Which system of equations has a solution of approximately \( (0.7,-1.4) \) ? \( \begin{array}{l}x-y=2 \text { and } x+2 y=-2 \\ x-2 y=-2 \text { and } x+y=2 \\ x-y=2 \text { and } x-2 y=-2 \\ x+y=2 \text { and } x+2 y=-2\end{array} \) Mark this and return

We start by examining each system of equations to see if it has an approximate solution of \( (0.7, -1.4) \). ### Option 1 The system is \[ x - y = 2 \quad \text{and} \quad x + 2y = -2. \] 1. Solve the first equation for \( y \): \[ y = x - 2. \] 2. Substitute into the second equation: \[ x + 2(x - 2) = -2 \quad \Longrightarrow \quad x + 2x - 4 = -2. \] 3. Combine like terms: \[ 3x - 4 = -2. \] 4. Solve for \( x \): \[ 3x = 2 \quad \Longrightarrow \quad x = \frac{2}{3} \approx 0.67. \] 5. Substitute \( x \) back to find \( y \): \[ y = \frac{2}{3} - 2 = -\frac{4}{3} \approx -1.33. \] The approximate solution is \( (0.67, -1.33) \), which is very close to \( (0.7, -1.4) \). ### Option 2 The system is \[ x - 2y = -2 \quad \text{and} \quad x + y = 2. \] 1. Solve the second equation for \( x \): \[ x = 2 - y. \] 2. Substitute into the first equation: \[ (2 - y) - 2y = -2 \quad \Longrightarrow \quad 2 - 3y = -2. \] 3. Solve for \( y \): \[ -3y = -4 \quad \Longrightarrow \quad y \approx 1.33. \] 4. Thus, \( x = 2 - 1.33 \approx 0.67 \). This gives the solution \( (0.67, 1.33) \), which does not match \( (0.7, -1.4) \). ### Option 3 The system is \[ x - y = 2 \quad \text{and} \quad x - 2y = -2. \] 1. From the first equation, \( x = y + 2 \). 2. Substitute into the second: \[ (y + 2) - 2y = -2 \quad \Longrightarrow \quad 2 - y = -2. \] 3. Solve for \( y \): \[ -y = -4 \quad \Longrightarrow \quad y = 4. \] 4. Then, \( x = 4 + 2 = 6 \). The solution here is \( (6, 4) \), which is not an approximate match. ### Option 4 The system is \[ x + y = 2 \quad \text{and} \quad x + 2y = -2. \] 1. Subtract the first equation from the second: \[ (x + 2y) - (x + y) = -2 - 2 \quad \Longrightarrow \quad y = -4. \] 2. Substitute \( y = -4 \) into \( x + y = 2 \): \[ x - 4 = 2 \quad \Longrightarrow \quad x = 6. \] This gives the solution \( (6, -4) \), which is not an approximate match. ### Conclusion Only **Option 1** produces a solution that is approximately \( (0.7, -1.4) \). \[ \boxed{x - y = 2 \quad \text{and} \quad x + 2y = -2.} \]