- Simplify the first equation to \(y = -x^2\).
- Rearrange the second equation to \(y = 8x + 16\).
- Equate the two expressions to obtain the quadratic equation \(x^2 + 8x + 16 = 0\).
- Factor the quadratic as \((x + 4)^2 = 0\) to find the solution: \(\boxed{-4}\).
Explanation
- Problem Analysis and Strategy
We are given a system of two equations:
1) \(x^2 + y + 10 = 10\)
2) \(8x + 16 - y = 0\)
We aim to find the value of \(x\) that satisfies this system. Our strategy is to express one variable in terms of the other and solve for \(x\).
- Simplifying and Expressing y in Terms of x
First, let's simplify the first equation:
\[x^2 + y + 10 = 10\]
Subtracting 10 from both sides gives:
\[x^2 + y = 0\]
We can express \(y\) in terms of \(x\) as:
\[y = -x^2\]
Now, look at the second equation:
\[8x + 16 - y = 0\]
Rearranging this to solve for \(y\), we get:
\[y = 8x + 16\]
Now that we have expressions for \(y\) from both equations, we set them equal to each other.
- Solving for x
Setting the two expressions for \(y\) equal:
\[-x^2 = 8x + 16\]
To solve this equation, move all terms to one side to form a standard quadratic equation:
\[x^2 + 8x + 16 = 0\]
We notice that this is a perfect square trinomial:
\[(x + 4)^2 = 0\]
Taking the square root of both sides:
\[x + 4 = 0\]
\[x = -4\]
Since the question asks for the value of \(x\), we have found it. We can optionally verify the corresponding value of \(y\): \(y = -(-4)^2 = -16\). Checking this in the second equation: \(8(-4) + 16 - (-16) = -32 + 16 + 16 = 0\). The values satisfy both equations.
Examples
Solving systems of equations is fundamental in fields like engineering and economics. For example, if you are modeling the intersection of two paths—one parabolic (e.g., the trajectory of a ball, \(y = -x^2\)) and one linear (e.g., the path of a moving object, \(y = 8x + 16\))—solving the system of equations helps you determine the exact point and time they collide. This ensures accurate prediction and safety in real-world scenarios.