Verified Answer :

  • Multiply the first equation by \(-5\) and the second equation by \(2\) to prepare for elimination: \(-10x - 15y = -160\) and \(10x + 8y = 118\).
  • Add the equations to eliminate \(x\) and solve for \(y\): \(-7y = -42 \Rightarrow y = 6\).
  • Substitute \(y = 6\) into the original equation \(2x + 3y = 32\) to solve for \(x\): \(2x + 18 = 32 \Rightarrow x = 7\).
  • The solution to the system of equations is \(\boxed{x=7, y=6}\).

Explanation

  1. Analysis of the System of Equations

We are given a system of two linear equations with two variables, \(x\) and \(y\):

  1. \(2x + 3y = 32\)
  2. \(5x + 4y = 59\)

Our goal is to solve for \(x\) and \(y\) using the elimination method. This involves manipulating the equations so that adding them together eliminates one of the variables, allowing us to solve for the other. We are specifically instructed to make the coefficient of \(x\) zero by multiplying the first equation by \(-5\) and the second by \(2\).

  1. Manipulating the Equations

First, we multiply the first equation, \(2x + 3y = 32\), by \(-5\):
\(-5(2x + 3y) = -5(32)\)
\(-10x - 15y = -160\)

Next, we multiply the second equation, \(5x + 4y = 59\), by \(2\):
\(2(5x + 4y) = 2(59)\)
\(10x + 8y = 118\)

  1. Eliminating and Solving for \(y\)

Now, we add the two resulting equations together:
\((\text{-}10x - 15y) + (10x + 8y) = -160 + 118\)
The \(x\)-terms cancel out:
\((-10x + 10x) + (-15y + 8y) = -42\)
\(-7y = -42\)

Solving for \(y\):
\(y = \frac{-42}{-7} = 6\)

  1. Solving for \(x\)

Now that we have the value of \(y\), we substitute it back into the original first equation, \(2x + 3y = 32\):
\(2x + 3(6) = 32\)
\(2x + 18 = 32\)
Subtracting 18 from both sides:
\(2x = 32 - 18\)
\(2x = 14\)
Solving for \(x\):
\(x = \frac{14}{2} = 7\)

  1. Verification and Final Conclusion

We have found the solution \(x = 7\) and \(y = 6\). To verify, we substitute these into the second equation:
\(5(7) + 4(6) = 35 + 24 = 59\).
Since the result matches the given constant, our solution is correct. Thus, the solution to the system is \(x = 7\) and \(y = 6\). The final result is \(\boxed{x=7, y=6}\).

Examples

Solving systems of equations is fundamental in fields like economics, engineering, and logistics. For instance, imagine a company that produces two products, A and B. The production of each requires specific amounts of two resources (e.g., labor and materials). If Product A uses 2 units of labor and 5 units of material, and Product B uses 3 units of labor and 4 units of material, with total availability being 32 units of labor and 59 units of material, we can model this as a system of equations to determine exactly how many units of each product to manufacture to use all available resources efficiently. Mastery of these methods enables effective planning and resource management in real-world scenarios.

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