Chapter 7.1, Problem 16E

Solution

To calculate: The solution of the system of equation by algebraically.

  $\begin{array}{c}y=x^3+x^2\\ y=-x^2\end{array}$

The solution of system of equation is $\left(0,0\right)$, and $\left(-2,-4\right)$ .

Given Information:

The given system of equation is,

  $\begin{array}{c}y=x^3+x^2\\ y=-x^2\end{array}$

Concept Used:

Substitute one equation value to in other equation and make the one variable equation and solve.

Calculation:

Consider the given system of equations,

  $\begin{array}{c}y=x^3+x^2\\ y=-x^2\end{array}$

Substitute second equation value $y=-x^2$ in the first equation.

  $\begin{array}{c}-x^2=x^3+x^2\\ x^3+2x^2=0\end{array}$

Take the common terms and equate to zero.

  $\begin{array}{c}{x}^2\left(x+2\right)=0\\ x=0,0,-2\end{array}$

Now, find the value of $y$ by substituting 0 for $x$ in the equation $y=-x^2$ .

  $\begin{array}{c}y=-{\left(0\right)}^2\\ =0\end{array}$

Find the value of $y$ by substituting $-2$ for $x$ in the equation $y=-x^2$ .

  $\begin{array}{c}y=-{\left(-2\right)}^2\\ =-4\end{array}$

Now, to support the solution, use the graphing calculator to draw the graph of the system.

  

Hence, the required solution is $\left(0,0\right)$ and $\left(-2,-4\right)$ .