Problem 10 - College Algebra and Calculus: An Applied Approach

Question

Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} 4 x-2 y+z &=8 \\\\-y+z &=4 \\ z &=2 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 3\), \(y = -2\), \(z = 2\).

Step 1: Solve for Z

The third equation of the system is \(z = 2\), which directly gives us the value of \(z\).

Step 2: Substitute Z into the second equation

Now, take the value of \(z\) and substitute it into the second equation: \(-y + z = 4\). This gives \(-y + 2 = 4\), which when simplified provides the value of \(y\).

Step 3: Substitute Y and Z into the first equation

Substitute the value of \(y\) and \(z\) into the first equation: \(4x - 2y + z = 8\). Solve the equation to find the value of \(x\).

Key Concepts

Back-Substitution

Back-substitution is like solving a puzzle by starting from the last piece and working your way back to the first one. This method is commonly used when solving a system of equations that has been arranged in a staircase or triangular form. In this exercise, back-substitution begins from the equation that has only the variable \(z\), making it easy to find its value right away.

**Steps in Back-Substitution:**

• Solve for the simplest variable: Begin with the equation that makes it easiest to find a variable. In the given system, it's the third equation, \(z = 2\), which directly provides the value of \(z\).
• Use the found values in other equations: Next, use the known value of \(z\) in the second equation to solve for \(y\). This results in \(-y + 2 = 4\), simplifying to find \(y\).
• Continue substituting upward: Finally, substitute the values of \(y\) and \(z\) into the first equation to solve for \(x\). This is how back-substitution works upward from the last equation to find all the variables.

Each step builds on the previous one, making it simple to solve the entire system! That’s the beauty of back-substitution.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system involves finding values for each variable that satisfy all the equations simultaneously.

In this exercise, we have a system of three linear equations:

• \(4x - 2y + z = 8\)
• \(-y + z = 4\)
• \(z = 2\)

A system of equations can tell us a lot depending on the situation:

• **Unique Solution:** One set of values for the variables that makes all equations true, as seen here using back-substitution.
• **No Solution:** The equations are contradictory and cannot be satisfied by any set of values.
• **Infinite Solutions:** There are endless solutions, often when the equations are multiples of each other.

Recognizing the type of system helps decide the appropriate method of solving it.

Solving Equations

Solving equations involves finding the value of unknown variables that make the equation true, a fundamental aspect of mathematics. In this problem, equations are solved sequentially using a special technique.

**Process of Solving Equations in the System:**

• Isolate variables: The method starts by isolating the easiest variable first, here \(z = 2\).
• Substitute back: Using the value of \(z\), substitute it into the equations above it to solve for other variables. This systematic approach reduces complexity progressively.
• Solve linearly: Once substitutions are made, it usually simplifies to basic arithmetic, like finding integer values that satisfy the equalities. For instance, finding \(y\) and \(x\) by substitution.

With this layered way, solving equations in systems becomes manageable and logical, breaking down complex problems into smaller, tractable parts.