We are given the following system of equations: {x−y=3(1)2x−y=−1(2) \begin{cases} x - y = 3 \quad \text{(1)} \\ 2x - y = -1 \quad \text{(2)} \end{cases} {x−y=3(1)2x−y=−1(2)
Subtract equation (1) from equation (2) to eliminate y y y: (2x−y)−(x−y)=−1−3 (2x - y) - (x - y) = -1 - 3 (2x−y)−(x−y)=−1−3 Simplify: 2x−y−x+y=−4 2x - y - x + y = -4 2x−y−x+y=−4 x=−4 x = -4 x=−4
Substitute x=−4 x = -4 x=−4 into equation (1): −4−y=3 -4 - y = 3 −4−y=3 Solve for y y y: −y=3+4 -y = 3 + 4 −y=3+4 −y=7 -y = 7 −y=7 y=−7 y = -7 y=−7
The solution to the system of equations is: x=−4,y=−7 \boxed{x = -4, \quad y = -7} x=−4,y=−7
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