Question

1. -/1 points solve the system by the method of substitution. use a graphi \

$$\begin{cases} y = -x \\\\ y = x^3 + 2x^2 + 7x \\end{cases}$$

(x, y) = (\boxed{ }) resources read it watch it

Explanation:

Step1: Substitute \( y = -x \) into the second equation

Since \( y = -x \) and \( y = x^3 + 2x^2 + 7x \), we set them equal: \( -x = x^3 + 2x^2 + 7x \)

Step2: Rearrange the equation

Bring all terms to one side: \( x^3 + 2x^2 + 7x + x = 0 \) which simplifies to \( x^3 + 2x^2 + 8x = 0 \)

Step3: Factor out \( x \)

Factor \( x \) from the left side: \( x(x^2 + 2x + 8) = 0 \)

Step4: Solve for \( x \)

Set each factor equal to zero. First, \( x = 0 \). For the quadratic \( x^2 + 2x + 8 = 0 \), the discriminant \( \Delta = 2^2 - 4\times1\times8 = 4 - 32 = -28 < 0 \), so no real solutions. Thus, \( x = 0 \)

Step5: Find \( y \)

Substitute \( x = 0 \) into \( y = -x \), we get \( y = 0 \)

Answer:

\((0, 0)\)