Problem 64 - A Graphical Approach to College Algebra

Question

Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} 3 x+6 y-3 z &=12 \\ -x-2 y+z &=16 \\ x+y-2 z &=20 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions in terms of \(z\) are: \(x = 36 + 3z\), \(y = -16 - z\), \(z = z\).

Step 1: Write the System of Equations

We start by clearly identifying the given system of equations: 1. \(3x + 6y - 3z = 12\) 2. \(-x - 2y + z = 16\) 3. \(x + y - 2z = 20\)

Step 2: Simplify the Equations

Notice that the first equation \(3x + 6y - 3z = 12\) can be simplified by dividing every term by 3, resulting in: \(x + 2y - z = 4\).

Step 3: Express System in Terms of x and y

With the simplified first equation, we now have: 1. \(x + 2y - z = 4\) 2. \(-x - 2y + z = 16\) 3. \(x + y - 2z = 20\). Add equations 1 and 2 to eliminate \(z\): \((x + 2y - z) + (-x - 2y + z) = 4+16\), which simplifies to \(0 = 20\). This equation is true regardless of \(x\), \(y\), and \(z\), indicating dependence and an infinite number of solutions.

Step 4: Solve for x and y in Terms of z

Substitute \(z = z\) into the simplified first equation: \(x + 2y - z = 4\) rewritten as \(x + 2y = z + 4\). Then rearrange equation 3 in terms of \(x\): \(x = 20 - y + 2z\).

Step 5: Substitute and Solve for x, y in Terms of z

Using the rearranged equation 3, substitute \(x = 20 - y + 2z\) into \(x + 2y = z + 4\): \((20 - y + 2z) + 2y = z + 4\), which yields \(20 + y + 2z = z + 4\). Solve for \(y\): \(y = -16 - z\).

Step 6: Express Final Solutions in Terms of z

Now substitute the expression for \(y\) back into \(x = 20 - y + 2z\): \(x = 20 + 16 + z + 2z\), simplifying this gives \(x = 36 + 3z\). Thus, the solutions are \(x = 36 + 3z\), \(y = -16 - z\), and \(z = z\).

Key Concepts

Linear Equations

Linear equations form the foundation of solving more complex systems. In simple terms, a linear equation is an equation involving two variables that creates a straight line when graphed. Consider the equations given: each represents a line in a three-dimensional space, where variables such as \(x\), \(y\), and \(z\) depict the directions along each axis.

• Example: The equation \(3x + 6y - 3z = 12\) is linear because it represents a plane in a three-dimensional space.

Linear equations maintain the principle of balance. That means if you add, subtract, multiply, or divide the same number from both sides of the equation, the equality remains true.
The trick when dealing with systems of linear equations is to use combinations of the equations to find solutions for the variables involved. In our exercise, we combined equations to simplify and see their relationships more clearly.

Algebraic Manipulation

Algebraic manipulation is crucial when working with systems of equations. It allows us to rearrange and simplify equations, which is essential to find solutions for unknown variables. Let's shed light on a few key steps in the algebraic process from our example:

• Simplification: The first equation \(3x + 6y - 3z = 12\) was divided by 3, simplifying it to \(x + 2y - z = 4\). This step makes further calculations easier and more manageable.
• Substitution: By expressing one variable in terms of another, for example, \(z = z\), you obtain expressions for \(x\) and \(y\) relative to \(z\).
• Rearrangement: Rearrange the equations to isolate variables, helping to uncover relationships between them. Example: Rearranging \(x + 2y = z + 4\) aids in finding \(y\).
  

Remember, effective algebraic manipulation requires practice and creativity in arranging equations to simplify and solve them.

Infinite Solutions

Encountering a system of equations with infinite solutions can seem confusing at first. It means that there are endless combinations of values for the variables that satisfy all equations simultaneously. A consistent and dependent system like this one indicates equations that are not independent and hence overlap in some way in their graphical representation.

• In the exercise, when we added equations 1 and 2, we obtained \(0 = 20\)—a true statement. This reflects that the equations are dependent and infinitely many solutions exist as there is no contradiction.
• The solution involves writing remaining variables, such as \(x\) and \(y\), in terms of a variable such as \(z\). This creates forms like \(x = 36 + 3z\) and \(y = -16 - z\), where \(z\) remains arbitrary.

Understanding systems yielding infinite solutions highlights the interconnectedness within the system of equations, promoting a deeper comprehension of linear algebra concepts.