Question

a. $x + 2y = 6$ and $x - y = 10$
b. $x + y = 6$ and $x - 2y = 10$
c. $x + 2y = 10$ and $x - y = 6$
d. $x + y = 6$ and $x - y = -10$

Explanation:

Step1: Analyze the blue line (y-intercept 6)

The blue line has a y - intercept of 6. Let's write its equation in slope - intercept form \(y = mx + b\), where \(b = 6\). From the graph, when \(x = 0\), \(y = 6\); when \(x = 2\), \(y = 4\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{4 - 6}{2 - 0}=\frac{- 2}{2}=-1\). So the equation is \(y=-x + 6\), which can be rewritten as \(x + y=6\).

Step2: Analyze the red line (intersection point and slope)

The intersection point of the two lines: Let's assume the intersection point is \((x,y)\). From the graph, when \(x=-2\), \(y = 8\) (since the lines cross at \(x=-2,y = 8\)). Let's check the equations in option D: \(x + y=6\) (we already know this is the blue line equation) and \(x - y=-10\). Substitute \(x=-2,y = 8\) into \(x - y\): \(-2-8=-10\), which satisfies the equation. Let's check the slope of the red line. From the equation \(x - y=-10\), we can rewrite it as \(y=x + 10\), so the slope \(m = 1\), which matches the positive slope of the red line in the graph.

Let's check other options:

• Option A: For \(x + 2y=6\), when \(x = 0\), \(y = 3

eq6\), so it's not the blue line.

• Option B: For \(x + y=6\) (blue line), but for \(x - 2y=10\), substitute \(x=-2,y = 8\): \(-2-2\times8=-2 - 16=-18

eq10\).

• Option C: For \(x + 2y=10\), when \(x = 0\), \(y = 5

eq6\), so it's not the blue line.

Answer:

D. \(x + y = 6\) and \(x - y=-10\)