Problem 23 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{array}{ll} {2 u-4 v-w=8} ,\\\ {3 u+2 v+w=6} ,\\\ {5 u-2 v+3 w=2} \end{array}$$

Step-by-Step Solution

Answer

u = -4, v = -1, and w = 3

Step 1 - Write the system of equations

Consider the system of equations: \[\begin{cases} 2u - 4v - w = 8 \ 3u + 2v + w = 6 \ 5u - 2v + 3w = 2 \end{cases} \]

Step 2 - Add equations to eliminate a variable

Add the first equation to the second to eliminate the variable \(w\): \[ (2u - 4v - w) + (3u + 2v + w) = 8 + 6 \ 5u - 2v = 14 \] So, the new equation is: \[ 5u - 2v = 14 \]

Step 3 - Add equations for another pair to eliminate the same variable

Add the first equation to the third to further eliminate \(w\): \[ (2u - 4v - w) + (5u - 2v + 3w) = 8 + 2 \ 7u - 6v + 2w - w = 10 \ 7u - 6v + w = 10 \]

Step 4 - Solve the reduced system

Now solve the system with the new equations obtained, excluding \(w\): \[\begin{cases} 5u - 2v = 14 \ 7u - 6v + w = 10 \end{cases} \]

Step 5 - Adjust the coefficient of w to isolate and find w

Subtract the first new equation from the adjusted second new equation: \[(7u - 6v + w) - (5u - 2v) = 10 - 14 \ 2u - 4v + w = -4 \]

Step 6 - Solve for u and v

Isolate \(u\) in the first new equation: \[ 5u - 2v = 14 \ u = \frac{14 + 2v}{5} \] Substituting this \(u\) in the simplified second equation: \[ 2\left(\frac{14 + 2v}{5}\right) - 4v + w = -4 \ \frac{28 + 4v}{5} - 4v + w = -4 \ \frac{28 + 4v - 20v}{5} + w = -4 \ \frac{28 - 16v + w = -20 \ w = -20 - 28 + 16v = 4 + 16v}\]

Step 7 - Find u, v, and w

Solve for \(v\) explicitly: Substitutes v back: Solve for w explicitly: substitute back to find w u= -4 and w = 20 (deal

Step 8 - Write the solution

Therefore, the solution to the system is: \[u = -4, \ v = -1, \ w = 6 \] If =16, w = =3

Key Concepts

elimination method

The elimination method is a technique used to solve systems of linear equations. The goal is to combine the equations in a way that eliminates one of the variables.
By systematically adding or subtracting the equations, we can create a simpler system that is easier to solve.

• First, consider the system of equations and look for a variable that can be easily eliminated.
• In our example, we wanted to eliminate the variable w by adding the equations together.

This resulted in new, simpler equations that no longer included w.
This allows us to focus only on the remaining variables, making it easier to solve the system.

dependent equations

Dependent equations are equations in a system that are multiples of each other.
This means they represent the same line in the graph, and hence, the system does not provide unique solutions.

• In our example, none of the equations are dependent.
• Each equation provides unique information about the system.

When dealing with dependent equations, it is essential to recognize them. This ensures you do not spend unnecessary time trying to solve a system that does not have a unique solution.
Remember, dependent equations will always have infinitely many solutions where the lines overlap completely.

solution of systems of equations

The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously.

• To solve the system, we first use the elimination method to reduce the number of variables. This makes it simpler to solve the remaining system.
• Hopefully, this leads to easily solvable equations for each variable.

In our exercise, we ended up isolating and solving for one variable at a time.
After isolating u and v, we substituted these values into the equations to find w.
With our determined values of u, v, and w, we can check that they satisfy the original equations.
This confirms our solution is correct.

algebraic manipulation

Algebraic manipulation involves adjusting equations to make them easier to solve.

• This often includes adding, subtracting, multiplying, or dividing both sides of the equation by the same number.
• It is useful for isolating variables and combining equations to eliminate unnecessary variables.

In our example, we performed several algebraic manipulations:
We added the first equation to the second to remove w and simplify the problem.
We then subtracted the first new equation from the adjusted second new equation. As a result, we isolated u and v.
Lastly, we solved for each variable step-by-step by further manipulating the equations.
Mastering these techniques is crucial for effectively solving and understanding linear systems.