- Substitute the linear equation \(y = x - 4\) into the quadratic equation \(x^2 + 2y = 7\).
- Simplify and factor the resulting equation to \(x^2 + 2x - 15 = 0\), which yields the roots \(x = 3\) and \(x = -5\).
- Calculate the corresponding \(y\) value for \(x = -5\) using the linear equation to get \(y = -9\).
- The other solution set is \(x = \boxed{-5}, y = \boxed{-9}\).
Explanation
- Problem Analysis
The goal is to find the secondary solution to the system of equations given by:
\[x^2 + 2y = 7\]
\[y = x - 4\]
We are provided with one solution set: \(x = 3, y = -1\). We know that this system involves a quadratic and a linear equation, which means there are at most two points of intersection. By substituting the expression for \(y\) from the second equation into the first, we can reduce the system to a single quadratic equation in terms of \(x\), which we can then solve to find all possible values of \(x\).
- Solving for x
To find the other solution, we substitute \(y = x - 4\) into the first equation \(x^2 + 2y = 7\):
\[x^2 + 2(x - 4) = 7\]
Expanding this equation gives:
\[x^2 + 2x - 8 = 7\]
Subtracting 7 from both sides to set the equation to zero:
\[x^2 + 2x - 15 = 0\]
We factor this quadratic equation:
\[(x - 3)(x + 5) = 0\]
The roots are \(x = 3\) and \(x = -5\). The value \(x = 3\) corresponds to the given solution \(y = -1\). Therefore, the other solution corresponds to \(x = -5\).
- Finding y and Verifying the Solution
Now that we have the second root \(x = -5\), we substitute this value back into the linear equation \(y = x - 4\) to find the corresponding value of \(y\):
\[y = -5 - 4 = -9\]
Thus, the other solution set is \(x = -5, y = -9\). We can check this:
For the first equation: \((-5)^2 + 2(-9) = 25 - 18 = 7\). This matches the given value of 7.
For the second equation: \(-9 = -5 - 4\). This also matches.
Both equations are satisfied, confirming the solution set is correct.
Examples
This type of problem is essential in physics and engineering when determining the intersection points of a projectile path (represented by a parabola) and a straight line. For instance, calculating when a ball's trajectory, defined by a quadratic function, meets a specific linear path allows scientists to predict collisions or interception points in sports or navigation. Solving systems like this provides the foundation for determining these critical points in space and time.