Chapter 7.1, Problem 15E

Solution

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

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

The solution of system of equation is $\left(0,0\right)$, and $\left(3,18\right)$ .

Given Information:

The given system of equation is,

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

Concept Used:

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

If the general equation is $a{x}^2+bx+c=0$ then root of the equation is defined as,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where, $a$, $b$, and $c$ are constant real nubers.

Calculation:

Consider the given system of equations,

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

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

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

Take the common terms and equate to zero.

  $\begin{array}{c}{x}^2\left(x-3\right)=0\\ x=0,3\end{array}$

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

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

Find the value of $y$ by substituting 3 for $x$ in the equation $y=2x^2$ .

  $\begin{array}{c}y=2{\left(3\right)}^2\\ =18\end{array}$

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 $\left(3,18\right)$ .