Solve the system of equations:
$\begin{array}{l}3 x+8 y=54 \\ 4 x+5 y=38\end{array}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
\[ \begin{cases} 3x + 8y = 54 \ 4x + 5y = 38 \end{cases} \]
Our objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. This is a classic algebraic problem that can be approached using methods like elimination, substitution, or matrix algebra. Given the coefficients, the elimination method is a very direct and efficient way to reach the solution.
To eliminate \(x\), we can make the coefficients of \(x\) in both equations identical.
\[ 4(3x + 8y) = 4(54) \implies 12x + 32y = 216 \]
\[ 3(4x + 5y) = 3(38) \implies 12x + 15y = 114 \]
Now we have a new system:
\[ \begin{cases} 12x + 32y = 216 \ 12x + 15y = 114 \end{cases} \]
Subtract the second equation from the first to eliminate \(x\):
\[ (12x - 12x) + (32y - 15y) = 216 - 114 \]
\[ 17y = 102 \]
Solving for \(y\):
\[ y = \frac{102}{17} = 6 \]
Now that we have the value of \(y\), substitute \(y = 6\) back into the first original equation to solve for \(x\):
\[ 3x + 8(6) = 54 \]
\[ 3x + 48 = 54 \]
Subtract 48 from both sides:
\[ 3x = 54 - 48 \]
\[ 3x = 6 \]
Dividing by 3:
\[ x = \frac{6}{3} = 2 \]
Thus, the solution to the system is \(x = 2\) and \(y = 6\). Verification:
For the first equation: \(3(2) + 8(6) = 6 + 48 = 54\).
For the second equation: \(4(2) + 5(6) = 8 + 30 = 38\).
Both equations are satisfied.
Imagine a small business where you sell two types of gift boxes. Box A costs \(\$3\) per unit plus \(\$8\) for decoration (totaling \(\$54\) for a certain number of boxes), and Box B costs \(\$4\) per unit plus \(\$5\) for decoration (totaling \(\$38\)). By solving this system of equations, you determine exactly how many of each box type you have sold, allowing you to manage your inventory and finances accurately. This mathematical skill is foundational in logistics, resource allocation, and daily economic planning.