https://file.aiquickdraw.com/mathbot/file/1727069820213rdf9r75x.jpg

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.

Solve the system of equations:
1. -x^2 - 3x + y = 0 (Equation 1)
2. -12x^2 - 3x + y = 0 (Equation 2)

Learn how to solve a system of quadratic equations step by step. This example covers elimination and substitution methods to solve two equations: -x^2 - 3x + y = 0 and -12x^2 - 3x + y = 0, suitable for Grades 9-12 students.

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