Given:
[tex]\begin{gathered} 2y=4x \\ x^2+y=3 \end{gathered}[/tex]
Required:
To solve the system of equations.
Explanation:
Consider
[tex]\begin{gathered} 2y=4x \\ \\ y=\frac{4x}{2} \\ \\ y=2x \end{gathered}[/tex]
Consider the secon equation
[tex]\begin{gathered} x^2+y=3 \\ \\ x^2+2x=3 \\ \\ x^2+2x-3=0 \\ \\ x^2+3x-x-3=0 \\ \\ x(x+3)-1(x+3)=0 \\ \\ (x+3)(x-1)=0 \\ \\ x=1,-3 \end{gathered}[/tex]
When x=1
[tex]\begin{gathered} y=2(1) \\ y=2 \end{gathered}[/tex]
When x= -3,
[tex]\begin{gathered} y=2(-3) \\ =-6 \end{gathered}[/tex]
Final Answer:
The option B is correct.
[tex](1,2)\text{ and }(-3,-6)[/tex]
