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# 3.9: Solve Multi-Step Equations Involving Decimals

Difficulty Level: Basic Created by: CK-12
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Practice Equations with Decimals
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Have you ever counted change? Take a look at this situation involving many different coins.

Sam found a bunch of change under his bed. He has a pile of quarters, a pile of dimes and a pile of nickels. He has the same number of quarters, dimes and nickels. When he adds it all up, he has eight dollars and eighty cents.

How many of each coin does Sam have?

To figure this out, you will need to write an equation and solve it. This Concept will show you how to work with equations that have decimals in them.

### Guidance

Did you know that you can solve equations with rational numbers in them? Do you know what a rational number is? What does this have to do with integers? Did you know that they are connected?

Integers include positive whole numbers (1, 2, 3, 4, 5, $\ldots$ ), their opposites (-1, -2, -3, -4, -5, $\ldots$ ), and zero.

Integers are rational numbers .

A rational number is any number that can be written as the ratio of two integers or you can think of this in fraction form. So, an integer such as -3, which can be written as the ratio $\frac{-3}{1}$ , is a rational number.

What are some other rational numbers?

A fraction, such as $\frac{1}{4}$ , can obviously be written as the ratio of two integers. So, fractions are rational numbers.

A terminating decimal, such as 0.1, is also rational because it can be written as the ratio $\frac{1}{10}$ . A repeating decimal, such as $0.\overline{3}$ , is rational because even though the digit 3 repeats over and over in the decimal form, it can be expressed as the ratio of two integers: $\frac{1}{3}$ .

All integers, fractions, terminating decimals and repeating decimals are rational numbers.

You can solve equations with other rational numbers in them.

Let’s start by looking at solving equations involving decimals.

You will use the same strategy to solve a multi-step equation that includes decimals that you would use to solve any other multi-step equation. You will first combine like terms or use the distributive property to simplify the equation. Then, you will use inverse operations to isolate the variable on one side of the equation.

Remember back, you will need to remember how to perform operations involving decimals to be effective at solving equations with decimals.

Solve for $x$ : $3x - 2.5x + 0.5 = 4.5$

First, subtract the like terms— $3x$ and $2.5x$ —on the left side of the equation. It may help to remember that $3x = 3.0x$ .

$3x - 2.5x + 0.5 &= 4.5\\(3.0x - 2.5x) + 0.5 &= 4.5\\0.5x + 0.5 &= 4.5$

Notice that 0.5 cannot be combined with $0.5x$ because they are not like terms.

Now, we can solve as we would solve any two-step equation.

The next step is to isolate the term with the variable, $0.5x$ , on one side of the equation. Since 0.5 is added to $0.5x$ , we should subtract 0.5 from both sides of the equation.

$0.5x + 0.5 &= 4.5\\0.5x + 0.5 - 0.5 &= 4.5 - 0.5\\0.5x + 0 &= 4.0\\0.5x &= 4$

Since $0.5x$ means $0.5 \cdot x$ , our next step is to divide each side of the equation by 0.5 to get the $x$ by itself on one side of the equation.

$0.5x &= 4\\\frac{0.5x}{0.5} &= \frac{4}{0.5}\\1x &= 8\\x &= 8$

The value of $x$ is 8.

Exactly, the trickiest part is to remember the rules for adding, subtracting, multiplying and dividing decimals. Once you remember those rules, you can apply the rules to working with the equations themselves.

#### Example A

$.7x = 4.90$

Solution:  $x = 7$

#### Example B

$.3x+10 = 31$

Solution:  $x = 70$

#### Example C

$.18x + .2x + 4 = 4.76$

Solution:  $x = 2$

Now let's go back to the dilemma from the beginning of the Concept.

Sam found a bunch of change under his bed. He has a pile of quarters, a pile of dimes and a pile of nickels. He has the same number of quarters, dimes and nickels. When he adds it all up, he has eight dollars and eighty cents.

We want to figure out how many of each coin Sam has collected. Let's use $c$ as the unknown for coin.

Here is our equation.

$.25c + .05c + .10c = 8.80$

Quarters are worth twenty-five cents, nickels are worth five cents and dimes are worth ten cents. Our operation is addition because Sam figured out the total amount of money.

Now that we have an equation, our next step is to combine like terms.

$.4c = 8.80$

Next, we divide both sides by .4.

$c = 22$

Sam has 22 of each type of coin.

### Guided Practice

Here is one for you to try on your own.

Solve for $x$ : $0.1(z - 4.2) = 0.48$

Solution

First you can see that we have parentheses in this equation. Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 0.1 and then subtract those products.

$0.1(z - 4.2) &= 0.48\\(0.1 \times z) - (0.1 \times 4.2) &= 0.48\\0.1z - 0.42 &= 0.48$

Now, solve as you would solve any two-step equation. To get $0.1z$ by itself on one side of the equation, add 0.42 to both sides.

$0.1z - 0.42 &= 0.48\\0.1z - 0.42 + 0.42 &= 0.48 + 0.42\\0.1z +(-0.42 + 0.42) &= 0.9\\0.1z + 0 &= 0.9\\0.1z &= 0.9$

To get $z$ by itself on one side of the equation, divide both sides by 0.1.

$0.1z &= 0.9\\\frac{0.1z}{0.1} &= \frac{0.9}{0.1}\\1z &= 9\\z &= 9$

The value of $z$ is 9.

### Explore More

Directions: Solve each equation to find the value of the variable.

1. $3.2n + 6.5n = 38.8$
2. $0.2(3 + p) = 4.6$
3. $0.09y - 0.08y = 1.2$
4. $.06x + .05x = .99$
5. $.9x = 81$
6. $.6x + 1 = 19$
7. $9.05x = 27.15$
8. $.16x + 3 = 3.48$
9. $2.3a + 4 = 15.5$
10. $2(a+4) + .5a = 23$
11. $.54y+.16y+.22y = 3.68$
12. $\frac{x}{.6}=.8$
13. $\frac{y}{.25}=9$
14. $.6x -.5x + 11 = 12.1$
15. $.26x + .18x = -3.08$

Basic

Dec 19, 2012

Feb 26, 2015