- Substitute the linear equation \(y = x - 3\) into the quadratic equation \(x^2 + 3y = 1\).
- Simplify the resulting expression to get the quadratic equation \(x^2 + 3x - 10 = 0\).
- Factor the equation into \((x + 5)(x - 2) = 0\) to find the two roots, \(x = 2\) and \(x = -5\).
- Calculate the y-value for the second root: \(y = -5 - 3 = -8\).
- The other solution set is \(\boxed{x = -5, y = -8}\).
Explanation
- Analysis of the System of Equations
We are given a system of two equations involving variables \(x\) and \(y\):
1) \(x^2 + 3y = 1\)
2) \(y = x - 3\)
We already know that one solution set is \(x = 2, y = -1\). Our goal is to find the other solution set \((x, y)\) for this system. Since the second equation expresses \(y\) directly in terms of \(x\), we can solve this by substitution.
- Substituting and Forming a Quadratic Equation
We substitute the second equation \(y = x - 3\) into the first equation \(x^2 + 3y = 1\):
\[x^2 + 3(x - 3) = 1\]
Expanding the expression, we get:
\[x^2 + 3x - 9 = 1\]
Subtracting 1 from both sides gives the quadratic equation in terms of \(x\):
\[x^2 + 3x - 10 = 0\]
- Solving the Quadratic Equation
To solve \(x^2 + 3x - 10 = 0\), we look for two numbers that multiply to \(-10\) and add up to \(3\). These numbers are \(5\) and \(-2\). Thus, we can factor the quadratic equation as:
\[(x + 5)(x - 2) = 0\]
This gives us two possible values for \(x\):
\(x_1 = 2\) and \(x_2 = -5\).
Since we already knew the solution corresponding to \(x = 2\), the second solution must correspond to \(x = -5\).
- Finding the Other Solution Set
Now that we have the second x-value, \(x = -5\), we find the corresponding y-value by substituting this into the linear equation \(y = x - 3\):
\[y = -5 - 3 = -8\]
So the second solution set is \(x = -5, y = -8\). We verify this in the first equation:
\((-5)^2 + 3(-8) = 25 - 24 = 1\).
This confirms the result is correct.
Examples
This method of solving systems of equations is fundamental in various fields, such as physics and engineering. For example, if you are calculating the point where an object moving along a curved path (defined by a parabola) intersects with an object moving along a straight line (linear path), you are essentially solving a system of equations like this. By finding these intersection points, you can determine where the objects meet or collide in real-world scenarios.