Chapter 9.9, Problem 5ST

To determine

To calculate: To solve the system of equations

Answer to Problem 5ST

Solutions are $\left(-9,-11\right),\left(1,-1\right)$

Explanation of Solution

Given information: System of equations $3x^2+2y^2=1;y=x-2$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}3x^2+2y^2=1----(1)\\ y=x-2------(2)\end{array}$

From equation (2),

  $y=x-2$

Substituting the values in (1):

  $\begin{array}{l}3x^2-2{\left(x-2\right)}^2=1\\ 3x^2-2\left(x^2+4-4x\right)=1\\ 3x^2-2x^2-8+8x=1\\ x^2+8x-9=0\\ x^2+9x-x-9=0\\ \left(x+9\right)\left(x-1\right)=0\\ x=-9,1\end{array}$

When $x=-9$

  $\begin{array}{l}y=-9-2\\ y=-11\end{array}$

When $x=1$

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

Conclusion:

Hence, solutions are $\left(-9,-11\right),\left(1,-1\right)$