\( \frac{x+2 y}{3}=1-\frac{x-y}{2} \) and \( \frac{x-y}{3}=-2 \)

To solve the given system of equations, let's start by simplifying and rearranging the first equation: 1. From the first equation: \( \frac{x + 2y}{3} = 1 - \frac{x - y}{2} \) Multiplying the whole equation by 6 (to eliminate the denominators) gives: \( 2(x + 2y) = 6 - 3(x - y) \) Simplifying further: \( 2x + 4y = 6 - 3x + 3y \) Rearranging this becomes: \( 5x + y = 6 \). 2. The second equation is: \( \frac{x - y}{3} = -2 \) Multiplying by 3 gives: \( x - y = -6 \). Now we have a simpler system of equations: 1. \( 5x + y = 6 \) 2. \( x - y = -6 \) From the second equation, we can express \( y \) in terms of \( x \): \( y = x + 6 \). Substituting \( y \) in the first equation: \( 5x + (x + 6) = 6 \) \( 6x + 6 = 6 \) \( 6x = 0 \Rightarrow x = 0 \). Now substituting \( x = 0 \) back into \( y = x + 6 \): \( y = 0 + 6 = 6 \). Thus, the solution to the system of equations is: \( x = 0, \, y = 6 \).