https://file.aiquickdraw.com/mathbot/file/17267952961450wq81ycm.jpg

The image shows two equations:

1. \( y = x^2 - 5x - 5 \)
2. \( y = 2x - 15 \)

To solve these, we can set the equations equal to each other and solve for \(x\):

\[
x^2 - 5x - 5 = 2x - 15
\]

Let's solve for \(x\). I'll begin by rearranging and solving this quadratic equation.The solutions for \(x\) are:

- Solution 1: \( x = 2 \)
- Solution 2: \( x = 5 \)

Next, we substitute these \(x\)-values into one of the original equations (e.g., \(y = 2x - 15\)) to find the corresponding \(y\)-values:

For \( x = 2 \):
\[
y = 2(2) - 15 = 4 - 15 = -11
\]

For \( x = 5 \):
\[
y = 2(5) - 15 = 10 - 15 = -5
\]

Thus, the full solutions are:
- Solution 1: \( (x = 2, y = -11) \)
- Solution 2: \( (x = 5, y = -5) \)

Would you like more details on this or have any questions?

Here are 5 related questions to explore further:
1. How do you solve quadratic equations using factoring?
2. Can you solve this problem graphically by plotting the two functions?
3. What is the significance of the intersection points of these curves?
4. How would you solve this system using substitution or elimination methods?
5. How can you verify the solutions algebraically?

**Tip:** Always double-check your work by substituting solutions back into the original equations.

Solve the system of equations: y = x^2 - 5x - 5 and y = 2x - 15

Learn how to solve the system of equations y = x^2 - 5x - 5 and y = 2x - 15 by setting the equations equal, solving the quadratic equation, and substituting the solutions back into one of the original equations. Suitable for students in Grades 8-10.

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