- Substitute \(y = 2x - 3\) from the linear equation into the product equation \(xy = 9\).
- Solve the resulting quadratic equation \(2x^2 - 3x - 9 = 0\) using the quadratic formula.
- Identify the secondary root \(x = -1.5\) and compute the corresponding \(y = -6\).
- The other solution set is \(x = \boxed{-1.5}, y = \boxed{-6}\).
Explanation
- Setting up the Quadratic Equation
We are given a system of two equations:
1) \(2x - y = 3\)
2) \(xy = 9\)
We already know one solution is \(x = 3, y = 3\). Our goal is to find the other pair \((x, y)\) that satisfies both equations. To do this, we can use the substitution method. From the first equation, we can isolate \(y\) as:
\[y = 2x - 3\]
Now, we substitute this expression for \(y\) into the second equation:
\[x(2x - 3) = 9\]
Expanding this equation gives:
\[2x^2 - 3x = 9\]
Rearranging it into the standard form for a quadratic equation (\(ax^2 + bx + c = 0\)):
\[2x^2 - 3x - 9 = 0\]
- Solving for x
To solve the quadratic equation \(2x^2 - 3x - 9 = 0\), we can use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
where \(a = 2\), \(b = -3\), and \(c = -9\).
First, calculate the discriminant:
\[\Delta = (-3)^2 - 4(2)(-9) = 9 + 72 = 81\]
Now, find the values of \(x\):
\[x = \frac{-(-3) \pm \sqrt{81}}{2(2)} = \frac{3 \pm 9}{4}\]
This gives us two solutions for \(x\):
1) \(x_1 = \frac{3 + 9}{4} = \frac{12}{4} = 3\)
2) \(x_2 = \frac{3 - 9}{4} = \frac{-6}{4} = -1.5\)
The first solution, \(x = 3\), corresponds to the known solution set where \(y = 3\). We are looking for the second solution.
- Finding the Corresponding y and Verifying the Solution
Now that we have the second value for \(x\), which is \(x = -1.5\), we substitute this back into our linear expression for \(y\):
\[y = 2x - 3\]
\[y = 2(-1.5) - 3 = -3 - 3 = -6\]
Finally, verify this solution in the original equations:
1) \(2x - y = 2(-1.5) - (-6) = -3 + 6 = 3\) (Satisfied)
2) \(xy = (-1.5)(-6) = 9\) (Satisfied)
The other solution set is \(x = -1.5, y = -6\).
Examples
This system of equations represents the intersection of a line and a hyperbola on a coordinate plane. In engineering or physics, such problems arise when finding points where two different paths cross, like a particle moving along a straight line trajectory intersecting a curved boundary defined by a hyperbolic function. By solving for these intersection points, we can determine the exact coordinates where events occur.