Verified Answer :

  • Multiply the first equation by 2 to get \(16x - 12y = -40\).
  • Add this to the second equation \(-16x + 7y = 30\) to eliminate \(x\), resulting in \(-5y = -10\).
  • Solve for \(y\), giving \(y = 2\).
  • Substitute \(y = 2\) into the first original equation to solve for \(x\), yielding \(x = -1\).
  • The solution to the system is \(\boxed{(-1, 2)}\).

Explanation

  1. Analysis of the System

We are given a system of two linear equations in two variables:
\[\begin{cases} 8x - 6y = -20 \quad (1) \ -16x + 7y = 30 \quad (2) \end{cases}\]
The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously using the elimination method. The elimination method involves manipulating the equations so that adding or subtracting them eliminates one of the variables, allowing us to solve for the remaining variable.

  1. Eliminating the Variable x

To eliminate the variable \(x\), we look at the coefficients of \(x\) in both equations, which are \(8\) and \(-16\). If we multiply the first equation by \(2\), the coefficient of \(x\) becomes \(16\), which is the additive inverse of \(-16\).
Multiplying equation (1) by \(2\):
\[2 \times (8x - 6y) = 2 \times (-20)\]
\[16x - 12y = -40 \quad (3)\]

  1. Solving for y

Now, we add equation (3) to equation (2) to eliminate \(x\):
\[(16x - 12y) + (-16x + 7y) = -40 + 30\]
The \(x\)-terms cancel out:
\[(16x - 16x) + (-12y + 7y) = -10\]
\[-5y = -10\]
Dividing both sides by \(-5\), we find:
\[y = 2\]

  1. Solving for x

With the value of \(y = 2\), we can substitute this into equation (1) to solve for \(x\):
\[8x - 6(2) = -20\]
\[8x - 12 = -20\]
Adding \(12\) to both sides:
\[8x = -8\]
Dividing by \(8\):
\[x = -1\]

  1. Verification and Final Result

Finally, we verify our solution \((x, y) = (-1, 2)\) by substituting the values into the second original equation:
\[-16(-1) + 7(2) = 16 + 14 = 30\]
The values satisfy both equations. Thus, the solution to the system is \(x = -1\) and \(y = 2\), which can be written as \(\boxed{(-1, 2)}\).

Examples

The elimination method is a powerful tool for solving systems of equations, which can be applied to real-world scenarios like balancing budgets or manufacturing. For instance, if you are buying items for a project where you know the total cost and the quantity of two different components, you can use a system of equations to determine the unit price of each component. By treating the unknown unit prices as variables, you can quickly find the exact costs and ensure your project stays within its financial constraints.

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