Answer:
Option c
Step-by-step explanation:
A system of equations are given to us. And we need to solve them . The given system is
[tex]\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}[/tex]
We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .
[tex]\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 } [/tex]
We got two values of x as 0 & 2 . Alternatively substituting these values we have ,
[tex]\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 } [/tex]
Thefore the required answer is ,
[tex]\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}[/tex]
