Chapter 9, Problem 22CT

To determine

To solve given equations.

Answer to Problem 22CT

The solutions are (-4, 6) (3, 13)

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2+2x-2\\ y=x+10\end{array}$

Here we shall substitute the value of $y$ in the first equation in order to solve.

Hence we get,

  $\begin{array}{l}\left(x+10\right)=x^2+2x-2\\ x^2+2x-2-x-10=0\\ x^2+x-12=0\end{array}$

Now we shall determine the values of $x.$

  $\begin{array}{l}x=\frac{-1\pm \sqrt{1^2-4\left(-12\right)}}{2}\\ x=\frac{-1\pm \sqrt{49}}{2}\\ x=-4,3\end{array}$

Now we shall put the values of $x$ in any given equation to determine the values of $y.$

We get $y=6,13$

Therefore the solutions are (-4, 6) (3, 13)