{Y=4x-x2 {2x+y=5

To analyze the system of equations you provided, we have: 1. \( Y = 4x - x^2 \) 2. \( 2x + y = 5 \) First, from Equation 2, we can express \( y \) in terms of \( x \): \[ y = 5 - 2x \] Now, we can substitute this expression for \( y \) into Equation 1: \[ 5 - 2x = 4x - x^2 \] Next, we'll rearrange the equation to collect all terms on one side: \[ x^2 - 6x + 5 = 0 \] Now, we can factor this quadratic equation: \[ (x - 1)(x - 5) = 0 \] Setting each factor to zero gives us the solutions for \( x \): \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \] Now we'll find the corresponding \( y \) values for each \( x \): 1. For \( x = 1 \): \[ y = 5 - 2(1) = 5 - 2 = 3 \] 2. For \( x = 5 \): \[ y = 5 - 2(5) = 5 - 10 = -5 \] Thus, the system of equations has two solutions: 1. \( (1, 3) \) 2. \( (5, -5) \) If you need further assistance or different analyses, let me know!