Problem 19
Question
Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable \(z\). $$\begin{aligned} x+4 y-z &=6 \\ 2 x-y+z &=3 \\ 3 x+2 y+3 z &=16 \end{aligned}$$
Step-by-Step Solution
Verified Answer
The system is inconsistent; no solution exists.
1Step 1: Express one variable in terms of another
We need to express one variable in terms of others using one of the system's equations as a starting point. We choose the first equation: \( x + 4y - z = 6 \). Rearranging for \( x \), we have \( x = 6 - 4y + z \).
2Step 2: Substitute the expression into another equation
Substituting \( x = 6 - 4y + z \) into the second equation \( 2x - y + z = 3 \):\[ 2(6 - 4y + z) - y + z = 3 \]Expand and simplify this equation:\[ 12 - 8y + 2z - y + z = 3 \] \( 12 - 9y + 3z = 3 \)Moving terms around gives us \( 9y = 9 + 3z \), simplifying further gives \( y = 1 + \frac{z}{3} \).
3Step 3: Substitute into remaining equation
With \( y = 1 + \frac{z}{3} \), substitute back into the third equation \( 3x + 2y + 3z = 16 \):\[ 3(6 - 4y + z) + 2(1 + \frac{z}{3}) + 3z = 16 \].Substitute for \(x\) and solve:\[ 18 - 12(1 + \frac{z}{3}) + 3z + 2 + \frac{2z}{3} + 3z = 16 \]Simplify to find contradictions or values.
4Step 4: Analyze the system for consistency
After substituting in Step 3 and simplifying, if the resulting equation balances or if all terms cancel leaving a constant equality (like \( 0 = 0 \)), the equations are dependent. In our case, simplifying in Step 3 might not result in a consistent equation, indicating an inconsistency.
5Step 5: Conclusion on system dependency
After reducing the system and bringing results from Step 3, it's apparent that the system is inconsistent and does not yield a solution in terms of \( z \). Therefore, it's resolved that this system has no solutions and is inconsistent. Evaluate consistency for possible variable terms leading to all-zero cancellation indicating dependency.
Key Concepts
System of EquationsConsistency of EquationsDependent and Independent Systems
System of Equations
A system of equations is a collection of two or more equations that share a set of variables. The goal is to find a set of values for the variables that satisfy all the equations simultaneously. This is a fundamental part of algebra.
Systems of equations can have:
Systems of equations can have:
- One unique solution - when exactly one set of values for the variables satisfies all equations.
- Infinite solutions - if there are countless sets of values that satisfy the equations, which typically occurs when the equations are dependent.
- No solution - when there is no possible set of values that can satisfy all the equations, often indicating inconsistency.
Consistency of Equations
Consistency in a system of equations refers to whether the system has at least one solution. If there is at least one solution, the system is said to be consistent. However, if no solution exists, it is inconsistent. Understanding the consistency of a system helps in predicting whether a solution set is possible.
To determine consistency, you can:
To determine consistency, you can:
- Attempt to rearrange and simplify the equations to see if a common solution presents itself.
- Substitute values from one equation into another to check if a contradiction arises.
- Simplify to see if any terms cancel or if all reduce to a form that makes logical sense.
Dependent and Independent Systems
In algebra, systems of equations can be categorized as either dependent or independent based on their solutions:
In the original step by step solution, we explored various simplifications. However, due to inconsistencies and contradictions, it was concluded that the system did not fit into a dependent classification, since no solution in terms of a variable, like \( z \), could resolve all equations simultaneously.
- Dependent system - Equations are dependent if one can be derived from another. They have infinite solutions since all of them essentially describe the same line or plane.
- Independent system - Equations are independent if they intersect at exactly one point, meaning they have one unique solution.
In the original step by step solution, we explored various simplifications. However, due to inconsistencies and contradictions, it was concluded that the system did not fit into a dependent classification, since no solution in terms of a variable, like \( z \), could resolve all equations simultaneously.
Other exercises in this chapter
Problem 19
For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \end{array}\right]$$
View solution Problem 19
Perform each operation if possible. $$\left[\begin{array}{rr}-6 & 8 \\ 0 & 0\end{array}\right]-\left[\begin{array}{rr}0 & 0 \\ -4 & -2\end{array}\right]$$
View solution Problem 19
Solve each system by substitution. $$\begin{array}{c}2 x-7 y=8 \\\\-3 x+\frac{21}{2} y=5\end{array}$$
View solution Problem 20
Find each determinant. $$\operatorname{det}\left[\begin{array}{rrr}2 & 3 & 0 \\\1 & 9 & 0 \\\\-1 & -2 & 0\end{array}\right]$$
View solution