- Add equations (1) and (2) to eliminate \(z\), resulting in \(3x - y = 26\).
- Add equations (2) and (3) to eliminate \(z\), resulting in \(2x + y = 19\).
- Solve the system of two equations for \(x\) and \(y\) by addition, yielding \(x = 9\) and \(y = 1\).
- Substitute \(x = 9\) and \(y = 1\) into one of the original equations to find \(z = 0\). The solution is \((x, y, z) = \boxed{(9, 1, 0)}\).
Explanation
- Problem Analysis
The problem asks to solve a system of three linear equations in three variables \(x\), \(y\), and \(z\):
- \(2x + y - 3z = 19\)
- \(x - 2y + 3z = 7\)
- \(3x + 4y - 3z = 31\)
The method requested is the addition (elimination) method, which involves combining equations to eliminate variables, reducing the system step-by-step until the variables are found.
- Eliminating z
We look for opportunities to eliminate a variable. Notice that equation (1) has \(-3z\) and equation (2) has \(+3z\). By adding equations (1) and (2), we eliminate \(z\):
\[(2x + y - 3z) + (x - 2y + 3z) = 19 + 7\]
\[3x - y = 26 \quad (\text{Equation 4})\]
Next, observe equation (2) has \(+3z\) and equation (3) has \(-3z\). By adding equations (2) and (3), we eliminate \(z\) again:
\[(x - 2y + 3z) + (3x + 4y - 3z) = 7 + 31\]
\[4x + 2y = 38\]
Dividing by 2, we simplify this to:
\[2x + y = 19 \quad (\text{Equation 5})\]
- Solving for x and y
Now we have a system of two equations with two variables:
- \(3x - y = 26\)
- \(2x + y = 19\)
Adding these two equations conveniently eliminates \(y\):
\[(3x - y) + (2x + y) = 26 + 19\]
\[5x = 45\]
\[x = 9\]
Now, substitute \(x = 9\) into equation (5) to find \(y\):
\[2(9) + y = 19\]
\[18 + y = 19\]
\[y = 1\]
- Solving for z and Conclusion
Finally, substitute the values \(x = 9\) and \(y = 1\) into original equation (2) to solve for \(z\):
\[x - 2y + 3z = 7\]
\[9 - 2(1) + 3z = 7\]
\[9 - 2 + 3z = 7\]
\[7 + 3z = 7\]
\[3z = 0\]
\[z = 0\]
We verify these values using equation (1): \(2(9) + 1 - 3(0) = 18 + 1 = 19\), which is correct. Using equation (3): \(3(9) + 4(1) - 3(0) = 27 + 4 = 31\), which is also correct. Thus, the solution is \((x, y, z) = (9, 1, 0)\).
Examples
Solving systems of linear equations is essential in fields like engineering and economics. For example, if you are calculating the cost of different components in a project where you know the total cost of three different combinations of those components, you can represent the system as linear equations to find the unit cost of each specific component. Using the elimination method allows you to systematically isolate and solve for each variable, ensuring accurate resource budgeting.