Chapter 9, Problem 1CLT

To determine

To solve: the system given below using substitution.

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

Answer to Problem 1CLT

  $\left(1,2\right),\left(-\frac{3}{2},\frac{3}{4}\right)$

Explanation of Solution

Consider the system of equations,

  $y=3-x^2$ ….. (1)

  $2\left(y-2\right)=x-1$ ….. (2)

Substitute $y=3-x^2$ from equation (1) into the equation (2),

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

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

This gives that,

  $\begin{array}{c}2x+3=0\\ 2x=-3\\ x=-\frac{3}{2}\end{array}$

Or

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

Now, first substitute $x=-\frac{3}{2}$ in $y=3-x^2$ ,

  $\begin{array}{c}y=3-{\left(-\frac{3}{2}\right)}^2\\ =3-\frac{9}{4}\\ =\frac{3}{4}\end{array}$

Now, substitute $x=1$ in $y=3-x^2$ ,

  $\begin{array}{c}y=3-{\left(1\right)}^2\\ =3-1\\ =2\end{array}$

Thus, the solutions of the given system are $\left(1,2\right),\left(-\frac{3}{2},\frac{3}{4}\right)$