Chapter 12.6, Problem 58SR

To determine

To find: The exact solution of each system of equations.

Answer to Problem 58SR

The solutions are $\left(2,4\right)$ and $\left(-1,1\right)$

Explanation of Solution

Given information:

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

Given

  $y=x+2$ -----(1)

  $y=x^2$ --------(2)

Equating equation-(1) and equation-(2)

We get

  $\begin{array}{l}\Rightarrow x^2=x+2\\ \Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\\ \Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x+1\right)\left(x-2\right)=0\end{array}$

It means

  $\begin{array}{l}x+1=0\\ \Rightarrow x=-1\end{array}$ or $\begin{array}{l}x-2=0\\ \Rightarrow x=2\end{array}$

On substituting value of $x=-1$ in equation −(1)

We get

  $\begin{array}{l}y=x+2\\ \Rightarrow y=-1+2\\ \Rightarrow y=1\end{array}$

Therefore,

  $\left(-1,1\right)$ is one of the solution of given equations.

On substituting value of $x=2$ in equation −(1)

  $\begin{array}{l}y=x+2\\ \Rightarrow y=2+2\\ \Rightarrow y=4\end{array}$

Therefore,

  $\left(2,4\right)$ is solution of given equation.

Therefore,

The solutions are $\left(2,4\right)$ and $\left(-1,1\right)$