Q. 27 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The point of intersection is $(0,1) \& \left(-\frac{2}{3},\frac{1}{3}\right)$.

Step 1. Given information

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

Step 2. Graphing each equation

$y=2x+1$

$2x^2+y^2=1$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ y=2x+1 .....\left(1\right) \\ 2x^2+y^2=1 .....\left(2\right) \\ Now substituting the value of y from eq.1 into eq.2 \\ we get \\ 2x^2+{(2x+1)}^2=12x^2+4x^2+4x+1=1 \\ 6x^2+4x=0 \\ x=0,-\frac{2}{3} \\ Now substituting value of x in eq. 1 we get \\ y=2\left(0\right)+1 \\ y=1 \\ y=2\left(-\frac{2}{3}\right)+1 \\ y=-\frac{1}{3} \end{gathered}$$