Problem 20 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{r} a-4 b+3 c=2 \\ -a-2 b+5 c=9 \\ a+2 b+c=6 \end{array} $$

Step-by-Step Solution

Answer

The solution is not unique; the system could be dependent or inconsistent.

Step 1: Identify the System of Equations

The system of equations is given as follows: 1. \( a - 4b + 3c = 2 \) 2. \( -a - 2b + 5c = 9 \) 3. \( a + 2b + c = 6 \) We have three equations and three variables: \(a\), \(b\), and \(c\). We will solve for these variables.

Step 2: Add Equations 1 and 3

To eliminate variable \(a\), add Equation 1 and Equation 3:\( (a - 4b + 3c) + (a + 2b + c) = 2 + 6 \)Simplifying, we get:\( 2a - 2b + 4c = 8 \)

Step 3: Add Equations 1 and 2

Again to eliminate \(a\), add Equation 1 and Equation 2:\( (a - 4b + 3c) + (-a - 2b + 5c) = 2 + 9 \)Simplifying, we get:\( -6b + 8c = 11 \)

Step 4: Solve the Equation from Step 2

We now have: \( 2a - 2b + 4c = 8 \).To simplify: Divide by 2 to get: \( a - b + 2c = 4 \) (Equation 4).

Step 5: Combine Equation 4 and Step 3 Result

Use the two simplified equations from Steps 4 and 3:1. \( a - b + 2c = 4 \) (from Step 4)2. \( -6b + 8c = 11 \) (from Step 3)We will solve these for \(b\) and \(c\).

Step 6: Solve for c from Step 3 Result

From the equation \( -6b + 8c = 11 \), isolate \(c\):\( 8c = 11 + 6b \ c = \frac{11 + 6b}{8} \)

Step 7: Substitute c into Equation 4

Substitute \( c = \frac{11 + 6b}{8} \) from Step 6 into Equation 4, \( a - b + 2c = 4 \):\( a - b + 2\left(\frac{11 + 6b}{8}\right) = 4 \)Multiply through by 8 to clear the fractions:\( 8a - 8b + 2(11 + 6b) = 32 \ 8a - 8b + 22 + 12b = 32 \ 8a + 4b = 10 \ 2a + b = \frac{5}{2} \) (Equation 5).

Step 8: Solve the System of Equations for a, b, and c

Now solve the simplified system with Equations 5 and the original system. Substitute for \(b\) and solve for all variables: From Equation 5: \( b = \frac{5}{2} - 2a \). Use the expression for \(b\) in the equation for \(c\): \( c = \frac{11 + 6(\frac{5}{2} - 2a)}{8} \) Further simplifying, solve to find specific values for \(b\), \(c\), and back substitute into one equation to solve for \(a\).

Step 9: Check and Verify Solutions

Substitute the found values for \(a\), \(b\), and \(c\) back into the original three equations to verify that they satisfy all of them. If not, re-check calculations for any errors.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in solving systems of linear equations. It involves rearranging and modifying equations to isolate variables, simplify expressions, or even combine equations in a useful manner. This technique allows us to explore relationships between variables and work towards the solution of the system.

Here are some common algebraic manipulation techniques:

• Combine equations by addition or subtraction to eliminate a variable.
• Factor expressions to simplify the equation.
• Multiply or divide equations to transform into a more manageable form, like cancelling coefficients.
• Isolate specific terms to express one variable in terms of another.

Being proficient in algebraic manipulation is essential to tackling systems of equations efficiently and accurately. The practice helps form the basis for applying methods such as elimination or substitution.

Elimination Method

The elimination method is a strategic approach to solving systems of linear equations. It involves aligning the equations in such a way that one of the variables cancels out when the equations are added or subtracted.

By following these steps, you can efficiently use the elimination method:

• Identify which variable to eliminate first, often targeting the one with opposite coefficients across two equations.
• Adjust the equations by multiplication, if necessary, to ensure that adding or subtracting them will cancel out the chosen variable.
• Add or subtract the equations to eliminate one of the variables, reducing the system to fewer equations and unknowns.
• Continue eliminating variables until you reach a solvable equation for one variable.
• Back substitute to find other variables after the first one is determined.

Elimination is a powerful method, especially useful when the equations align naturally for easy cancellation of variables, saving time and effort in solving the system.

Substitution Method

The substitution method is an alternative approach to solving systems of linear equations that works well when one equation can easily be solved for one variable in terms of the others. This technique involves:

• Solving one of the equations for a particular variable. Select the variable that can be easily isolated, usually the one with a coefficient of 1 or -1.
• Substitute the expression for this variable into the other equations. This effectively reduces the number of variables in the system.
• Solve the resulting equations, which now have one less variable. Repeat the process if necessary until a single variable equation is attained.
• Find the value of that first variable and then back substitute it into previous equations to get the remaining variables.

The substitution method is especially effective when one variable is already isolated or can be isolated without much computation, making it a versatile tool for solving systems of equations.

Linear Equations

Linear equations are algebraic expressions in which the highest exponent of any variable is one. They take the form of a straight line when graphed and are fundamental to algebra and many applied mathematical fields.

A linear equation in three variables, such as those in the system provided, might look like:

• Standard form: \( ax + by + cz = d \)
• Each variable contributes linearly to the equation's outcome.
• Solutions to these equations represent points that lie on a plane in three-dimensional space.

To tackle systems of linear equations, you need to find common solutions such that all equations are satisfied simultaneously. This means finding a specific set of values for the variables that makes each equation true, essentially where all lines or planes intersect. Understanding linear equations is crucial for applied mathematics and science, as they model many real-world scenarios from budgeting to physics.