{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!