If $x = \frac{2}{5}y + 3$, how does $y$ change when $x$ increases from $1$ to $2$?

### Concept & Strategy To observe how $y$ behaves in response to $x$, we must rearrange the linear equation to make $y$ the independent subject. Then, substitution will reveal the boundary values. ### Step-by-Step Solution * **Given:** $x = \frac{2}{5}y + 3$ * **Calculation:** Let's isolate $y$. * Subtract $3$ from both sides: $x - 3 = \frac{2}{5}y$ * Multiply both sides by $\frac{5}{2}$: $y = \frac{5}{2}(x - 3)$ * Now, evaluate $y$ at the given bounds for $x$: * When $x = 1$: $$y = \frac{5}{2}(1 - 3)$$ $$y = \frac{5}{2}(-2) = -5$$ * When $x = 2$: $$y = \frac{5}{2}(2 - 3)$$ $$y = \frac{5}{2}(-1) = -\frac{5}{2}$$ * **Deduction:** The value of $y$ moves from $-5$ to $-2.5$. Moving from a lower number ($-5$) to a higher number on the number line ($-2.5$) constitutes an increase. ### Exam Strategy & Shortcut Instead of algebraically isolating $y$, plug the $x$ values directly into the original equation. For $x=1$: $1 = 0.4y + 3 \implies -2 = 0.4y \implies y = -5$. For $x=2$: $2 = 0.4y + 3 \implies -1 = 0.4y \implies y = -2.5$. Recognize that $-2.5$ is greater than $-5$, meaning it is an increase. ### Common Pitfall A prevalent error is selecting "decreases" because the final fraction $-\frac{5}{2}$ "looks" like a smaller quantity than the integer $5$. Students forget that for negative numbers, a smaller absolute value means the number is actually mathematically larger (closer to zero). ### Final Answer Therefore, the correct answer is **$y$ increases from $-5$ to $-\frac{5}{2}$**.