 && (\approx \text{Combine like terms})\\ {\color{red}(\cancel{3})} \frac{7}{\cancel{3}} y &= 584 {\color{red}(3)} && (\approx \text{Multiply both sides by the denominator in the fraction})\\ 7y &= 1752 && (\approx \text{Simplify})\\ \frac{\cancel{7} y}{{\color{red}\cancel{7}}} &= \frac{1752}{{\color{red}7}} && (\approx \text{Divide both sides by the numerator in the fraction})\\ y &= 250.28 && (\approx \text{Simplify})\end{align*}

So candidate 2 received 250 votes. Candidate 1 would then receive \begin{align*}584 - 250 = 334\end{align*} votes.

Example A

\begin{align*}\frac{1}{3}t+5 &= -1\\ \frac{1}{3}t+5 {\color{red}-5} &= -1 {\color{red}-5} && (\approx \text{Subtract} \ 5 \ \text{from both sides to isolate the variable})\\ \frac{1}{3}t &= -6 && (\approx \text{Simplify})\\ ({\color{red}\cancel{3}}) \frac{1}{\cancel{3}}t &= -6 ({\color{red}3}) && (\approx \text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ t &= -18 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}t = -18\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{3}t+5 &= -1\\ \frac{1}{3} ({\color{red}-18})+5 &= -1\\ -6+5 &= -1\\ -1 &= -1 \ \ Y\end{align*}

Example B

\begin{align*}\frac{3}{4}x-3 &= 2\\ \frac{3}{4}x-3 {\color{red}+3} &= 2 {\color{red}+3} && (\approx \text{Add} \ 3 \ \text{to both sides to isolate the variable})\\ \frac{3}{4}x &= 5 && (\approx \text{Simplify})\\ ({\color{red}\cancel{4}}) \frac{3}{4}x &= 5({\color{red}4})&& (\approx \text{Multiply both sides by the denominator} \ ({\color{red}4}) \ \text{in the fraction}\\ 3x &= 20 && (\approx \text{Simplify})\\ \frac{\cancel{3} x}{{\color{red}\cancel{3}}} &= \frac{20}{{\color{red}3}}\\ x &= \frac{20}{3}\end{align*}

Therefore \begin{align*}y = 2\end{align*}.

\begin{align*}\text{Check} &\\ \frac{3}{4} x-3 &= 2\\ \frac{3}{4} \left(\frac{20}{3}\right)-3 &= 2\\ \frac{20}{4}-3 &= 2\\ 5-3 &= 2\\ 2 &= 2 \ \ Y\end{align*}

Example C

\begin{align*}\frac{2}{5} x-4 &= -\frac{1}{5}x+8\\ \frac{2}{5}x {\color{red}+\frac{1}{5}x}-4 &= -\frac{1}{5}x {\color{red}+\frac{1}{5}x}+8 && (\approx \text{Add} \ \frac{1}{5}x \ \text{to both sides of the equal sign to combine variables})\\ \frac{3}{5} x-4 &= 8 && (\approx \text{Simplify})\\ \frac{3}{5}x-4 {\color{red}+4} &= 8 {\color{red}+4} && (\approx \text{Add} \ 4 \ \text{to both sides of the equation to isolate the variable})\\ \frac{3}{5}x &= 12 && (\approx \ \text{Simplify})\\ ({\color{red}\cancel{5}}) \frac{3}{\cancel{5}}x &= 12({\color{red}5}) && (\approx \ \text{Multiply both sides by the denominator } ({\color{red}5}) \text{ in the fraction})\\ 3x &= 60 && (\approx \text{Simplify})\\ \frac{\cancel{3}x}{\cancel{3}} &= \frac{60}{3} && (\approx \text{Divide both sides by the numerator } ({\color{red}3}) \text{ in the fraction})\\ x &= 20 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = 20\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{5}x-4 &= -\frac{1}{5}x+8\\ \frac{2}{5} ({\color{red}20})-4 &=- \frac{1}{5} ({\color{red}20})+8\\ \frac{40}{5}-4 &=-\frac{20}{5}+8\\ 8-4 &= -4+8\\ 4 &= 4 \ \ Y\end{align*}

Vocabulary

Fraction
A fraction is a part of a whole consisting of a numerator divided by a denominator. For example, if a pizza is cut into eight slices and you ate 3 slices, you would have eaten \begin{align*}\frac{3}{8}\end{align*} of the pizza. \begin{align*}\frac{3}{8}\end{align*} is a fraction with 3 being the numerator and 8 being the denominator.
Least Common Denominator
The least common denominator or lowest common denominator is the smallest number that all of the denominators (or the bottom numbers) can be divided into evenly. For example with the fractions \begin{align*}\frac{1}{2}\end{align*} and \begin{align*}\frac{1}{3}\end{align*}, the smallest number that both 2 and 3 will divide into evenly is 6. Therefore the least common denominator is 6.

Guided Practice

1. Solve for the variable in the problem \begin{align*}\frac{2}{3}x=12\end{align*}.
2. Solve for the variable in the problem \begin{align*}\frac{3}{4}x-5=19\end{align*}.
3. Solve for the variable in the problem \begin{align*}\frac{1}{4}x-3=\frac{2}{3}x\end{align*}.

1. \begin{align*}\frac{2}{3}x &= 12\\ ({\color{red}\cancel{3}}) \frac{2}{3}x &= 12 ({\color{red}3}) && (\approx \text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ 2x &= 36 && (\approx \text{Simplify})\\ \frac{\cancel{2} x}{{\color{red}\cancel{2}}} &= \frac{36}{{\color{red}2}} && (\approx \text{Divide both sides by the numerator} \ ({\color{red}2}) \ \text{in the fraction})\\ x &= 18 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = 18\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{3}x &= 12\\ \frac{2}{3} ({\color{red}18}) &= 12\\ \frac{36}{3} &= 12\\ 12 &= 12 \ \ Y\end{align*}

