1. \( y=x^{2}-5 x+1 \) 2. \( 3 x+y-3=0 \) 3. \( y(y)=3 x+y+1 \)

It looks like you have provided three equations. Let's analyze each one: 1. **Quadratic Equation**: \[ y = x^2 - 5x + 1 \] This is a quadratic equation in terms of \(x\). It represents a parabola that opens upwards. 2. **Linear Equation**: \[ 3x + y - 3 = 0 \quad \text{or} \quad y = -3x + 3 \] This is a linear equation in terms of \(x\) and \(y\). It represents a straight line with a slope of -3 and a y-intercept of 3. 3. **Another Equation**: \[ y(y) = 3x + y + 1 \] This equation seems to have a typo or is not standard. If we interpret \(y(y)\) as \(y^2\), then it becomes: \[ y^2 = 3x + y + 1 \] Rearranging gives: \[ y^2 - y - 3x - 1 = 0 \] This is also a quadratic equation, but in terms of \(y\). ### Solving the System of Equations If you want to find the points of intersection between these equations, we can solve them simultaneously. 1. **Substituting the Linear Equation into the Quadratic**: Substitute \(y = -3x + 3\) into the first equation: \[ -3x + 3 = x^2 - 5x + 1 \] Rearranging gives: \[ x^2 - 2x - 2 = 0 \] Now, we can use the quadratic formula to solve for \(x\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \] Thus, the solutions for \(x\) are: \[ x_1 = 1 + \sqrt{3}, \quad x_2 = 1 - \sqrt{3} \] 2. **Finding Corresponding \(y\) Values**: Now, substitute these \(x\) values back into the linear equation to find \(y\): For \(x_1 = 1 + \sqrt{3}\): \[ y_1 = -3(1 + \sqrt{3}) + 3 = -3 - 3\sqrt{3} + 3 = -3\sqrt{3} \] For \(x_2 = 1 - \sqrt{3}\): \[ y_2 = -3(1 - \sqrt{3}) + 3 = -3 + 3\sqrt{3} + 3 = 3\sqrt{3} \] ### Summary of Intersection Points The points of intersection between the first two equations are: 1. \((1 + \sqrt{3}, -3\sqrt{3})\) 2. \((1 - \sqrt{3}, 3\sqrt{3})\) If you want to explore the third equation further or find intersections with it, please clarify how you would like to proceed!