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# 2.4: Equations with Decimals

Difficulty Level: At Grade Created by: CK-12

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?

### Guidance

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 because of the decimals. 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 you can see that the answer is \begin{align*}x=1\end{align*}.

When you have decimals in an equation, you can get rid of them by multiplying by 10 (if one decimal place), 100 (if two decimal places), or 1000(if three decimal places). Then, solve the equation as usual. If you feel confident with adding, subtracting, multiplying, and dividing decimals, you can also solve equations with decimals by isolating the variable and solving without first removing the decimals.

#### Example A

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

You can think about this problem with the balance method. 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 \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.

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 \ \ \end{align*}

#### Example B

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

This time try solving the problem using algebra tiles. Multiply this problem by 10 to get rid of the decimals.

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

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

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

Simplify and rearrange and you get:

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

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

Note: since the numbers were so large with this problem, the balance 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.

#### Example C

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

You can again use the balance 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 \ \ \end{align*}

#### Concept Problem Revisited

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*}.

### 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. 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 \ \ \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 \ \ \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 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*}

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 \ \ \end{align*}

### Practice

Use the balance model 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*}

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Date Created:
Dec 19, 2012