2. \begin{align*}\frac{3}{4}x-5 &= 19\\ \frac{3}{4}x-5 {\color{red}+5} &= 19 {\color{red}+5} && (\approx \text{Add} \ 5 \ \text{to both sides of the equal sign to isolate the variable})\\ \frac{3}{4}x &= 24 && (\approx Simplify)\\ ({\color{red}\cancel{4}}) \frac{3}{\cancel{4}}x &= 24({\color{red}4}) && (\approx \text{Multiply both sides by the denominator} ({\color{red}4}) \ \text{in the fraction})\\ 3x &= 96 && (\approx \text{Simplify})\\ \frac{\cancel{3} x}{{\color{red}\cancel{3}}}&=\frac{96}{{\color{red}3}} && (\approx \text{Divide both sides by numberator } ({\color{red}2}) \text{ in the fraction})\\ x &= 32 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = 32\end{align*}.

\begin{align*}\text{Check} &\\ \frac{3}{4}x-5 &= 19\\ \frac{3}{4} ({\color{red}32})-5 &= 19\\ \frac{96}{4}-5 &= 19\\ 24-5 &= 19\\ 19 &= 19 \ \ Y\end{align*}

3. \begin{align*}\frac{1}{4}w-3 &= \frac{2}{3}w\\ \frac{1}{4}w-3 {\color{red}+3} &= \frac{2}{3}w {\color{red}+3} && (\approx \text{Add} \ 3 \ \text{to both sides of the equal sign to start})\\ \frac{1}{4}w &= \frac{2}{3}w+3 && (\approx \text{Simplify})\\ \frac{1}{4}w {\color{red}-\frac{2}{3}w} &= \frac{2}{3}w {\color{red}-\frac{2}{3}w}+3 && (\approx \text{Subtract} \ \frac{2}{3}w \ \text{from both sides of the equal sign to get variables on same side})\\ \frac{1}{4}w-\frac{2}{3}w &= 3 && (\approx \text{Simplify})\end{align*}

In order to simplify further, you need to make the fractions all have the same denominator. In order to do this you need to find the least common denominator (LCD). If you look at the last equation \begin{align*}\left(\frac{1}{4}w-\frac{2}{3}w=3 \right)\end{align*} you will see that the denominators are 4, 3, and 1. The equation can really be seen as \begin{align*}\left(\frac{1}{{\color{blue}4}}w-\frac{2}{{\color{blue}3}}w=\frac{3}{{\color{blue}1}}\right)\end{align*}.

So what is the least common denominator for the numbers 1, 3, and 4? Well all three numbers will divide into 12 evenly \begin{align*}({\color{blue}4} \times 3 = 12, {\color{blue}3} \times 4 = 12\end{align*}, and \begin{align*}{\color{blue}1} \times 12 = 12)\end{align*}. Therefore you need to multiply \begin{align*}\frac{1}{{\color{blue}4}}w\end{align*} by \begin{align*}\frac{3}{3},-\frac{2}{{\color{blue}3}}w\end{align*}, by \begin{align*}\frac{4}{4}\end{align*}, and \begin{align*}\frac{3}{{\color{blue}1}}\end{align*} by \begin{align*}\frac{12}{12}\end{align*} in order to carry on with this problem and solve for the variable. Notice that by multiplying by \begin{align*}\frac{3}{3}\end{align*}, you are really just multiplying by 1!

\begin{align*}{\color{magenta}\left(\frac{3}{3}\right)} \frac{1}{4}w- {\color{magenta}\left(\frac{4}{4}\right)} \frac{2}{3}w &= {\color{magenta} \left(\frac{12}{12}\right)} 3 && (\approx \text{Multiply by the LCD})\\ \frac{3}{12}w-\frac{8}{12}w &= \frac{36}{12} && (\approx \text{Simplify})\end{align*}

Since all the denominators are the same (12), we can simplify further:

\begin{align*}3w-8w &= 36 && (\approx \text{Combine like terms})\\ -5w &= 36 && (\approx \text{Simplify})\\ \frac{-5w}{{\color{red}-5}} &= \frac{36}{{\color{red}-5}} && (\approx \text{Divide by} \ -5 \ \text{to solve for the variable})\\ w &= -\frac{36}{5} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}w = -\frac{36}{5}\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{4}w-3 &= \frac{2}{3}w\\ \frac{1}{4} \left({\color{red}\frac{-36}{5}} \right)-3 &= \frac{2}{3} \left({\color{red}\frac{-36}{5}}\right)\\ \frac{-36}{20} -3 &= \frac{-72}{15}\\ \frac{-108}{60}-\frac{180}{60} &= \frac{-288}{60}\\ \frac{-288}{60} &= \frac{-288}{60} \ \ Y\end{align*}

Summary

You have expanded your problem solving techniques for solving equations with one variable to include problems that involve fractions. You have learned that the solution steps are the same as those you have learned previously in that whatever you do to one side of the equal sign, you must do to the other. As well, your goal still remains to isolate the variables onto one side of the equation and then solve for the variable.

With fractions, however, there is the added step of multiplying and dividing the equation by the numerator and denominator in order to solve for the variable. As well, if your fractions do not have the same denominator, you must first find the least common denominator (LCD) before combining like terms or combining constants.

Problem Set

Solve for the variable in each of the following problems.

1. \begin{align*}\frac{1}{3}p=5\end{align*}
2. \begin{align*}\frac{3}{7}j=8\end{align*}
3. \begin{align*}\frac{2}{5}b+4=6\end{align*}
4. \begin{align*}\frac{2}{3}x-2=1\end{align*}
5. \begin{align*}\frac{1}{3}x+3=-3\end{align*}

Solve for the variable in each of the following problems.

1. \begin{align*}\frac{1}{8}k+\frac{2}{3}=5\end{align*}
2. \begin{align*}\frac{1}{6}c+\frac{1}{3}=-2\end{align*}
3. \begin{align*}\frac{4}{5}x+3=\frac{2}{3}\end{align*}
4. \begin{align*}\frac{3}{4}x-\frac{2}{5}=\frac{1}{2}\end{align*}
5. \begin{align*}\frac{1}{4}t+\frac{2}{3}=\frac{1}{2}\end{align*}

Solve for the variable in each of the following problems.

