Solve the simultaneous equations.
$\begin{array}{l}
9x + 4y = 11 \\
2x + y = 3
\end{array}$
$x = \quad y = $
We are presented with a system of two linear equations in two variables, \(x\) and \(y\):
\[ \begin{cases} 9x + 4y = 11 \quad (1) \ 2x + y = 3 \quad (2) \end{cases} \]
Our goal is to find the unique pair \((x, y)\) that satisfies both equations simultaneously. We can use the elimination method, which involves manipulating the equations to eliminate one of the variables.
To eliminate \(y\), we observe the coefficients of \(y\) in the two equations: 4 in equation (1) and 1 in equation (2). If we multiply equation (2) by \(-4\), the coefficient of \(y\) will become \(-4\), which will cancel out when we add it to equation (1).
Multiplying equation (2) by \(-4\):
\[-4 \times (2x + y) = -4 \times 3\]
\[-8x - 4y = -12 \quad (3)\]
Now, add equation (1) and equation (3):
\[(9x + 4y) + (-8x - 4y) = 11 + (-12)\]
\[(9x - 8x) + (4y - 4y) = -1\]
\[x = -1\]
Now that we have the value of \(x\), we can substitute it into equation (2) to find the value of \(y\):
\[2(-1) + y = 3\]
\[-2 + y = 3\]
Adding 2 to both sides gives:
\[y = 3 + 2\]
\[y = 5\]
Thus, the solution to the system is \(x = -1\) and \(y = 5\).
To ensure our solution is correct, we substitute \(x = -1\) and \(y = 5\) back into the first equation:
\[9(-1) + 4(5) = -9 + 20 = 11\]
The result matches the original equation. Therefore, the solution is indeed \(x = -1\) and \(y = 5\). The final answer is \(\boxed{x = -1, y = 5}\).
Systems of linear equations are incredibly useful in everyday life, such as in business logistics or personal budgeting. For example, if you buy 9 packs of blue pens and 4 packs of red pens for \(\$11\), and a friend buys 2 packs of blue pens and 1 pack of red pens for \(\$3\), you can use this system of equations to determine the individual price of each pack of pens. Solving these equations helps you manage resources, calculate costs, and make informed financial decisions.