Problem 16 - College Algebra

Question

For the following exercises, solve each system by substitution. $$ \begin{aligned} 4 x+6 y+9 z &=0 \\\\-5 x+2 y-6 z &=3 \\ 7 x-4 y+3 z &=-3 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = -1, y = \frac{1}{2}, z = -\frac{1}{3}\).

Step 1: Choose an Equation and Solve for a Variable

Let's take the second equation, \(-5x + 2y - 6z = 3\), and solve for \(y\). To isolate \(y\), begin by rearranging the equation: \(2y = 5x + 6z + 3\), then divide everything by 2 to get: \(y = \frac{5}{2}x + 3z + \frac{3}{2}\).

Step 2: Substitute into Another Equation

With \(y\) expressed in terms of \(x\) and \(z\), replace \(y\) in the third equation \(7x - 4y + 3z = -3\):\(7x - 4(\frac{5}{2}x + 3z + \frac{3}{2}) + 3z = -3\). Expand the terms: \(7x - 10x - 12z - 6 + 3z = -3\). Simplify the equation: \(-3x - 9z = 3\). Divide the entire equation by -3: \(x + 3z = -1\). So, express \(x\) as \(x = -3z - 1\).

Step 3: Substitute Back to Find Another Variable

Use the expression for \(x\) from Step 2 in the equation found in Step 1 for \(y\): \(y = \frac{5}{2}(-3z - 1) + 3z + \frac{3}{2}\). Simplify: \(y = \frac{-15z}{2} - \frac{5}{2} + 3z + \frac{3}{2}\). Combine like terms: \(y =\frac{-15z + 6z}{2} + \frac{-5 + 3}{2}\). \(y =\frac{-9z}{2} - 1\).

Step 4: Substitute into the Original to Check Solution

Now that we have expressions for \(x, y, z\), substitute \(x = -3z - 1\) and \(y = \frac{-9z}{2} - 1\) into the first equation, \(4x + 6y + 9z = 0\):\(4(-3z - 1) + 6(\frac{-9z}{2} - 1) + 9z = 0\).Simplify: \(-12z - 4 - 27z - 6 + 9z = 0\), combine like terms: \(-30z - 10 = 0\). This implies \(z = -\frac{1}{3}\).

Step 5: Find Values for All Variables

With \(z = -\frac{1}{3}\), substitute back to find \(x\) and \(y\). For \(x\):\(x = -3(-\frac{1}{3}) - 1 = 0 - 1 = -1\). For \(y\):\(y = \frac{-9(-\frac{1}{3})}{2} - 1 = \frac{3}{2} - 1 = \frac{1}{2}\). Thus the solution to the system is \(x = -1, y = \frac{1}{2}, z = -\frac{1}{3}\).

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. This method is particularly useful for solving linear and non-linear equations by reducing the number of variables one at a time. Here’s a simple breakdown of how it works:
1. Choose one of the equations in the system and solve for one variable in terms of the others. This means expressing one variable as an equation that includes the remaining variables.
2. Substitute this expression into the other equations in the system. This substitution helps eliminate the chosen variable from those equations, allowing you to solve for another variable.
3. After solving one variable, backtrack by substituting its value into previously derived expressions to find the other variables.
By repeatedly using these steps, you can solve a complex system with ease. The substitution method is powerful because it simplifies multi-variable equations into a more manageable form.

Algebraic Expressions

An algebraic expression consists of numbers, variables, and operations like addition, subtraction, multiplication, and division. In the context of the substitution method, algebraic expressions play a crucial role in isolating variables.
For example, in our problem, the expression for 'y' in terms of 'x' and 'z' is derived as:
\( y = \frac{5}{2}x + 3z + \frac{3}{2} \)
Here, this expression acts as a tool to replace 'y' in other equations. Understanding these expressions is vital. They allow one to transform and simplify complex systems into forms that are easier to manage. This creates a pathway to access the hidden solution within a system.

Variable Isolation

The core step in the substitution method revolves around isolating one variable from an equation. This process involves rearranging algebraic expressions to express one variable entirely in terms of others. Here, isolation of variables reveals relationships between them, which are then substituted into other equations.
To exemplify, look at how 'y' was isolated:
Starting with \(-5x + 2y - 6z = 3\), rearrange to express 'y':
\[ 2y = 5x + 6z + 3 \]
Dividing all terms by 2 gives:
\[ y = \frac{5}{2}x + 3z + \frac{3}{2} \]
This isolation step is critical as it simplifies the equation, pruning out one variable which makes solving the rest of the system straightforward.

Linear Equations

Linear equations form the backbone of many algebraic problems, especially in systems of equations. A linear equation is an equation of the first degree, meaning the variables are not raised to any power other than one. These equations can be written in the standard form: \( Ax + By + Cz = D \).
In solving systems using the substitution method, linear equations maintain consistency, allowing various operations like addition, subtraction, and substitution to be performed seamlessly. This property simplifies solving these systems.
When applying the substitution method on systems of linear equations, we recognize a distinct advantage. The linear nature means the algebraic expressions formed during substitution remain manageable and leads to elegant solutions for each variable in the system.