1. \begin{align*}\frac{1}{3}x+\frac{1}{4}x=1\end{align*}
2. \begin{align*}\frac{1}{5}d+\frac{2}{3}d=\frac{5}{3}\end{align*}
3. \begin{align*}\frac{1}{2}x-1=\frac{1}{3}x\end{align*}
4. \begin{align*}\frac{1}{3}x-\frac{1}{2}=\frac{3}{4}x\end{align*}
5. \begin{align*}\frac{2}{3}j-\frac{1}{2}=\frac{3}{4}j+\frac{1}{3}\end{align*}

Solve for the variable...

\begin{align*}\frac{1}{3}p &=5 && (\approx \text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ ({\color{red}\cancel{3}}) \frac{1}{\cancel{3}}p&=5({\color{red}3}) && (\approx \text{Simplify})\\ p &= 15\end{align*}

Therefore \begin{align*}p = 15\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{3}p &= 5\\ \frac{1}{3} ({\color{red}15}) &= 5\\ \frac{15}{3} &= 5\\ 5 &= 5 \ \ Y\end{align*}

\begin{align*}\frac{2}{5}b+4 &= 6\\ \frac{2}{5}b+4 {\color{red}-4} &= 6 {\color{red}-4} && (\approx \text{Subtract} \ 4 \ \text{from both sides of the equal sign to isolate the variable})\\ \frac{2}{5}b &= 2 && (\approx \text{Simplify})\\ ({\color{red}\cancel{5}}) \frac{2}{\cancel{5}} b &= 2({\color{red}5}) && (\approx \text{Multiply both sides by the denominator} \ ({\color{red}5}) \ \text{in the fraction})\\ 2b &= 10 && (\approx \text{Simplify})\\ \frac{\cancel{2} b}{{\color{red}\cancel{2}}}&=\frac{10}{\color{red}2} && (\approx \text{Divide both sides by the numerator} \ ({\color{red}2}) \ \text{in the fraction})\\ b &= 5 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}b = 5\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{5}b+4 &= 6\\ \frac{2}{5}({\color{red}5})+4 &= 6\\ \frac{10}{5}+4 &= 6\\ 2+4 &= 6\\ 6 &= 6 \ \ Y\end{align*}

\begin{align*}\frac{1}{3}x+3 &= -3\\ \frac{1}{3}x+3 {\color{red}-3} &= -3 {\color{red}-3} && (\approx \text{Subtract} \ 3 \ \text{from both sides of the equal sign to isolate the variable})\\ \frac{1}{3}x &= -6 && (\approx \text{Simplify})\\ ({\color{red}\cancel{3}}) \frac{1}{\cancel{3}}x &= -6 ({\color{red}3}) && (\approx \text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ x &= -18 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = -18\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{3}x+3 &= -3\\ \frac{1}{3} ({\color{red}-18})+3 &= -3\\ \frac{-18}{3}+3 &= -3\\ -6+3 &= -3\\ -3 &= -3 \ \ Y\end{align*}

Solve for the variable...

1. \begin{align*}\frac{1}{8}k+\frac{2}{3}=5\end{align*} First, find the LCD for 8, 3, and 1. Since it is 24, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second fraction by \begin{align*}\frac{8}{8}\end{align*}, and the third number by \begin{align*}\frac{24}{24}\end{align*}.

\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{8}k+\left({\color{red}\frac{8}{8}}\right) \frac{2}{3}&=\left({\color{red}\frac{24}{24}}\right)5 && (\approx \text{Multiply by the LCD})\\ \frac{3}{24}k+\frac{16}{24} &= \frac{120}{24} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3k+16 &= 120\\ 3k+16 {\color{red}-16} &= 120 {\color{red}-16} && (\approx \text{Subtract} \ 16 \ \text{from both sides of the equal sign to isolate the variable})\\ 3k &= 104 && (\approx \text{Simplify})\\ \frac{\cancel{3}k}{{\color{red}\cancel{3}}}&=\frac{104}{{\color{red}3}} && (\approx \text{Divide both sides by} \ 3 \ \text{to solve for the variable})\\ k &= 34.7 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}k = 34.7\end{align*}.

\begin{align*}\text{Check}& \\ \frac{1}{8}k+\frac{2}{3} &= 5\\ \frac{1}{8} (34.7)+\frac{2}{3} &= 5\\ \frac{34.7}{8}+\frac{2}{3} &= 5\\ 4.33+0.67 &= 5\\ 5 &= 5 \ \ Y\end{align*}

1. \begin{align*}\frac{4}{5}x+3=\frac{2}{3}\end{align*} First, find the LCD for 5, 1, and 3. Since it is 15, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second number by \begin{align*}\frac{15}{15}\end{align*}, and the third fraction by \begin{align*}\frac{5}{5}\end{align*}.

\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{4}{5}x+ \left({\color{red}\frac{15}{15}}\right)3 &= \left({\color{red}\frac{5}{5}}\right) \frac{2}{3} && (\approx \text{Multiply by the LCD})\\ \frac{12}{15}x+\frac{45}{45} &= \frac{10}{15} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}12x+45 &= 10\\ 12x+45 {\color{red}-45} &= 10 {\color{red}-45} && (\approx \text{Subtract} \ 45 \ \text{from both sides of the equal sign to isolate the variable})\\ 12x &= -35 && (\approx \text{Simplify})\\ \frac{12x}{{\color{red}12}} &= \frac{-35}{{\color{red}12}} && (\approx \text{Divide both sides by} \ 12 \ \text{to solve for the variable})\\ x &= -2.92 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = -2.92\end{align*}.

\begin{align*}\text{Check} &\\ \frac{4}{5}x+3 &= \frac{2}{3}\\ \frac{4}{5} ({\color{red}-2.92}) +3 &= \frac{2}{3}\\ \frac{-11.68}{5}+3 &= \frac{2}{3}\\ -2.33+3 &= 0.67\\ 0.67 &= 0.67 \ \ Y\end{align*}

1. \begin{align*}\frac{1}{4}t+\frac{2}{3}=\frac{1}{2}\end{align*} First, find the LCD for 4, 3, and 2. Since it is 12, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second fraction by \begin{align*}\frac{4}{4}\end{align*}, and the third fraction by \begin{align*}\frac{6}{6}\end{align*}.

