4:01 .11 Previous 33 Next Unit Pretes Submit Test Reader Tools (i) Info Save \& Exit 33 Select the correct answer. How many solutions does this system of equations have? \[ \begin{array}{l} y=2 x+5 \\ y=x^{3}+4 x^{2}+x+2 \end{array} \] A. no solution B. 1 solution C. 2 solutions D. 3 solutions

To determine how many solutions the system of equations has, we need to find the number of real solutions to the following system: \[ \begin{cases} y = 2x + 5 \\ y = x^3 + 4x^2 + x + 2 \end{cases} \] ### Step 1: Set the Equations Equal to Each Other Since both equations equal \( y \), we can set them equal to each other: \[ 2x + 5 = x^3 + 4x^2 + x + 2 \] ### Step 2: Rearrange the Equation Rearrange the equation to bring all terms to one side: \[ x^3 + 4x^2 - x - 3 = 0 \] ### Step 3: Analyze the Cubic Equation To determine the number of real roots of the cubic equation \( x^3 + 4x^2 - x - 3 = 0 \), we can use the following methods: **1. Rational Root Theorem:** - Possible rational roots are \( \pm1, \pm3 \). - Testing these, none satisfy the equation, indicating no rational roots. **2. Sign Changes:** Evaluate the function \( f(x) = x^3 + 4x^2 - x - 3 \) at various points: - \( f(-5) = -23 \) - \( f(-4) = 1 \) (Sign change from negative to positive → one root between -5 and -4) - \( f(-1) = 1 \) - \( f(0) = -3 \) (Sign change from positive to negative → one root between -1 and 0) - \( f(1) = 1 \) (Sign change from negative to positive → one root between 0 and 1) **3. Conclusion:** Since there are three sign changes, the cubic equation has **three real roots**. Therefore, the original system of equations has **three solutions**. ### Final Answer **D. 3 solutions**