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Describe and correct any error a student may have made when solving 0.15(y−0.2)=2−0.5(1−y)0.15(y−0.2)=2−0.5(1−y).

Format it like this:

The student first made an error when writing _________ (mistake) the expression on the ____ (which side) of that equation should be ____ (corrected expression) the correct answer is _____ (final correct answer) 

The student first made an error when writing **100(0.15y - 0.3) = 100(1.5 + 0.5y)** (mistake) on the **right-hand side** of the equation. The correct expression should be **100(1.5 + 0.5y) = 150 + 50y**.

Therefore, the error occurs after expanding the 100 on both sides, but prior to simplifying the terms correctly.

The correct answer is **\( y = -\frac{180}{35} \), which simplifies to \( y = -\frac{36}{7} \) or approximately \( y = -5.14 \)**.

Would you like a detailed breakdown of the correct solution or explanations of each step?

Here are 5 related questions you might find helpful:
1. How do you solve equations with decimals effectively?
2. What is the process for expanding brackets in an equation?
3. How can we identify when to multiply both sides of an equation by a constant?
4. What methods can help check your final answer for accuracy?
5. How can errors in solving linear equations be avoided?

**Tip:** Always check the distribution and signs carefully, especially when working with negative numbers.

Learn how to solve the linear equation 0.15(y−0.2) = 2−0.5(1−y) using the distributive property and solving for variables step-by-step. Ideal for students in Grades 8-10, this problem covers core algebraic concepts and solving techniques.

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