\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{4}t+\left({\color{red}\frac{4}{4}}\right)\frac{2}{3} &= \left({\color{red}\frac{6}{6}}\right)\frac{1}{2} && (\approx \text{Multiply by the LCD})\\ \frac{3}{12}t+\frac{8}{12} &= \frac{6}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3t+8 &= 6\\ 3t+8 {\color{red}-8} &= 6 {\color{red}-8} && (\approx \text{Subtract} \ 8 \ \text{from both sides of the equal sign to isolate the variable})\\ 3t &= -2 && (\approx \text{Simplify})\\ \frac{3t}{{\color{red}3}} &= \frac{-2}{{\color{red}3}} && (\approx \text{Divide both sides by} \ 3 \ \text{to solve for the variable})\\ x &= \frac{-2}{3} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = \frac{-2}{3}\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{4}t+\frac{2}{3} &=\frac{1}{2}\\ \frac{1}{4} \left({\color{red}\frac{-2}{3}}\right)+\frac{2}{3} &= \frac{1}{2}\\ \frac{-2}{12}+\frac{2}{3} &= \frac{1}{2}\\ -0.17+0.67 &= 0.50\\ 0.50 &= 0.50 \ \ Y\end{align*}

Solve for the variable...

1. \begin{align*}\frac{1}{3}x+\frac{1}{4}x=1\end{align*} First, find the LCD for 3, 4, and 1. Since it is 12, multiply the first fraction by \begin{align*}\frac{4}{4}\end{align*}, the second fraction by \begin{align*}\frac{3}{3}\end{align*}, and the third number by \begin{align*}\frac{12}{12}\end{align*}.

\begin{align*}\left({\color{red}\frac{4}{4}}\right)\frac{1}{3}x+\left({\color{red}\frac{3}{3}}\right)\frac{1}{4}x &= \left({\color{red}\frac{12}{12}}\right)1 && (\approx \text{Multiply by the LCD})\\ \frac{4}{12}x+\frac{3}{12}x &= \frac{12}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}4x+3x &= 12\\ 7x &= 12 && (\approx \text{Combine like terms})\\ \frac{7x}{{\color{red}7}} &= \frac{12}{{\color{red}7}} && (\approx \text{Divide both sides by} \ 7 \ \text{to solve for the variable})\\ x &= \frac{12}{7} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = \frac{12}{7}\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{3}x +\frac{1}{4}x &= 1\\ \frac{1}{3} \left({\color{red}\frac{12}{7}}\right)+\frac{1}{4} \left({\color{red}\frac{12}{7}}\right) &= 1\\ \frac{12}{21}+\frac{12}{28} &= 1\\ 0.57+0.43 &= 1\\ 1 &= 1 \ \ Y\end{align*}

1. \begin{align*}\frac{1}{2}x-1=\frac{1}{3}x\end{align*} First, find the LCD for 2, 1, and 3. Since it is 6, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second number by \begin{align*}\frac{6}{6}\end{align*}, and the third fraction by \begin{align*}\frac{2}{2}\end{align*}.

\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{2}x- \left({\color{red}\frac{6}{6}}\right)1 & = \left ( {\color{red}\frac{2}{2}} \right ) \frac{1}{3}x && (\approx \text{Multiply by the LCD})\\ \frac{3}{6}x-\frac{6}{6} &= \frac{2}{6}x && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3x-6 &= 2x\\ 3x-6 {\color{red}+6} &= 2x {\color{red}+6} && (\approx \text{Subtract} \ 6 \ \text{from both sides of the equal sign})\\ 3x &= 2x+6 && (\approx \text{Simplify})\\ 3x {\color{red}-2x} &= 2x {\color{red}-2x}+6 && (\approx \text{Subtract} \ 2x \ \text{from both sides of the equal sign to get variables on same side})\\ x &= 6\end{align*}

Therefore \begin{align*}x = 6\end{align*}.

\begin{align*}\text{Check}\\ \frac{1}{2}x-1 &= \frac{1}{3}x\\ \frac{1}{2} ({\color{red}6})-1 &= \frac{1}{3} ({\color{red}6})\\ \frac{6}{2}-1 &= \frac{6}{3}\\ 3-1 &= 2\\ 2 &= 2 \ \ Y\end{align*}

1. \begin{align*}\frac{2}{3}j-\frac{1}{2}=\frac{3}{4}j+\frac{1}{3}\end{align*} First, find the LCD for 3, 2, 4,and 3. Since it is 12, multiply the first fraction by \begin{align*}\frac{4}{4}\end{align*}, the second fraction by \begin{align*}\frac{6}{6}\end{align*}, the third fraction by \begin{align*}\frac{3}{3}\end{align*}, and the fourth fraction by \begin{align*}\frac{4}{4}\end{align*}.

\begin{align*}\left( {\color{red}\frac{4}{4}}\right)\frac{2}{3}j-\left({\color{red}\frac{6}{6}}\right) \frac{1}{2} &= \left({\color{red}\frac{3}{3}}\right) \frac{3}{4}j+\left({\color{red}\frac{4}{4}}\right)\frac{1}{3} && (\approx \text{Multiply by the LCD})\\ \frac{8}{12}j-\frac{6}{12}&=\frac{9}{12}j+\frac{4}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}8j-6 &= 9j+4\\ 8j-6 {\color{red}+6} &= 9j+4 {\color{red}+6} && (\approx \text{Add} \ 6 \ \text{to both sides of the equal sign})\\\ 8j &= 9j+10 && (\approx \text{Simplify})\\ 8j {\color{red}-9j} &= 9j {\color{red}-9j} + 10 && (\approx \text{Subtract} \ 9j \ \text{from both sides of the equal sign to get variables on same side})\\ -j &= 10\end{align*}

Therefore \begin{align*}j = -10\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{3}j-\frac{1}{2} &= \frac{3}{4}j+\frac{1}{3}\\ \frac{2}{3} ({\color{red}-10})-\frac{1}{2} &= \frac{3}{4} ({\color{red}-10})+\frac{1}{3}\\ \frac{-20}{3}-\frac{1}{2} &= \frac{-30}{4}+\frac{1}{3}\\ -6.67-0.50 &= -7.50+0.33\\ -7.2 &= -7.2 \ \ Y\end{align*}

## Solving Linear Equations with Fractions and Decimals that Apply the Distributive Property

Introduction

If you recall, the distributive property is a mathematical way of grouping terms. As you learned in a previous lesson, 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. In this lesson, you will use the distributive property with linear equations and fractions. The same rules apply. You have to multiply what is on the outside of the brackets by what is on the inside of the brackets and this is your first step in solving equations with one variable where there are brackets.

