Q. 29 - Precalculus Enhanced with Graphing Utilities

Question

Graph each equation of the system. Then solve the system to find the points of intersection.

$\left\{\begin{array}{l}x+y+1=0\\ x^2+y^2+6y-x=-5\end{array}\right.$

Step-by-Step Solution

Answer

The point of intersection is $\left(0,-1\right) \& \left(\frac{5}{2},-\frac{7}{2}\right)$.

Step 1. Given information

$\left\{\begin{array}{l}x+y+1=0\\ x^2+y^2+6y-x=-5\end{array}\right.$

Step 2. Graphing each equation

$x+y+1=0$

$x^2+y^2+6y-x=-5$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ x+y+1=0 .....\left(1\right) \\ {x}^2+y^2+6y-x=-5 .....\left(2\right) \\ Now substituting the value of y from eq.1 into eq.2 \\ we get \\ {x}^2+{(-x-1)}^2+6(-x-1)-x=-5 \\ {x}^2+x^2+1+2x-6x-6-x=-5 \\ x=0,\frac{5}{2} \\ Now substituting value of x in eq. 1 we get \\ y=-1 \\ y=-\frac{7}{2} \end{gathered}$$