To solve the system of equations algebraically, we need to identify the points where the two equations intersect. Given the system:
1. [tex]\( y = -x - 10 \)[/tex]
2. [tex]\( y = x^2 - 5x - 6 \)[/tex]
Let's solve these step-by-step:
### Step 1: Set the equations equal to each other
Since both equations are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ -x - 10 = x^2 - 5x - 6 \][/tex]
### Step 2: Rearrange to form a quadratic equation
Move all the terms to one side of the equation:
[tex]\[ x^2 - 5x - 6 + x + 10 = 0 \][/tex]
[tex]\[ x^2 - 4x + 4 = 0 \][/tex]
### Step 3: Factor the quadratic equation
The quadratic equation [tex]\( x^2 - 4x + 4 \)[/tex] factors neatly:
[tex]\[ (x - 2)^2 = 0 \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Set each factor equal to zero:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
### Step 5: Find the corresponding [tex]\( y \)[/tex] value
Having found [tex]\( x = 2 \)[/tex], substitute this back into one of the original equations to find [tex]\( y \)[/tex]. We'll use the first equation:
[tex]\[ y = -x - 10 \][/tex]
[tex]\[ y = -2 - 10 \][/tex]
[tex]\[ y = -12 \][/tex]
So, the solution to the system is [tex]\( (x, y) = (2, -12) \)[/tex].
### Step 6: Check for any additional solutions
Since the quadratic equation [tex]\( (x - 2)^2 = 0 \)[/tex] has a repeated root, [tex]\( x = 2 \)[/tex] is the only solution. Thus, there are no additional solutions.
### Conclusion:
The solution to the given system of equations is:
[tex]\[ \boxed{(2, -12) \ \text{none}} \][/tex]
This means there is one intersection point at [tex]\( (2, -12) \)[/tex], and no other intersection points.
