Problem 24 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} &4 a+b+c=3,\\\ &2 a-b+c=6,\\\ &2 a+2 b-c=-9 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( a = -\frac{1}{2} \), \( b = -1 \), \( c = 6 \).

Step 1: Arrange the equations

First, let's rewrite the system of equations for clarity:\[ \begin{aligned} 4a + b + c &= 3, \ 2a - b + c &= 6, \ 2a + 2b - c &= -9 \end{aligned} \]

Step 2: Eliminate one variable

We will eliminate one variable (e.g., \(c\)) by adding and subtracting the equations.Add the first and second equations:\[ (4a + b + c) + (2a - b + c) = 3 + 6 \]This simplifies to:\[ 6a + 2c = 9 \tag{4} \]Next, add the second and third equations:\[ (2a - b + c) + (2a + 2b - c) = 6 + (-9) \]This simplifies to:\[ 4a + b = -3 \tag{5} \]

Step 3: Solve for one variable

Using the new equations from Step 2, we can express one variable in terms of another.From equation (5):\[ 4a + b = -3 \]Solve for \( b\):\[ b = -3 - 4a \tag{6} \]

Step 4: Substitute back

Substitute equation (6) into equation (4):\[ 6a + 2c = 9 \]We need another equation for substitution. Let's take equation (3) and use it directly with substitution:\[ 2a + 2b - c = -9 \]Substitute \( b = -3 - 4a\):\[ 2a + 2(-3 - 4a) - c = -9 \]Simplify:\[ 2a - 6 - 8a - c = -9 \]\[ -6a - 6 - c = -9 \]\[ -6a - c = -3 \tag{7} \]

Step 5: Solve for the second variable

Now solve equation (7) for \( c\):\[ -6a - c = -3 \]\[ c = -6a + 3 \tag{8} \]

Step 6: Substitute the value of c

Substitute \( c = -6a + 3 \) back into equation (1):\[ 4a + b + c = 3 \]Use \( b = -3 - 4a\) and \( c = -6a + 3\):\[ 4a + (-3 - 4a) + (-6a + 3) = 3 \]Simplify:\[ 4a - 3 - 4a - 6a + 3 = 3 \]\[ -6a = 3 \]\[ a = -\frac{1}{2} \]

Step 7: Find values of b and c

With \( a = -\frac{1}{2} \), find \( b \) and \( c \):\[ b = -3 - 4(-\frac{1}{2}) = -3 + 2 = -1 \]\[ c = -6(-\frac{1}{2}) + 3 = 3 + 3 = 6 \]

Key Concepts

Dependent Equations

Dependent equations are equations that describe the same line in a graph. These equations have infinitely many solutions because every point on the line is a solution. In our given system, the equations are different and describe distinct lines. Thus, they are **not dependent**. To check this, one can transform the equations to see if they can be derived from each other through multiplication or addition and subtraction.

Substitution Method

The substitution method is a common technique to solve systems of linear equations. Here is a step-by-step breakdown:

• Solve one of the equations for one variable in terms of the other variables.
• Substitute this expression into the other equations.
• This reduces the system to a simpler form, often a single equation with one variable.
• Solve the simplified equation.
• Substitute back to find the values of the remaining variables.

The method allows for a systematic reduction of variables and simplifies solving complex systems.

Variable Elimination

Variable elimination involves adding or subtracting equations to eliminate one of the variables. Here's how we applied it:

• Added the first and second equations: \(4a + b + c + 2a - b + c = 3 + 6\) simplifies to \(6a + 2c = 9\) (New Equation 4).
• Added the second and third equations: \(2a - b + c + 2a + 2b - c = 6 + (-9)\) simplifies to \(4a + b = -3\) (New Equation 5).

This reduced our initial three-variable system to simplified two-variable systems. By systematically eliminating variables, solving the system becomes more manageable.

Step-by-Step Solution

To solve systems of equations, a clear step-by-step approach ensures accuracy:

• **Step 1:** Rewrite the equations clearly for easy manipulation.
• **Step 2:** Eliminate one variable, typically using addition or subtraction of equations.
• **Step 3:** Solve for one variable using the simplified equations.
• **Step 4:** Substitute back to find the other variables.
• **Step 5:** Verify by substituting back into the original equations.

In our particular system, after eliminating \(c\), we simplified to two equations in \(a\) and \(b\). We then solved for \(b\) in terms of \(a\) and substituted back to eliminate remaining variables systematically. The final values for the variables were found: \(a = -1/2, b = -1, c = 6 \). This methodical progression ensures that nothing is overlooked, leading to the correct solution.