Watch This

Guidance

Pens are $9 per dozen and pencils are$6 per dozen. Janet need to buy a half dozen of each for school. How much is the total cost of her purchase?

If first let’s write down what you know:

Let \begin{align*}x =\end{align*} total cost

Cost of pens: $9/dozen Cost of pencils:$6/dozen

Janet needs one half dozen of each.

The total cost would therefore be:

\begin{align*}\frac{1}{2}(\9+\6) &= x\\ \frac{\ 9}{2}+\frac{\ 6}{2} &= x\\ \4.50+\3.00 &= x\\ \7.50 &= x\end{align*}

Therefore Janet would need \$7.50 to buy these supplies.

Example A

\begin{align*}\frac{2}{5}(d+4) &= 6\\ \frac{2}{5}d+\frac{8}{5} &= 6 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 5, 5, and 1. Since it is 5, multiply the last number by \begin{align*}\frac{5}{5}\end{align*}, to get the same denominator.

\begin{align*}\frac{2}{5}d+\frac{8}{5} &= \left({\color{red}\frac{5}{5}}\right)6\\ \frac{2}{5}d+\frac{8}{5} &= \frac{30}{5} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}2d+8 &= 30\\ 2d+8 {\color{red}-8} &= 30 {\color{red}-8} && (\approx \text{Subtract} \ 8 \ \text{from both sides of the equal sign to isolate the variable})\\ 2d &= 22 && (\approx \text{Simplify})\\ \frac{2d}{{\color{red}2}}&=\frac{22}{{\color{red}2}} && (\approx \text{Divide by} \ 2 \ \text{to solve for the variable})\\ d &= 11\end{align*}

Therefore \begin{align*}d = 11\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{5}(d+4) &= 6\\ \frac{2}{5} ({\color{red}11} +4) &=6\\ \frac{2}{5}(15) &= 6\\ \frac{30}{5} &= 6\\ 6 &= 6 \ \ Y\end{align*}

Example B

\begin{align*}\frac{1}{4}(3x+7) &= -2\\ \frac{3}{4}x+\frac{7}{4} &= -2 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by \begin{align*}\frac{4}{4}\end{align*}, to get the same denominator.

\begin{align*}\frac{3}{4}x+\frac{7}{4} &= \left({\color{red}\frac{4}{4}}\right)-2\\ \frac{3}{4}x+\frac{7}{4} &= \frac{-8}{4} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3x+7 &= -8\\ 3x+7 {\color{red}-7} &= -8 {\color{red}-7} && (\approx \text{Subtract} \ 7 \ \text{from both sides of the equal sign to isolate the variable})\\ 3x &= -15 && (\approx \text{Simplify})\\ \frac{3x}{{\color{red}3}} &= \frac{-15}{{\color{red}3}} && (\approx \text{Divide by} \ 3 \ \text{to solve for the variable})\\ x &= -5\end{align*}

Therefore \begin{align*}x = -5\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{4}(3x+7) &= -2\\ \frac{1}{4} (3 ({\color{red}-5})+7) &= -2\\ \frac{1}{4}(-15+7) &= -2\\ \frac{1}{4}(-8) &= -2\\ \frac{-8}{4} &= -2\\ -2 &= -2 \ \ Y\end{align*}

Example C

\begin{align*}\frac{1}{3}(x-2) &= -\frac{2}{3}(2x+4)\\ \frac{1}{3}x-\frac{2}{3} &= -\frac{4}{3}x-\frac{8}{3} && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}x-2 &= -4x-8\\ x {\color{red}+4x} -2 &= -4x {\color{red}+4x}-8 && (\approx \text{Add} \ 4x \ \text{to both sides of the equal sign to combine variables})\\ 5x-2 &= -8 && (\approx \text{Simplify})\\ 5x-2 {\color{red}+2} &= -8 {\color{red}+2} && (\approx \text{Add} \ 2 \ \text{to both sides of the equation to isolate the variable})\\ 5x &= -6 && (\approx \text{Simplify})\\ \frac{5x}{{\color{red}5}} &= \frac{-6}{{\color{red}5}} && (\approx \text{Divide both sides by} \ 5 \ \text{to solve for the variable})\\ x&=\frac{-6}{5} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-6}{5}\end{align*}.

\begin{align*}\text{Check}&\\ \frac{1}{3}(x-2) &= -\frac{2}{3}(2x+4)\\ \frac{1}{3} \left( {\color{red}\left(\frac{-6}{5}\right)} -2 \right) &= -\frac{2}{3} \left(2 {\color{red}\left(\frac{-6}{5}\right)}+4\right)\\ 0.33(-1.2-2) &= -0.67(2(-1.2)+4)\\ 0.33(-3.2) &= -0.67(1.6)\\ -1.1 &= -1.1 \ \ Y\end{align*}

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: \begin{align*}{\color{red}\frac{2}{3}} ({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.

Guided Practice

1. Solve for the variable in the problem \begin{align*}\frac{1}{2}(5x+3)=1\end{align*}.
2. Solve for the variable in the problem \begin{align*}\frac{2}{3}(9x-6)=2\end{align*}.
3. Solve for the variable in the problem \begin{align*}\frac{2}{3}(3x+9)=\frac{1}{4}(2x+5)\end{align*}.

