- Express the given system of linear equations in standard form: \(2x - 5y = -2\) and \(-4x + 6y = 1\).
- Use the elimination method by multiplying the first equation by 2 to align the \(x\) coefficients and then add them to find the value of \(y\): \(y = \frac{3}{4}\).
- Substitute the value of \(y\) back into the original equation to solve for \(x\), which yields the final result: \(\boxed{\frac{7}{8}}\).
Explanation
- Problem Analysis
We are given the following system of linear equations:
1) \(2x = 5y - 2\)
2) \(6y = 1 + 4x\)
Our goal is to find the value of \(x\) that satisfies both equations simultaneously. To solve this system, we can either use the method of substitution or the method of elimination. We will use the elimination method by aligning the terms involving \(x\) and \(y\) in both equations.
- Solving for y using Elimination
First, rewrite the equations in standard form \(Ax + By = C\):
From equation (1), we have:
\[2x - 5y = -2\]
From equation (2), we have:
\[-4x + 6y = 1\]
To eliminate \(x\), we can multiply the first equation by 2 so that the coefficient of \(x\) matches the magnitude of the coefficient of \(x\) in the second equation:
\[2 \times (2x - 5y) = 2 \times (-2)\]
\[4x - 10y = -4\]
Now, add this result to the second equation:
\[(4x - 4x) + (-10y + 6y) = -4 + 1\]
\[-4y = -3\]
Solving for \(y\), we divide both sides by -4:
\[y = \frac{-3}{-4} = \frac{3}{4}\]
- Solving for x
Now that we have the value of \(y\), we substitute it back into the first equation to find \(x\):
\[2x = 5\left(\frac{3}{4}\right) - 2\]
\[2x = \frac{15}{4} - 2\]
Convert 2 to a fraction with a denominator of 4:
\[2x = \frac{15}{4} - \frac{8}{4}\]
\[2x = \frac{7}{4}\]
Finally, divide by 2:
\[x = \frac{7}{4 \times 2} = \frac{7}{8}\]
The value of \(x\) is \(\frac{7}{8}\).
Examples
This system of equations can be used to determine the equilibrium point between two different supply and demand curves. For example, if the first equation represents the relationship between the quantity of a product and its price in one market, and the second represents it in another, finding the intersection point tells us the price point where supply meets demand in both markets simultaneously. Mastering these algebraic techniques is essential for fields like economics, engineering, and logistics where balancing multiple constraints is common.