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We are given that \( x < 2 \) and the equation \( y = 6 - 5x \). Let's address each part of the problem:

### (a) Find the range of the values of \( y \).

Since \( x < 2 \), we need to determine the range of \( y \) based on this condition.

- Start by plugging the boundary value \( x = 2 \) into the equation for \( y \):

\[
y = 6 - 5(2) = 6 - 10 = -4
\]

This is the value of \( y \) when \( x = 2 \). But since \( x \) must be strictly less than 2, \( y \) will be slightly greater than \(-4\) as \( x \to 2 \). 

Now consider what happens as \( x \to -\infty \). Plugging this into the equation:

\[
y = 6 - 5x \to 6 + 5\infty = +\infty
\]

So, as \( x \to -\infty \), \( y \to +\infty \).

Thus, the range of \( y \) is \( (-4, +\infty) \).

### (b) If \( y \) is a negative integer, write down all possible values of \( y \).

For \( y \) to be a negative integer, let's rewrite the equation:

\[
y = 6 - 5x
\]

We want to find when \( y \) is a negative integer, so:

\[
6 - 5x = y \quad \text{where} \quad y \in \mathbb{Z}^-
\]

To find possible values, we solve the inequality \( 6 - 5x < 0 \):

\[
6 < 5x
\]
\[
x > \frac{6}{5}
\]

Since \( x < 2 \), we now calculate the possible negative integer values of \( y \). Testing values of \( x \) between \( \frac{6}{5} \) and 2:

1. \( x = 1.5 \) gives:
   \[
   y = 6 - 5(1.5) = 6 - 7.5 = -1.5 \quad \text{(not an integer)}
   \]
2. \( x = 1.6 \) gives \( y = -2 \), but since the exact range isn't met for further integer

It is given that x < 2 and y = 6 - 5x. (a) Find the range of the values of y. (b) If y is a negative integer, write down all possible values of y.

Explore the problem where x < 2 and y = 6 - 5x. Learn how to find the range of y and determine all possible negative integer values of y. This problem covers inequalities and linear equations, suitable for Grades 9-10 students.

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