Problem 13 - A Graphical Approach to College Algebra

Question

$$\begin{array}{r} x+4 y-z=6 \\ 2 x-y+z=3 \\ 3 x+2 y+3 z=16 \end{array}$$

Step-by-Step Solution

Answer

Solve by eliminating variables successively to reduce complexity, manipulate equations to express each variable in terms of others, and finally obtain concrete numerical solutions for all variables.

Step 1: Write the System of Equations

First, write down the system of equations: 1. \( x + 4y - z = 6 \)2. \( 2x - y + z = 3 \)3. \( 3x + 2y + 3z = 16 \).

Step 2: Eliminate One Variable

We choose to eliminate \(z\). From equations 1 and 2, solve for \(z\): - From equation 1: \( z = x + 4y - 6 \).- From equation 2: \( z = 2x - y - 3 \).Set these equal to eliminate \(z\): \( x + 4y - 6 = 2x - y - 3 \).

Step 3: Simplify and Solve for One Variable

Rearrange the equation from Step 2 to solve for \(x\): \( x + 4y - 6 = 2x - y - 3 \) This becomes:\[ 5y - x = 3 \] Reorganize to find:\( x = 5y - 3 \).

Step 4: Substitute Back into Another Equation

Substitute \(x = 5y - 3\) back into equation 3:\( 3(5y - 3) + 2y + 3z = 16 \).Simplify:\( 15y - 9 + 2y + 3z = 16 \)which reduces to:\( 17y + 3z = 25 \).

Step 5: Solve for Remaining Variables

From the simplified equation: \( 17y + 3z = 25 \), solve for \(z\):\( 3z = 25 - 17y \)\( z = \frac{25 - 17y}{3} \).Now, substitute back into \(x = 5y - 3\) and use \(z\) to solve for both \(x\) and \(y\).

Step 6: Final Substitution

Plug the expression for \(z\) back into one of the original equations to find \(y\), then use it to find \(x\) and \(z\). For simplification, continue solving until specific values for all variables are derived from the relationships found.

Key Concepts

System of Equations

A system of equations involves two or more equations with multiple variables. Solving such systems requires finding values for each variable that satisfy all equations simultaneously. In this case, we have three equations, each involving the variables \(x\), \(y\), and \(z\). These types of problems model real-world scenarios where multiple conditions or constraints need to be addressed at the same time.
To approach a system effectively, we must methodically manipulate the equations. This often involves employing strategies like elimination or substitution to reduce complexity and solve for unknowns.

Variable Elimination

Variable elimination is a crucial technique used to simplify systems of equations. The goal is to remove one variable to make the system easier to solve.
In our problem, we chose to eliminate the variable \(z\). This is done by expressing \(z\) from two equations and setting them equal, which allows \(z\) to disappear, leaving an equation with fewer variables.

• From equation (1): \(z = x + 4y - 6\)
• From equation (2): \(z = 2x - y - 3\)

Setting them equal gives us \(x + 4y - 6 = 2x - y - 3\), which helps eliminate \(z\). This technique simplifies the equation, allowing for subsequent steps to focus on solving for the remaining variables.

Equation Substitution

Equation substitution involves replacing a variable with an expression obtained from another equation, which is useful for solving systems of equations.
At this stage, we solve for one variable and substitute it back into another equation to continue simplifying the system. From \(x + 4y - 6 = 2x - y - 3\), we solve for \(x\) which simplifies to \(x = 5y - 3\).
Substituting \(x = 5y - 3\) back into equation (3) helps us progress further:
\(3(5y - 3) + 2y + 3z = 16\), leading to the simplified equation \(17y + 3z = 25\). By substitution, we modularize the problem, allowing us to handle one part of the system at a time.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to solve for unknowns. This process requires careful handling of terms to effectively isolate variables.
After reducing our equations, we simplified \(17y + 3z = 25\) to solve for \(z\):

• \(3z = 25 - 17y\)
• \(z = \frac{25 - 17y}{3}\)

With \(z\) expressed in terms of \(y\), we continue manipulating equations by substituting into expressions for \(x\) and \(y\), until we find concrete values for all variables. Such manipulation is necessary in algebra to transition from complex systems to clear, solvable terms.