- Rewrite the equations as \(y = x^2 + 5x - 10\) (a parabola) and \(y = -5x + 1\) (a line).
- Equate them to find \(x^2 + 10x - 11 = 0\), which has a discriminant of 144, indicating 2 solutions.
- Factor the equation to find \(x = 1\) and \(x = -11\), then substitute to find the points \((1, -4)\) and \((-11, 56)\).
- Final solutions: \(\boxed{(1, -4), (-11, 56)}\).
Explanation
- Identifying the curves
The given system of equations is:
\[ \begin{array}{l} 10 + y = 5x + x^2 \quad (1) \ 5x + y = 1 \quad \quad \quad \quad \quad (2) \end{array} \]
To analyze these, we express both equations in the slope-intercept or standard functional form for \(y\).
For the first equation, we subtract 10 from both sides:
\[y = x^2 + 5x - 10\]
This is a quadratic equation, which represents a parabola.
For the second equation, we subtract \(5x\) from both sides:
\[y = -5x + 1\]
This is a linear equation, which represents a line.
- Analyzing the number of solutions
To find the solutions, we set the two expressions for \(y\) equal to each other, since any point that satisfies both equations must have the same \(y\)-coordinate:
\[x^2 + 5x - 10 = -5x + 1\]
Adding \(5x\) and subtracting 1 from both sides to form a standard quadratic equation:
\[x^2 + 10x - 11 = 0\]
We determine the number of solutions by examining the discriminant \(D = b^2 - 4ac\), where \(a=1, b=10,\) and \(c=-11\):
\[D = 10^2 - 4(1)(-11) = 100 + 44 = 144\]
Since the discriminant \(D = 144\) is greater than zero, there are 2 distinct real solutions to the system.
- Finding the solutions
We solve the quadratic equation \(x^2 + 10x - 11 = 0\) by factoring or using the quadratic formula. The equation factors as:
\[(x - 1)(x + 11) = 0\]
This gives us two possible values for \(x\):
\(x_1 = 1\) and \(x_2 = -11\).
Now, we substitute these \(x\)-values into the linear equation \(y = -5x + 1\) to find the corresponding \(y\)-values:
For \(x_1 = 1\):
\[y_1 = -5(1) + 1 = -4\]
For \(x_2 = -11\):
\[y_2 = -5(-11) + 1 = 55 + 1 = 56\]
Thus, the solutions to the system are \((1, -4)\) and \((-11, 56)\).
Examples
In engineering and physics, systems of equations like this one are commonly used to find the intersection points of different paths. For example, if a projectile follows a parabolic trajectory given by \(y = x^2 + 5x - 10\) and an object moves along a linear path defined by \(y = -5x + 1\), finding the solution to the system identifies the exact moments and positions where these two paths intersect. This is essential for collision detection, optimizing trajectories, and ensuring safety in automated navigation systems.