1. \begin{align*}\frac{1}{2} (5x+3) &=1 \\ \frac{5}{2}x+\frac{3}{2}&=1 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by \begin{align*}\frac{2}{2}\end{align*}, to get the same denominator.

\begin{align*}\frac{5}{2}x+\frac{3}{2} &= 1 {\color{red}\left(\frac{2}{2}\right)}\\ \frac{5}{2}x+\frac{3}{2} &= \frac{2}{2} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}5x+3 &= 2\\ 5x+3 {\color{red}-3} &= 2 {\color{red}-3} && (\approx \text{Subtract} \ 3 \ \text{from both sides of the equal sign to isolate the variable})\\ 5x &= -1 && (\approx \text{Simplify})\\ \frac{5x}{{\color{red}5}} &= \frac{-1}{{\color{red}5}} && (\approx \text{Divide both sides by the} \ 5 \ \text{to solve for the variable})\\ x &= \frac{-1}{5} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-1}{5}\end{align*}.

\begin{align*}\text{Check} &\\ \frac{1}{2} (5x+3) &= 1\\ \frac{1}{2} \left(5 \left({\color{red}\frac{-1}{5}}\right)+3\right) &= 1\\ \frac{1}{2}(-1+3) &= 1\\ \frac{1}{2}(2) &= 1\\ 1 &= 1 \ \ Y\end{align*}

2. \begin{align*}\frac{2}{3} (9x-6) &= 2\\ \frac{18}{3}x-\frac{12}{3} &= 2 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 3, 3, and 1. Since it is 3, multiply the last number by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.

\begin{align*}\frac{18}{3}x-\frac{12}{3} &= 2 {\color{red}\left(\frac{3}{3}\right)}\\ \frac{18}{3}x-\frac{12}{3} &= \frac{6}{3} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}18x-12 &= 6\\ 18x-12 {\color{red}+12} &= 6 {\color{red}+12} && (\approx \text{Add} \ 12 \ \text{to both sides of the equal sign to isolate the variable})\\ 18x &= 18 && (\approx \text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{18}{{\color{red}18}} && (\approx \text{Divide both sides by the} \ 18 \ \text{to solve for the variable})\\ x &= 1 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x=1\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{3}(9x-6) &= 2\\ \frac{2}{3}(9({\color{red}1})-6) &= 2\\ \frac{2}{3}(3) &= 2\\ 2 &= 2 \ \ Y\end{align*}

3. \begin{align*}\frac{2}{3}(3x+9) &= \frac{1}{4}(2x+5)\\ \frac{6}{3}x+\frac{18}{3} &= \frac{2}{4}x+\frac{5}{4} && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 3, 3, and 4, 4. Since it is 12, multiply the first two fractions by \begin{align*}\frac{4}{4}\end{align*} and the second two fractions by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.

\begin{align*}\left({\color{red}\frac{4}{4}}\right) \frac{6}{3}x+\left({\color{red}\frac{4}{4}}\right) \frac{18}{3} &= \left({\color{red}\frac{3}{3}}\right) \frac{2}{4}x+\left({\color{red}\frac{3}{3}}\right) \frac{5}{4}\\ \frac{24}{12}x+\frac{72}{12} &= \frac{6}{12}x+\frac{15}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}24x+72 &= 6x+15\\ 24x+72 {\color{red}-72} &= 6x+15 {\color{red}-72} && (\approx \text{Subtract} \ 72 \ \text{from both sides of the equal sign to isolate the variable})\\ 24x &= 6x-57 && (\approx \text{Simplify})\\ 24x {\color{red}-6x} &= 6x {\color{red}-6x} - 57 && (\approx \text{Subtract} \ 6x \ \text{from both sides of the equal sign to get variables on same side})\\ 18x &= -57 && (\approx \text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{-57}{{\color{red}18}} && (\approx \text{Divide both sides by} \ 18 \ \text{to solve for the variable})\\ x &= \frac{-57}{18} && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-57}{18}\end{align*}.

\begin{align*}\text{Check} &\\ \frac{2}{3} (3x+9) &= \frac{1}{4}(2x+5)\\\ \frac{2}{3} \left(3 \left( {\color{red}\frac{-57}{18}}\right)+9\right) &= \frac{1}{2} \left(2 \left({\color{red}\frac{-57}{18}} \right)+5\right)\\ \frac{2}{3}(-9.5+9) &= \frac{1}{4}(-6.33+5)\\ \frac{2}{3}(-0.5) &= \frac{1}{4}(-1.33)\\ -0.33 &= -0.33 \ \ Y\end{align*}

Summary

In this final lesson on solving linear equations with one variable, you have learned how to solve equations using the distributive property. The same rules apply with decimals and fractions as apply with whole numbers in that the brackets must be removed as the first step in solving these types of problems.

Always remember, 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 then solve the equation by combining like terms; move constants to one side of the equal sign, variables to the other side of the equal sign, then isolate the variable to find the solution.

Problem Set

Solve for the variable in each of the following problems.

1. \begin{align*}\frac{1}{2} (x+5)=6\end{align*}
2. \begin{align*}\frac{1}{4}(g+2)=8\end{align*}
3. \begin{align*}0.4(b+2)=2\end{align*}
4. \begin{align*}0.5(r-12)=4\end{align*}
5. \begin{align*}\frac{1}{4}(x-16)=7\end{align*}

Solve for the variable in each of the following problems.

1. \begin{align*}26.5-k=0.5(50-k)\end{align*}
2. \begin{align*}2(1.5c+4)=-1\end{align*}
3. \begin{align*}-\frac{1}{2}(3x-5)=7\end{align*}
4. \begin{align*}0.35+0.10(m-1)=5.45\end{align*}
5. \begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)=\frac{1}{2}\end{align*}

Solve for the variable in each of the following problems.

1. \begin{align*}\frac{1}{2}x-3 (x+4)=\frac{2}{3}\end{align*}
2. \begin{align*}-\frac{5}{8}x+x=\frac{1}{8}\end{align*}
3. \begin{align*}0.4(12-d)=18\end{align*}
4. \begin{align*}0.25(x+3)=0.4(x-5)\end{align*}
5. \begin{align*}\frac{2}{3}(t-2)=\frac{3}{4}(t+2)\end{align*}

Solve for the variable...

\begin{align*}\frac{1}{2}(x+5) &=6\\ \frac{1}{2}x+\frac{5}{2} &= 6 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by \begin{align*}\frac{2}{2}\end{align*} to get the same denominator.

\begin{align*}\frac{1}{2}x+\frac{5}{2}&=\left({\color{red}\frac{2}{2}}\right)6\\ \frac{1}{2}x+\frac{5}{2}&=\frac{12}{2} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}x+5 &= 12\\ x+5{\color{red}-5}&= 12{\color{red}-5} && (\approx \text{Subtract 5 from both sides of the equal sign to isolate the variable})\\ x &=7 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = 7\end{align*}.

