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Let's define the variables for this problem:

Let \( x \) be the number of bottles of shampoo Miko bought.
Since she bought one more bottle of shower foam than shampoo, the number of bottles of shower foam will be \( x + 1 \).

- The mass of one bottle of shower foam is 1.4 kg, so the total mass of the shower foam will be \( 1.4(x + 1) \).
- The mass of one bottle of shampoo is 1.6 kg, so the total mass of the shampoo will be \( 1.6x \).

The total mass of both the shampoo and shower foam is 10.4 kg, so we can set up the following equation:

\[
1.4(x + 1) + 1.6x = 10.4
\]

Now, let's solve this equation:

\[
1.4(x + 1) + 1.6x = 10.4
\]
\[
1.4x + 1.4 + 1.6x = 10.4
\]
\[
3x + 1.4 = 10.4
\]
\[
3x = 10.4 - 1.4
\]
\[
3x = 9
\]
\[
x = 3
\]

So, Miko bought 3 bottles of shampoo. Therefore, the number of bottles of shower foam is:

\[
x + 1 = 3 + 1 = 4
\]

Miko bought **4 bottles of shower foam**.

Would you like more details or explanations on this solution?

Here are five related questions for further practice:
1. If Miko had purchased 2 more bottles of shampoo, how much would the total mass be?
2. How many bottles would Miko need to buy if the total mass was 12 kg instead of 10.4 kg?
3. What is the total mass of only the shampoo Miko bought?
4. If the mass of one bottle of shower foam increased to 1.5 kg, how would that affect the total number of bottles she could buy for 10.4 kg?
5. How much more does one bottle of shampoo weigh compared to one bottle of shower foam?

**Tip**: Always define your variables clearly when setting up equations to avoid confusion and ensure a smooth solution process.

At the supermarket, a bottle of shower foam had a mass of 1.4 kg while a bottle of shampoo had a mass of 1.6 kg. Miko bought supplies for half a year and they had a mass of 10.4 kg. She purchased one more bottle of shower foam than shampoo. How many bottles of shower foam did she buy?

This problem involves solving a linear equation to determine how many bottles of shower foam and shampoo Miko bought. Using key algebraic concepts and a step-by-step approach, students will learn to solve equations that involve multiple variables. Suitable for students in Grades 6-8.

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