- Multiply the first equation by 2 to align the coefficient of \(x\) with the second equation.
- Subtract the second equation from the result to eliminate \(x\) and solve for \(y=2\).
- Substitute \(y=2\) into the first equation to solve for \(x=2.5\).
- The correct solution is \(\boxed{x=2.5, y=2}\).
Explanation
- Analysis of the System of Equations
We are given the system of linear equations:
\[2x + 3y = 11 \quad \text{(Equation 1)}\]
\[4x - 2y = 6 \quad \text{(Equation 2)}\]
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. We will use the elimination method to solve this system.
- Eliminating the Variable \(x\)
To eliminate one variable, we can make the coefficients of \(x\) the same in both equations. Multiplying Equation 1 by 2 gives:
\[2(2x + 3y) = 2(11)\]
\[4x + 6y = 22 \quad \text{(Equation 3)}\]
Now, we have a new system:
\[4x + 6y = 22\]
\[4x - 2y = 6\]
Subtracting Equation 2 from Equation 3 allows us to eliminate \(x\):
\[(4x + 6y) - (4x - 2y) = 22 - 6\]
\[4x - 4x + 6y + 2y = 16\]
\[8y = 16\]
Dividing by 8, we find:
\[y = 2\]
- Solving for \(x\) and Final Conclusion
Now that we have the value for \(y\), we substitute it back into Equation 1 to find \(x\):
\[2x + 3(2) = 11\]
\[2x + 6 = 11\]
Subtracting 6 from both sides:
\[2x = 5\]
Dividing by 2:
\[x = 2.5\]
The solution to the system is \((x, y) = (2.5, 2)\). Note that none of the provided options (a, b, c, d) match this result, indicating there may be a typo in the provided choices. Regardless, the correct values are \(\boxed{x=2.5, y=2}\).
Examples
Systems of linear equations are fundamental in many real-world applications, such as resource allocation or finding equilibrium points in physics and economics. For example, if you are running a business, you might use such systems to determine the quantities of two different products to produce based on constraints on labor and material costs. Solving these systems ensures that all constraints are met simultaneously, leading to optimal decisions. In this case, the method of elimination allowed us to efficiently find the exact values that balance the two given linear constraints.