Solve the system of equations:
$\begin{array}{l}
x+y-3=0 \\
2 x-5 y+1=0
\end{array}$
We are given a system of two linear equations with two variables \(x\) and \(y\):
1) \(x + y - 3 = 0\)
2) \(2x - 5y + 1 = 0\)
The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since both equations represent lines on a 2D plane, their solution is the point of intersection. We will use the method of substitution to solve this system.
From the first equation, \(x + y - 3 = 0\), we can isolate \(x\) in terms of \(y\):
\[x = 3 - y\]
Next, substitute this expression for \(x\) into the second equation:
\[2(3 - y) - 5y + 1 = 0\]
Now, expand and simplify the equation:
\[6 - 2y - 5y + 1 = 0\]
Combining like terms:
\[7 - 7y = 0\]
\[7y = 7\]
\[y = 1\]
With \(y = 1\) determined, we find \(x\) by substituting back into our expression for \(x\):
\[x = 3 - 1 = 2\]
Thus, the solution is \((x, y) = (2, 1)\).
To ensure the solution \((2, 1)\) is correct, we substitute these values back into both original equations:
Equation 1: \(2 + 1 - 3 = 3 - 3 = 0\) (Verified)
Equation 2: \(2(2) - 5(1) + 1 = 4 - 5 + 1 = 0\) (Verified)
The solution satisfies both equations, confirming that the intersection point is indeed \((2, 1)\). The final answer is \(\boxed{{x=2, y=1}}\).
Systems of linear equations are useful in everyday life for optimization and decision-making. For example, if you are comparing two phone plans: Plan A might cost \(\$30\) per month plus \(\$0.10\) per minute, while Plan B might cost \(\$20\) per month plus \(\$0.20\) per minute. By setting up a system of equations where \(x\) is the number of minutes and \(y\) is the total cost, you can find the exact number of minutes where the costs of both plans are equal, helping you decide which plan is more economical based on your usage.