Problem 43 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} b+2 c=7-a \\ a+c=2(4-b) \\ 2 a+b+c=9 \end{array}\right. $$

Step-by-Step Solution

Answer

Solution: \( a = 2, b = 3, c = 0 \). System is consistent and independent.

Step 1: Simplify the First Equation

Rearrange the first equation to isolate one of the variables. The given equation is: \[ b + 2c = 7 - a \] Isolate \( b \): \[ b = 7 - a - 2c \]

Step 2: Simplify the Second Equation

Rearrange the second equation to express one variable in terms of others. The given equation is:\[ a + c = 2(4 - b) \]This expands to: \[ a + c = 8 - 2b \] Rearrange to isolate \( a \):\[ a = 8 - 2b - c \]

Step 3: Substitute in the Third Equation

Now use your expressions for \( a \) and \( b \) and substitute them into the third equation. The third equation is:\[ 2a + b + c = 9 \]First, substitute \( b = 7 - a - 2c \) into this equation:\[ 2a + (7 - a - 2c) + c = 9 \]This simplifies to:\[ a + c + 7 = 9 \]So:\[ a + c = 2 \]

Step 4: Consistency Check

Notice that equation from step 3 is \( a + c = 2 \), which matches the equation obtained in Step 2: \[ a + c = 8 - 2b \]. Since the system remains consistent, we use these equations to solve for the specific values of variables.

Step 5: Solve for Variables

Use \( a + c = 2 \) and \( a = 8 - 2b - c \). Substitute \( a = 2 - c \) into \( a = 8 - 2b - c \). Equate the expressions for \( a \):\[ 2 - c = 8 - 2b - c \]Solving for \( b \):\[ 2 = 8 - 2b \]\[ 2b = 8 - 2 \]\[ 2b = 6 \] \[ b = 3 \] Now substitute \( b = 3 \) back into the equation \( a + c = 2 \):From \( a = 8 - 2b - c \), using \( b = 3 \), we get:\[ a = 8 - 6 - c = 2 - c \]Since \( a + c = 2 \): Substitute \( a = 2 - c \):\[ 2 - c + c = 2 \]Which holds true. Hence, consistent solution.

Step 6: Final Solution

The solutions are \( a = 2 - c \), \( b = 3 \), and from equation \( a + c = 2 \), solve for \( c = 2 - a \). If using \( c = 0 \), then \( a = 2 \). With \( a = 2 \) and \( c = 0 \), validate back in third equation: \[ 2(2) + 3 + 0 = 9 \] So solutions are consistent and valid.

Key Concepts

Inconsistent Systems

An inconsistent system of linear equations is one that has no solutions. When we have two or more equations that contradict each other, we label the system as inconsistent. Such contradictions arise when simplifying or solving the equations results in a false statement like \( 0 = 5 \).
If you ever encounter such an outcome while solving the system, it means there is no point that satisfies all the equations simultaneously. In the exercise we discussed, we never ended up with a contradiction, which means our system is consistent. When working with systems, always double-check by comparing your equations at various steps to ensure no unexpected contradictions arise.

Dependent Equations

Dependent equations occur when one equation in the system is a multiple or a linear combination of another. Essentially, this means some equations do not offer new information because they are a transformation of others.
When equations are mathematically equivalent, they describe the same line or plane, resulting in infinitely many solutions along this geometric entity. However, if such an equation does not manifest in the problem, like in our case, then we have a consistent system with either a single unique solution or intersection when combined with independent equations.

• Realizing that some equations do not add new information is crucial when determining the number of solutions.
• Always simplify and compare equations to identify cases of dependency.

Substitution Method

The substitution method is a very practical approach to solve systems of equations, particularly when the equations are easy to manipulate algebraically. This method involves solving one of the equations for one variable and then replacing it in the other equations.
In our given problem, we started by expressing one variable in terms of others. For instance, isolating \( b \) from the first equation and then substituting it into another equation helps reduce complexity.

• The main benefits of substitution are simplicity and the ability to reduce the number of variables in an equation, making it easier to handle.
• By reducing variables step-by-step, you proceed until only a single equation with a single variable remains, which can then be solved directly.
• Finally reinserting this solution back into the previously rearranged expressions gives the remaining values for the variables.

Ensuring each step involves correct algebraic manipulation is key to maintaining the integrity of the original equations throughout this process.