which equation can pair with x - y = -2 to create a consistent and dependent system?\n6x + 2y = 15\n-3x + 3y…

Explanation:

Step1: Recall the condition for consistent - dependent system

A consistent - dependent system of linear equations has infinitely many solutions, which means the two equations represent the same line. For two linear equations (a_1x + b_1y=c_1) and (a_2x + b_2y = c_2), they are the same line if (\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}). The given equation is (x - y=-2), which can be written as (1x+( - 1)y=-2), so (a_1 = 1), (b_1=-1), (c_1=-2).

Step2: Check each option

Option 1: (6x + 2y = 15)

Here (a_2 = 6), (b_2 = 2), (c_2 = 15). (\frac{a_1}{a_2}=\frac{1}{6}), (\frac{b_1}{b_2}=-\frac{1}{2}), since (\frac{1}{6}\neq-\frac{1}{2}), this is not the correct equation.

Option 2: (-3x + 3y = 6)

Here (a_2=-3), (b_2 = 3), (c_2 = 6). (\frac{a_1}{a_2}=-\frac{1}{3}), (\frac{b_1}{b_2}=-\frac{1}{3}), (\frac{c_1}{c_2}=-\frac{2}{6}=-\frac{1}{3}). Since (\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=-\frac{1}{3}), this equation represents the same line as (x - y=-2).

Option 3: (-8x-3y = 2)

Here (a_2=-8), (b_2=-3), (c_2 = 2). (\frac{a_1}{a_2}=-\frac{1}{8}), (\frac{b_1}{b_2}=\frac{1}{3}), since (-\frac{1}{8}\neq\frac{1}{3}), this is not the correct equation.

Option 4: (4x-4y = 6)

Here (a_2 = 4), (b_2=-4), (c_2 = 6). (\frac{a_1}{a_2}=\frac{1}{4}), (\frac{b_1}{b_2}=\frac{1}{4}), (\frac{c_1}{c_2}=-\frac{2}{6}=-\frac{1}{3}), since (\frac{1}{4}\neq-\frac{1}{3}), this is not the correct equation.

Answer:

B. (-3x + 3y = 6)