The image shows two equations from a system of equations, labeled "Example 6":

1. $-x^2 - 3x + y = 0$ (Equation 1)
2. $-12x^2 - 3x + y = 0$ (Equation 2)

Let's solve this system of equations step by step.

Step 1: Subtract Equation 1 from Equation 2

Subtract the two equations to eliminate $y$: $\left(-12x^2 - 3x + y\right) - \left(-x^2 - 3x + y\right) = 0$ Simplify: $-12x^2 + x^2 = -11x^2$ The other terms cancel out: $-11x^2 = 0$

Step 2: Solve for $x$

Dividing both sides by $-11$, we get: $x^2 = 0$ So, $x = 0$.

Step 3: Substitute $x = 0$ into one of the original equations

Using Equation 1: $-(0)^2 - 3(0) + y = 0$ Simplify: $y = 0$

Final Answer:

The solution to the system is: $x = 0, \, y = 0$

Would you like more details or have any questions?

Here are 5 related questions you can explore:

1. What would happen if one of the constants in the equations changed?
2. How would the solution change if we added a third variable, $z$, to the equations?
3. Can you graph these equations and show how the solution appears visually?
4. What are the implications if $x^2 + 3x$ was factored differently in both equations?
5. How can we verify the solution by substituting the values back into both equations?

Tip: Always check your solution by substituting the values back into the original equations to ensure they satisfy both.