1. The problem states that \(\alpha = \{(x, y) : y = -2x + 3\} \subseteq \mathbb{R} \times \mathbb{R}\). This means \(\alpha\) is the set of all points \((x, y)\) in the plane where \(y\) is defined by the linear function \(y = -2x + 3\). 2. The formula given is a linear equation in slope-intercept form \(y = mx + b\), where \(m = -2\) is the slope and \(b = 3\) is the y-intercept. 3. Important rules for linear functions: - The slope \(m\) tells us how steep the line is and the direction it goes. Here, \(m = -2\) means the line falls 2 units vertically for every 1 unit it moves horizontally to the right. - The y-intercept \(b = 3\) is the point where the line crosses the y-axis (at \(x=0\)). 4. To understand the set \(\alpha\), we can find some points: - When \(x=0\), \(y = -2(0) + 3 = 3\), so \((0, 3)\) is on the line. - When \(x=1\), \(y = -2(1) + 3 = 1\), so \((1, 1)\) is on the line. - When \(x=-1\), \(y = -2(-1) + 3 = 5\), so \((-1, 5)\) is on the line. 5. The set \(\alpha\) includes all points \((x, y)\) where \(y\) satisfies the equation \(y = -2x + 3\). This is a straight line in the plane with slope \(-2\) and y-intercept \(3\). Final answer: \(\alpha = \{(x, y) \in \mathbb{R}^2 : y = -2x + 3\}\) represents a line with slope \(-2\) and y-intercept \(3\).