Chapter 9.8, Problem 36PPE

To determine

To determine thesolution of the equations.

Answer to Problem 36PPE

The solutions are (-1, 6) (3, 18)

Explanation of Solution

Given information:

Given equations:

  $\begin{array}{l}y=x^2+x+6\\ y=2x^2-x+3\end{array}$

Here we shall use substitution method.

Now we shall put the value of $y$ from second equation into first equation. We get,

  $\begin{array}{l}2x^2-x+3=x^2+x+6\\ 2x^2-x^2-x-x+3-6=0\\ x^2-2x-3=0\\ x^2-3x+x-3=0\\ x\left(x-3\right)+\left(x-3\right)=0\\ \left(x+1\right)\left(x-3\right)=0\\ x=-1,3\end{array}$

now we shall put the values of $x$ in the given equation to find the values of $y$ .

We get $y=6,18$

Hence the solutions are (-1, 6) (3, 18)

The equations are solved by method of substitution.