Chapter 11, Problem 30CCSR

To determine

Find the solutions of the system of equations $y=x^2-x-6\text{and}y=2x-2$

Answer to Problem 30CCSR

The solutions are: $(4,6)\text{and}(-1,-4)$

Explanation of Solution

Given:

The system of equations $y=x^2-x-6\text{and}y=2x-2$

Concept Used:

The system of equations $y=x^2-x-6\text{and}y=2x-2$

Zero-Product Property: If we multiply two (or more) things together and the result is equal to zero, then we know that  at least one  of those things that we multiplied must also have been equal to zero.

Calculation:

Substitute the value of y from the 2^nd equation $y=2x-2$ to the first equation $y=x^2-x-6$ .

  $\begin{array}{l}y=x^2-x-\mathrm{6...........1}\\ y=2x-\mathrm{2.................2}\\ 2x-2=x^2-x-6[\text{Substitute }y=2x-2]\\ x^2-x-2x-6+2=2x-2x-2+2[\text{Subtract }2x\text{and 2}\text{add to both sides ] }\\ x^2-3x-4=0\\ (x-4)(x+1)=0\end{array}$

Now use zero product property:

  $\begin{array}{l}{x}^2-3x-4=0\\ (x-4)(x+1)=0\\ x^2-3x-4=0\\ (x-4)(x+1)=0\\ x-4=0\text{and}x+1=0[\text{Zero-Product Property}]\\ x=4\text{and}x=-1\end{array}$The solutions are: $(4,6)\text{and}(-1,-4)$

  $\begin{array}{l}x=4\Rightarrow y=2x-2\Rightarrow y=2·4-2=6\Rightarrow (4,6)\\ x=-1\Rightarrow y=2x-2\Rightarrow y=2·-1-2=-4\Rightarrow (-1,-4)\end{array}$

Thus, the solutions are: $(4,6)\text{and}(-1,-4)$