- Align the equations by multiplying them to match the coefficients of one variable (multiplying by 3 and 4 gives 12 as the common coefficient).
- Subtract the equations to eliminate the variable \(y\) and solve for \(x = 9\).
- Substitute \(x = 9\) back into one of the original equations to solve for \(y = 5\).
- The final solution to the system is \(\boxed{x=9, y=5}\).
Explanation
- Analysis of the System
We are presented with a system of two linear equations in two variables, \(x\) and \(y\):
\[3x + 4y = 47\]
\[5x + 3y = 60\]
Our objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. A system like this represents two lines on a coordinate plane, and their solution corresponds to the point where these lines intersect. To solve this, we can use algebraic methods such as substitution or elimination. Here, we will use the elimination method, which involves manipulating the equations to eliminate one variable.
- Eliminating a Variable
To eliminate the variable \(y\), we make the coefficients of \(y\) in both equations identical. The coefficients of \(y\) are 4 and 3, respectively. The least common multiple of 4 and 3 is 12. We multiply the first equation by 3 and the second equation by 4:
- Multiply the first equation by 3:
\[3(3x + 4y) = 3(47) \implies 9x + 12y = 141\]
- Multiply the second equation by 4:
\[4(5x + 3y) = 4(60) \implies 20x + 12y = 240\]
Now we have:
\[9x + 12y = 141\]
\[20x + 12y = 240\]
- Solving for x
Subtract the first resulting equation from the second one to eliminate the \(y\) term:
\[(20x + 12y) - (9x + 12y) = 240 - 141\]
\[11x = 99\]
Now, solve for \(x\) by dividing both sides by 11:
\[x = \frac{99}{11} = 9\]
The value of \(x\) is 9.
- Solving for y
Now that we have the value of \(x = 9\), we substitute it back into the original first equation (\(3x + 4y = 47\)) to find the value of \(y\):
\[3(9) + 4y = 47\]
\[27 + 4y = 47\]
Subtract 27 from both sides:
\[4y = 47 - 27\]
\[4y = 20\]
Divide by 4:
\[y = 5\]
The value of \(y\) is 5.
- Verification and Conclusion
To ensure the solution is correct, we substitute \(x = 9\) and \(y = 5\) into the second original equation (\(5x + 3y = 60\)):
\[5(9) + 3(5) = 45 + 15 = 60\]
The values satisfy the second equation as well. Thus, the solution to the system is \(x = 9\) and \(y = 5\), which can be written as \(\boxed{x=9, y=5}\).
Examples
Systems of linear equations are extremely common in economics, engineering, and logistics. For example, imagine a café owner buying two types of coffee beans. If they buy 3 kg of Arabica and 4 kg of Robusta for \$47, and 5 kg of Arabica and 3 kg of Robusta for \$60, they can use this system of equations to determine the price per kilogram for each type of bean. By solving the system, the owner finds that Arabica costs \$9 per kg and Robusta costs \$5 per kg, helping them manage their inventory and pricing strategy effectively.