Verified Answer :

  • Multiply the first equation by 2 to align the coefficients of \(y\): \(6x + 2y = 28\).
  • Add this new equation to the second equation (\(x - 2y = 0\)) to eliminate \(y\), resulting in \(7x = 28\).
  • Solve for \(x\) to get \(x = 4\), then substitute this value back into an original equation to find \(y = 2\).
  • The solution is \(\boxed{x=4, y=2}\).

Explanation

  1. Analysis of the System

The given system of linear equations is:
\[ \begin{cases} 3x + y = 14 & \text{(Equation 1)} \ x - 2y = 0 & \text{(Equation 2)} \end{cases} \]
Our objective is to solve this system using the elimination method, which involves manipulating the equations so that one variable cancels out when they are added or subtracted. Here, we can observe that Equation 1 has a term \(y\) and Equation 2 has a term \(-2y\). If we multiply Equation 1 by \(2\), the coefficients of \(y\) will become additive inverses (\(2\) and \(-2\)), allowing us to eliminate \(y\) by adding the equations.

  1. Eliminating the Variable y

To eliminate \(y\), we multiply Equation 1 by \(2\):
\[2 \times (3x + y) = 2 \times 14 \implies 6x + 2y = 28 \quad \text{(Equation 3)}\]
Now, we add Equation 3 to Equation 2:
\[(6x + 2y) + (x - 2y) = 28 + 0\]
Grouping like terms:
\[(6x + x) + (2y - 2y) = 28\]
\[7x = 28\]
Dividing by \(7\), we find:
\[x = 4\]

  1. Substitution and Verification

Now that we have found the value of \(x\), we substitute \(x = 4\) into Equation 2 to solve for \(y\):
\[4 - 2y = 0\]
Rearranging to solve for \(y\):
\[2y = 4\]
\[y = 2\]
Thus, the solution to the system is \(x = 4\) and \(y = 2\). We verify this by substituting these values into Equation 1:
\[3(4) + 2 = 12 + 2 = 14\]
The solution satisfies both equations perfectly. Therefore, the values are \(x = 4\) and \(y = 2\).

Examples

Systems of linear equations are frequently used in everyday shopping. Suppose you buy 3 apples and 1 orange for \$14, and an apple costs twice as much as an orange in a different bundle deal (or you find that for a specific set of prices, 1 apple costs the same as 2 oranges). By setting up a system of equations, you can determine the price of each individual fruit. This algebraic approach allows for efficient budget planning and decision-making when dealing with multi-item purchases.

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