To solve the given system of equations:
1. [tex]\( 3x + y = -3 \)[/tex]
2. [tex]\( (x + 1)^2 - 4(x + 1) - 6 = y \)[/tex]
we will follow these steps:
### Step 1: Express y from the First Equation
From the first equation:
[tex]\[ y = -3 - 3x \][/tex]
### Step 2: Substitute y in the Second Equation
Substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ (x + 1)^2 - 4(x + 1) - 6 = -3 - 3x \][/tex]
### Step 3: Simplify the Equation
Expand and simplify the left-hand side:
[tex]\[ (x + 1)^2 - 4(x + 1) - 6 = x^2 + 2x + 1 - 4x - 4 - 6 \][/tex]
[tex]\[ x^2 - 2x - 9 = -3 - 3x \][/tex]
Next, move all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 9 + 3 + 3x = 0 \][/tex]
[tex]\[ x^2 + x - 6 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Solve the quadratic equation [tex]\( x^2 + x - 6 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex]:
For [tex]\( x^2 + x - 6 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-6)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{-1 \pm \sqrt{{1 + 24}}}}{2} \][/tex]
[tex]\[ x = \frac{{-1 \pm 5}}{2} \][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{{-1 + 5}}{2} = 2 \][/tex]
[tex]\[ x = \frac{{-1 - 5}}{2} = -3 \][/tex]
### Step 5: Find Corresponding y-values
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex] back into [tex]\( y = -3 - 3x \)[/tex] to find the corresponding [tex]\( y \)[/tex]-values:
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -3 - 3(2) = -3 - 6 = -9 \][/tex]
For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -3 - 3(-3) = -3 + 9 = 6 \][/tex]
### Step 6: Determine the Solution with [tex]\( y > 0 \)[/tex]
We found two solutions:
- [tex]\( (x, y) = (2, -9) \)[/tex]
- [tex]\( (x, y) = (-3, 6) \)[/tex]
Among these, we need to identify the solution with [tex]\( y > 0 \)[/tex]. The solution [tex]\( (-3, 6) \)[/tex] has [tex]\( y = 6 \)[/tex], which is greater than 0.
### Conclusion
The value of [tex]\( y \)[/tex] for the solution where [tex]\( y > 0 \)[/tex] is:
[tex]\[ y = 6 \][/tex]
