Problem 16 - Algebra 2

Question

Solve each system of equations. \(4 a-2 b+8 c=30\) \(a+2 b-7 c=-12\) \(2 a-b+4 c=15\)

Step-by-Step Solution

Answer

The system has infinitely many solutions with \(a = \frac{18}{5}\), \(b = \frac{-12}{5}\), and \(c=0\).

Step 1: Understand the Given System of Equations

We have the following system of equations:1. \(4a - 2b + 8c = 30\)2. \(a + 2b - 7c = -12\)3. \(2a - b + 4c = 15\).We need to find the values of \(a\), \(b\), and \(c\) that satisfy all three equations simultaneously.

Step 2: Eliminate Variable b from Equations 1 and 3

To eliminate \(b\) from the equations, we'll first consider equations 1 and 3:- Multiply equation 2 by 2: \(2(a + 2b - 7c = -12)\) which gives \(2a + 4b - 14c = -24\).- Now, subtract this from equation 1: \((4a - 2b + 8c) - (2a + 4b - 14c) = 30 - (-24)\).- This results in \(2a - 6b + 22c = 54\), or simplified, \(a - 3b + 11c = 27\).

Step 3: Eliminate Variable b Again Using Equations 2 and 3

Next, we use equations 2 and 3 to eliminate \(b\):- Multiply equation 3 (\(2a - b + 4c = 15\)) by 2: \(4a - 2b + 8c = 30\).- Subtract this from equation 2: \((a + 2b - 7c) - (2a - b + 4c) = -12 - 15\).- This simplifies to \(-a + 3b - 11c = -27\).- Notice that if we solve for \(b\), the equation is balanced such that \(a - 3b + 11c = 27\) holds.

Step 4: Solve for one variable

Now we have two simplified equations from the above steps, solve them:1. \(a - 3b + 11c = 27\)2. \(-a + 3b - 11c = -27\)Adding these equations:- \((a - 3b + 11c) + (-a + 3b - 11c) = 27 - 27\)- This results in \(0 = 0\), indicating a simplified system with dependent equations.- To proceed, we need to substitute one variable freely, e.g., let \(c = k\), solve for \(a\) and \(b\) in terms of \(k\).

Step 5: Solve for a and b using substitution

Substituting \(c = 0\) for simplicity:1. Plugging \(c = 0\) back into \(4a - 2b + 8(0) = 30\): \(4a - 2b = 30\).- Solve: \(2a - b = 15\), or \(b = 2a - 15\).2. Substituting into \(a + 2b - 7(0) = -12\) yields:- \(a + 2(2a - 15) = -12\).- Simplify: \(a + 4a - 30 = -12\) leads to \(5a = 18\).- Thus, \(a = \frac{18}{5}\).- Substituting back, \(b = 2\left(\frac{18}{5}\right) - 15 = \frac{-12}{5}\).

Key Concepts

Linear Equations

Linear equations are the building blocks of algebra. They can be simplified to form a straight line when graphed on a coordinate plane. A linear equation usually takes the form of \(ax + by + cz = d\) where \(a, b, c\) are coefficients for the variables \(x, y, z\) respectively, and \(d\) is the constant. In our exercise, we dealt with the linear equations:

• \(4a - 2b + 8c = 30\)
• \(a + 2b - 7c = -12\)
• \(2a - b + 4c = 15\)

Each equation represents a plane in three-dimensional space. The purpose is to find a common point that lies on all the planes. Thus, solving these equations means you are seeking values for \(a, b, c\) such that all the equations hold true simultaneously.

Variable Elimination

Variable elimination is a strategic method used to simplify a system of equations by removing one of the variables. By focusing on reducing the number of variables step by step, the complexity of solving the system decreases.
For instance, in the provided solution, we eliminated variable \(b\) by manipulating Equations 1 and 3. By multiplying and subtracting terms:

• Multiply equation 2 by 2 and subtract from equation 1, resulting in a new, simpler equation: \(a - 3b + 11c = 27\).
• Similarly, employ equations 2 and 3 to achieve another simplified expression: \(-a + 3b - 11c = -27\).

Elimination is powerful because it reduces a multi-variable problem to a simpler form and thus helps in systematically finding solutions.

Dependent Equations

Dependent equations occur when one equation can be derived from another, revealing that they are not independent. In this problem, simplifying the equations led to a final equality of \(0 = 0\), illustrating dependence.
This means the equations do not define distinct lines but rather overlap in some manner, suggesting infinite solutions for some variables. The presence of a dependent system requires assigning a free variable value, such as setting \(c = k\), to compute others.
Recognizing dependent equations is crucial in identifying relationships between variables, ensuring a correct path to solving the system.

Substitution Method

The substitution method is a straightforward approach to solving systems of equations, especially effective when a dependent relationship among the equations exists.
In substitution, you solve for one variable in terms of another and then replace it into the other equations. For instance, by letting \(c = 0\) in the provided system, we could reform one equation as \(b = 2a - 15\).
Replacing \(b\) in a simpler form helps to directly solve for other variables.

• After solving \(b\) in terms of \(a\), substitute back to find the independent variable values.
• This sequential substitution simplifies solving for multiple unknowns without extensive algebraic manipulation.

By simplifying the equations through substitution, it becomes easier to identify specific solutions quickly and logically.