- Rewrite the system as \(x + y = 5\) and \(xy = 6\).
- Substitute \(y = 5 - x\) into the second equation to get the quadratic equation \(x^2 - 5x + 6 = 0\).
- Factor the quadratic equation to find the roots \(x = 2\) and \(x = 3\).
- Determine the corresponding \(y\) values to identify the solution sets \((2, 3)\) and \((3, 2)\).
- Conclude that the other solution set is \(x=3, y=2\).
- Final Answer: \(\boxed{x=3, y=2}\)
Explanation
- Problem Analysis
We are given the following system of equations involving two variables \(x\) and \(y\):
\[x + y - 5 = 0\]
\[xy = 6\]
We can rewrite the first equation as \(x + y = 5\). Our objective is to find both solution sets for \((x, y)\) and specifically identify the one that is not \(x=2, y=3\). This is a system of equations where the sum and product of the two variables are known, which is a classic scenario for finding the roots of a quadratic equation.
- Substitution and Quadratic Formulation
We can express \(y\) in terms of \(x\) using the first equation:
\[y = 5 - x\]
Now, we substitute this expression for \(y\) into the second equation, \(xy = 6\):
\[x(5 - x) = 6\]
Expanding the left side, we obtain:
\[5x - x^2 = 6\]
To solve this, we rearrange the terms into a standard quadratic equation form \(ax^2 + bx + c = 0\):
\[x^2 - 5x + 6 = 0\]
- Solving for the Variables
We solve the quadratic equation \(x^2 - 5x + 6 = 0\) by factoring. We look for two numbers that multiply to \(6\) and add up to \(-5\). These numbers are \(-2\) and \(-3\). Thus, the equation can be factored as:
\[(x - 2)(x - 3) = 0\]
This gives us two possible values for \(x\):
\[x = 2 \quad \text{or} \quad x = 3\]
Now, we find the corresponding \(y\) values using the relationship \(y = 5 - x\):
- For \(x = 2\), \(y = 5 - 2 = 3\).
- For \(x = 3\), \(y = 5 - 3 = 2\).
Thus, the two solutions are \((2, 3)\) and \((3, 2)\). Since \((2, 3)\) was provided in the problem, the other solution set is \(x=3, y=2\).
Examples
In many real-world scenarios, such as designing physical structures, you might need to find two dimensions (like length and width) when the perimeter and area are known. For a rectangular garden with a perimeter of \(10\) meters (\(2(l+w) = 10 \implies l+w=5\)) and an area of \(6\) square meters (\(l \times w = 6\)), this exact algebraic method helps determine that the dimensions must be \(3\) meters and \(2\) meters. Solving such systems is fundamental in engineering, economics, and logistics.