Question

which graph represents the system of equations below? \\(\

$$\begin{cases}-2x + y = 9\\\\\\frac{1}{2}x + y = 4\\end{cases}$$

\\) a. graph b. graph (other options partially shown)

Explanation:

Step1: Convert to slope - intercept form

For the first equation \(-2x + y=9\), we can rewrite it as \(y = 2x+9\). The slope \(m_1 = 2\) and the \(y\) - intercept \(b_1=9\).
For the second equation \(\frac{1}{2}x + y = 4\), we can rewrite it as \(y=-\frac{1}{2}x + 4\). The slope \(m_2=-\frac{1}{2}\) and the \(y\) - intercept \(b_2 = 4\).

Step2: Analyze the \(y\) - intercepts

The first line \(y = 2x + 9\) has a \(y\) - intercept of \(9\) (it crosses the \(y\) - axis at \((0,9)\)), and the second line \(y=-\frac{1}{2}x + 4\) has a \(y\) - intercept of \(4\) (it crosses the \(y\) - axis at \((0,4)\)). Also, the first line has a positive slope (\(m = 2\)) so it is increasing, and the second line has a negative slope (\(m=-\frac{1}{2}\)) so it is decreasing.

Looking at the graphs:

• In option A, one line has a \(y\) - intercept around \(6\) (not \(9\)) and the other around \(1\) (not \(4\)), so it is incorrect.
• In option B, one line (the increasing one) has a relatively high \(y\) - intercept (close to \(9\)) and the other (the decreasing one) has a \(y\) - intercept close to \(4\), and their slopes match (one increasing with a steep slope, one decreasing with a gentle slope).

Answer:

B