Chapter 7.1, Problem 18E

To determine

To find:The solution of system of equations $\{\begin{array}{c}y=x^3-3x^2+4\\ y=-2x+4\end{array}$ by method of substitution and check solution graphically.

Answer to Problem 18E

Thesolution of the system of equations $\{\begin{array}{c}y=x^3-3x^2+4\\ y=-2x+4\end{array}$ are $\left(0,4\right)$ , $\left(2,0\right)$ and $\left(1,2\right)$ .

Explanation of Solution

Given information:

The given system of equationsis $\{\begin{array}{c}y=x^3-3x^2+4\\ y=-2x+4\end{array}$ .

Calculation:

Substitute $-2x+4$ for $y$ in the equation $y=x^3-3x^2+4$ .

  $\begin{array}{c}-2x+4=x^3-3x^2+4\\ x^3-3x^2+2x=0\\ x\left(x^2-3x+2\right)=0\\ x\left(x-2\right)\left(x-1\right)=0\end{array}$

Solve the above equation as follows:

  $\begin{array}{c}x=0,\left(x-2\right)=0,\left(x-1\right)=0\\ x=0,1,2\end{array}$

Substitute 0 for $x$ in equation $y=-2x+4$ to find $y$ .

  $\begin{array}{c}y=-2\left(0\right)+4\\ =4\end{array}$

Substitute 1 for $x$ in equation $y=-2x+4$ to find $y$ .

  $\begin{array}{c}y=-2\left(1\right)+4\\ =2\end{array}$

Substitute 2 for $x$ in equation $y=-2x+4$ to find $y$ .

  $\begin{array}{c}y=-2\left(2\right)+4\\ =0\end{array}$

So, the solution of the system of equation are $\left(0,4\right)$ , $\left(2,0\right)$ and $\left(1,2\right)$ .

Verify the solution using graphical utility:

Draw the graph for the first equation $y=x^3-3x^2+4$ and the second equation $y=-2x+4$ using graphical utility.

  

From the above graph, the coordinates of the intersection pointsare $\left(0,2\right)$ , $\left(\sqrt{3},2-3\sqrt{3}\right)$ and $\left(-\sqrt{3},2+3\sqrt{3}\right)$ . It satisfied the result calculated algebraically.

Therefore, the solution of the system of equations $\{\begin{array}{c}y=x^3-3x^2+4\\ y=-2x+4\end{array}$ are $\left(0,4\right)$ , $\left(2,0\right)$ and $\left(1,2\right)$ .