which system of equations can be used to find the roots of the equation $12x^{3}-5x = 2x^{2}+x +…

which system of equations can be used to find the roots of the equation $12x^{3}-5x = 2x^{2}+x + 6$?\n$\begin{cases}y = 12x^{3}-5x\\y = 2x^{2}+x + 6end{cases}$\n$\begin{cases}y = 12x^{3}-5x + 6\\y = 2x^{2}+xend{cases}$\n$\begin{cases}y = 12x^{3}-2x^{2}-6x\\y = 6end{cases}$\n$\begin{cases}y = 12x^{3}-2x^{2}-6x - 6\\y = 0end{cases}$

which system of equations can be used to find the roots of the equation $12x^{3}-5x = 2x^{2}+x + 6$?\n$\begin{cases}y = 12x^{3}-5x\\y = 2x^{2}+x + 6end{cases}$\n$\begin{cases}y = 12x^{3}-5x + 6\\y = 2x^{2}+xend{cases}$\n$\begin{cases}y = 12x^{3}-2x^{2}-6x\\y = 6end{cases}$\n$\begin{cases}y = 12x^{3}-2x^{2}-6x - 6\\y = 0end{cases}$

Answer

Explanation:

Step1: Rewrite the given equation

The given equation is (12x^{3}-5x = 2x^{2}+x + 6). We want to set it equal to zero. Subtract (2x^{2}+x + 6) from both sides: (12x^{3}-2x^{2}-6x - 6=0).

Step2: Create a system of equations

We can create a system of equations by setting (y) equal to the left - hand side of the equation set equal to zero and (y = 0). So the system of equations is (\begin{cases}y=12x^{3}-2x^{2}-6x - 6\y = 0\end{cases})

Answer:

(\begin{cases}y=12x^{3}-2x^{2}-6x - 6\y = 0\end{cases})