\begin{align*}\text{Check}\\ \frac{1}{2}(x+5)&=6\\ \frac{1}{2}({\color{red}7}+5)&=6\\ \frac{1}{2}(12)&=6\\ 6 &= 6 \ \ Y\end{align*}

\begin{align*}0.4(b+2)&=2\\ 0.4b+0.8&=2 && (\approx \text{Apply the distributive property to remove the brackets})\\ 0.4b+0.8{\color{red}-0.8}&=2{\color{red}-0.8} && (\approx \text{Subtract 0.8 from both sides of the equation to isolate the variable})\\ 0.4b &=1.2 && (\approx \text{Simplify})\\ \frac{0.4}{{\color{red}0.4}}b&= \frac{1.2}{{\color{red}0.4}} && (\approx \text{Divide both sides by the 0.4 to solve for the variable})\\ b&=3\end{align*}

Therefore \begin{align*}b = 3\end{align*}.

\begin{align*}\text{Check}\\ 0.4(b+2)&=2\\ 0.4({\color{red}3}+2)&=2\\ 0.4(5)&=2\\ 2 &= 2 \ \ Y\end{align*}

\begin{align*}\frac{1}{4}(x-16)&=7\\ \frac{1}{4}x-\frac{16}{4}&=7 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by \begin{align*}\frac{4}{4}\end{align*} to get the same denominator.

\begin{align*}\frac{1}{4}x-\frac{16}{4}&=\left({\color{red}\frac{4}{4}}\right)7\\ \frac{1}{4}x-\frac{16}{4}&=\frac{28}{4} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}x-16 &=28\\ x-16{\color{red}+16}&=28{\color{red}+16} && (\approx \text{Add 16 to both sides to isolate and solve for the variable})\\ x &=44 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = 44\end{align*}.

\begin{align*}\text{Check}\\ \frac{1}{4}(x-16)&=7\\ \frac{1}{4}({\color{red}44}-16)&=7\\ \frac{1}{4}(28)&=7\\ 7 &= 7 \ \ Y\end{align*}

Solve for the variable...

\begin{align*}26.5-k &=0.5(50-k)\\ 26.5-k &= 25-0.50k && (\approx \text{Apply the distributive property to remove the brackets})\\ 26.5{\color{red}-25}-k&=25{\color{red}-25}-0.50k && (\approx \text{Subtract 25 from both sides})\\ 1.5-k&=-0.50k && (\approx \text{ Simplify})\\ 1.5-k{\color{red}+k}&=-0.50k{\color{red}+k} && (\approx \text{Add} \ k \ \text{to both sides of the equal sign to combine the variables})\\ 1.5 &= 0.50k && (\approx \text{Simplify})\\ \frac{1.5}{\color{red}0.50}&=\frac{0.50k}{{\color{red}0.50}} && (\approx \text{Divide both sides by 0.50 to solve for the variable})\\ k &=3\end{align*}

Therefore \begin{align*}k = 3\end{align*}.

\begin{align*}\text{Check}\\ 26.5-k &=0.5(50-k)\\ 26.5-{\color{red}3}&=0.5(50-{\color{red}3})\\ 23.5 &= 0.5(47)\\ 23.5 &= 23.5 \ \ Y\end{align*}

\begin{align*}-\frac{1}{2}(3x-5)&=7\\ -\frac{3}{2}x+\frac{5}{2}&=7 && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

First, find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by \begin{align*}\frac{2}{2}\end{align*} to get the same denominator.

\begin{align*}-\frac{3}{2}x+\frac{5}{2}&=\left({\color{red}\frac{2}{2}}\right)7\\ -\frac{3}{2}x+\frac{5}{2}&=\frac{14}{2} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}-3x+5 &=14\\ -3x+5{\color{red}-5}&=14{\color{red}-5} && (\approx \text{Subtract 5 from both sides of the equal sign to isolate the variable})\\ -3x &= -9 && (\approx \text{Simplify})\\ \frac{-3x}{{\color{red}-3}}&=\frac{9}{{\color{red}-3}} && (\approx \text{Divide both sides by -3 to solve for the variable})\\ x &=-3 && (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = -3\end{align*}.

\begin{align*}\text{Check}\\ -\frac{1}{2}(3x-5)&=7\\ -\frac{1}{2}(3({\color{red}-3})-5) &=7\\ -\frac{1}{2}(-9-5)&=7\\ -\frac{1}{2}(-14)&=7\\ 7 &= 7 \ \ Y\end{align*}

\begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)&=\frac{1}{2}\\ \frac{1}{4}+\frac{2}{3}t+\frac{2}{3}&=\frac{1}{2} && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

First, find the LCD for 4, 3, 3, and 2. Since it is 12, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second and third fractions by \begin{align*}\frac{4}{4}\end{align*}, and the fourth fraction by \begin{align*}\frac{6}{6}\end{align*}, to get the same denominator.

\begin{align*}\left({\color{red}\frac{3}{3}}\right)\frac{1}{4}+\left({\color{red}\frac{4}{4}}\right)\frac{2}{3}t+\left({\color{red}\frac{4}{4}}\right)\frac{2}{3}&=\left({\color{red}\frac{6}{6}}\right)\frac{1}{2}\\ \frac{3}{12}+\frac{8}{12}t+\frac{8}{12}&=\frac{6}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3+8t+8&=6\\ 8t+11&=6 && (\approx \text{Combine like terms (constants)})\\ 8t+11{\color{red}-11}&=6{\color{red}-11} && (\approx \text{Subtract 11 from both sides of the equal sign to isolate the variable})\\ 8t&=-5 && (\approx \text{Simplify})\\ \frac{8t}{{\color{red}8}}&=\frac{-5}{{\color{red}8}} && (\approx \text{Divide both sides by 8 to solve for the variable})\\ t&=\frac{-5}{8}&& (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = \frac{-5}{8}\end{align*}.

\begin{align*}\text{Check}\\ \frac{1}{4}+\frac{2}{3}(t+1)=\frac{1}{2}\\ \frac{1}{4}+\frac{2}{3}\left({\color{red}\frac{-5}{8}}+1\right)&=\frac{1}{2}\\ \frac{1}{4}+\frac{2}{3}\left(\frac{3}{8}\right)&=\frac{1}{2}\\ \frac{1}{4}+\frac{1}{4}& =\frac{1}{2}\\ \frac{1}{2}&=\frac{1}{2} \ \ Y\end{align*}

Solve for the variable...

