Solve the system of equations by the elimination method \( \left\{\begin{array}{l}4 x+y=5 \\ 8 x+3 y=12\end{array}\right. \) \( (x, y)=\square \)

To solve the system of equations using the elimination method, we can start by manipulating the first equation. From \( 4x + y = 5 \), we can express \( y \) in terms of \( x \): \( y = 5 - 4x \). Now, we can substitute this expression for \( y \) into the second equation \( 8x + 3y = 12 \): \( 8x + 3(5 - 4x) = 12 \). Expanding this gives: \( 8x + 15 - 12x = 12 \). Combining like terms results in: \( -4x + 15 = 12 \). Subtracting 15 from both sides yields: \( -4x = -3 \). Dividing both sides by -4 gives us: \( x = \frac{3}{4} \). Now we can substitute \( x = \frac{3}{4} \) back into the expression for \( y \): \( y = 5 - 4\left(\frac{3}{4}\right) = 5 - 3 = 2 \). So the solution to the system of equations is \( (x, y) = \left(\frac{3}{4}, 2\right) \).