Chapter 7.1, Problem 32E

To determine

To find the solution for the system of equations $\{\begin{array}{c}y=-x\\ y=x^3+3x^2+2x\end{array}$ by the method of substitution.

Answer to Problem 32E

  $\left(0,0\right)$

Explanation of Solution

Given information:

The system of equation given is $\{\begin{array}{c}y=-x\\ y=x^3+3x^2+2x\end{array}$ .

Calculations:

Let $y=-x$ … equation (1) and $y=x^3+3x^2+2x$ …equation (2).

First, substitute the expression of yfrom equation (1) in equation (2) and then solve the single variable equation in terms of x.

  $\begin{array}{l}\Rightarrow \left(-x\right)=x^3+3x^2+2x\\ \Rightarrow x^3+3x^2+3x=0\\ \Rightarrow x\left(x^2+3x+3\right)=0\end{array}$

Solve the above equation for x and get,

  $\Rightarrow x=0$ and, $x^2+3x+3$ =0 … equation (3)

For equation (3), to find out the roots, use the quadratic formula:

  $x=\frac{-3\pm \sqrt{1^2-4\times 3}}{2}=\frac{-3\pm \sqrt{-11}}{2}$ .

Since the discriminant is negative that is., -11, therefore, equation (3) has no real roots.

Now back-substitute the real value of x in equation (1) to further obtain the corresponding valueof y.

For $x=0,$

  $\Rightarrow y=-\left(0\right)=0$

Therefore,the system of equations $\{\begin{array}{c}y=-x\\ y=x^3+3x^2+2x\end{array}$ have only one real solution that is., $\left(0,0\right)$ .