1. \begin{align*}\frac{1}{2}(x-3)(x+4)&=\frac{2}{3}\\ \frac{1}{2}x-3x-12&=\frac{2}{3} && (\approx \text{Apply the distributive property to remove the brackets})\end{align*}

First, find the LCD for 2, 1, 1, and 3. Since it is 6, multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second and third numbers by \begin{align*}\frac{6}{6}\end{align*}, and the fourth fraction by \begin{align*}\frac{2}{2}\end{align*}, to get the same denominator.

\begin{align*}\left({\color{red}\frac{3}{3}}\right)\frac{1}{2}x-\left({\color{red}\frac{6}{6}}\right)3x-\left({\color{red}\frac{6}{6}}\right)12&=\left({\color{red}\frac{2}{2}}\right)\frac{2}{3}\\ \frac{3}{6}x-\frac{18}{6}x-\frac{72}{6}&=\frac{4}{6} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3x-18x-72 &=4\\ -15x-72&=4 && (\approx \text{Combine like terms})\\ -15x-72{\color{red}+72}&=4{\color{red}+72} && (\approx \text{Add 72 to both sides of the equal sign to isolate the variable})\\ -15x &=76 && (\approx \text{Simplify})\\ \frac{-15x}{{\color{red}-15}}&=\frac{76}{{\color{red}-15}} && (\approx \text{Divide both sides by -15 to solve for the variable})\\ x&=\frac{-76}{15}&& (\approx \text{Simplify})\end{align*}

Therefore \begin{align*}x = \frac{-76}{15}\end{align*}.

\begin{align*}\text{Check}\\ \frac{1}{2}(x-3)(x+4)&=\frac{2}{3}\\ \frac{1}{2}\left({\color{red}\frac{-76}{15}}\right)-3\left(\left({\color{red}\frac{-76}{15}}\right)+4\right)&=\frac{2}{3}\\ \left({\color{red}\frac{-38}{15}}\right)-3\left({\color{red}\frac{-16}{15}}\right)&=\frac{2}{3}\\ \frac{-38}{15}+\frac{48}{15}&=\frac{2}{3}\\ \frac{10}{15}&=\frac{2}{3}\\ \frac{2}{3}&=\frac{2}{3} \ \ Y\end{align*}

\begin{align*}0.4(12-d)&=18\\ 4.8-0.4d &=18 && (\approx \text{Apply the distributive property to remove the brackets })\\ 4.8{\color{red}-4.8}-0.4d &= 18{\color{red}-4.8} && (\approx \text{Subtract 4.8 from both sides of the equal sign to isolate the variable})\\ -0.4d &= 13.2 && (\approx \text{Simplify})\\ \frac{-0.4d}{{\color{red}-0.4}}&=\frac{13.2}{{\color{red}-0.4}} && (\approx \text{Divide both sides by -0.4 to solve for the variable})\\ d &= -33\end{align*}

Therefore \begin{align*}d = -33\end{align*}.

\begin{align*}\text{Check}\\ 0.4(12-d)&=18\\ 0.4(12-({\color{red}-33}))&=18\\ 0.4(45)&= 18\end{align*}

\begin{align*}\frac{2}{3}(t-2)&=\frac{3}{4}(t+2)\\ \frac{2}{3}t-\frac{4}{3}&=\frac{3}{4}t+\frac{6}{4} && (\approx \text{Apply the distributive property to remove the brackets })\end{align*}

First, find the LCD for 3, 3, 4, and 4. Since it is 12, multiply the first two fractions by \begin{align*}\frac{4}{4}\end{align*}, and the second two fractions by \begin{align*}\frac{3}{3}\end{align*} to get the same denominator.

\begin{align*}\left({\color{red}\frac{4}{4}}\right)\frac{2}{3}t-\left({\color{red}\frac{4}{4}}\right)\frac{4}{3}&=\left({\color{red}\frac{3}{3}}\right)\frac{3}{4}t+\left({\color{red}\frac{3}{3}}\right)\frac{6}{4}\\ \frac{8}{12}t-\frac{16}{12}&=\frac{9}{12}t+\frac{18}{12} && (\approx \text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}8t-16&=9t+18\\ 8t{\color{red}-8t}-16&=9t{\color{red}-8t}+18 && (\approx \text{Subtract} \ 8t \ \text{from both sides to have variables on same side of equal sign})\\ -16 &=t+18 && (\approx \text{Simplify})\\ -16{\color{red}-18}&=t+18{\color{red}-18} && (\approx \text{Subtract 18 from both sides of the equal sign to solve for the variable})\\ -34&=t\end{align*}

Therefore \begin{align*}t = -34\end{align*}.

\begin{align*}\text{Check}\\ \frac{2}{3}(t-2)&=\frac{3}{4}(t+2)\\ \frac{2}{3}({\color{red}-34}-2)&=\frac{3}{4}({\color{red}-34}+2)\\ \frac{2}{3}(-36)&=\frac{3}{4}(-32)\\ -24 &= -24 \ \ Y\end{align*}

## Summary

In this lesson, you have extended your learning with linear equations in one variable. You still worked with learning 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. In this lesson, decimals and fractions were introduced. With decimals, the calculations become more complex. Normally, you would multiply by 10, 100, or 1000 to get rid of the decimal. Then you could follow the same rules and the same steps in problem-solving as you learned in the previous lesson. Remember that whatever you do to one side of the equal sign, you do the other in order for the equation to stay in balance.

With fractions, it is important to have a common denominator. To do this, you learned to find the least common denominator (LCD) and then simplify your equation once all terms have the common denominator. From here, isolating and solving for the variable uses the same procedure.

As far as the methods are concerned, you spent little time with algebra tiles as they are more difficult to use with fractions and decimals. Using the balance (seesaw) method or solving using algebra steps is both helpful and efficient.

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