Problem 32 - Algebra and Trigonometry with Analytic Geometry

Question

$$ \left\\{\begin{array}{r} x+2 z=1 \\ 2 y-z=4 \\ x y z=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((0, \frac{9}{4}, \frac{1}{2})\), \((9, 0, -4)\), and \((1, 2, 0)\).

Step 1: Analyze the System of Equations

We are given a system of three equations: 1. \( x + 2z = 1 \) 2. \( 2y - z = 4 \) 3. \( xyz = 0 \) The first two are linear equations, while the third implies that at least one of \(x\), \(y\), or \(z\) must be zero. Let's explore each case to determine the point where the entire system is satisfied.

Step 2: Assume x = 0

First, assume \( x = 0 \). Substitute \( x = 0 \) into the first equation: \[ 0 + 2z = 1 \] This gives \( z = \frac{1}{2} \).Substituting \( z = \frac{1}{2} \) into the second equation: \[ 2y - \frac{1}{2} = 4 \] Solving this equation for \( y \) gives:\[ 2y = 4 + \frac{1}{2} = \frac{9}{2} \]\[ y = \frac{9}{4} \]Check the third equation with \( xyz = 0 \), which is satisfied since \( x = 0 \). Thus, one solution is \((0, \frac{9}{4}, \frac{1}{2})\).

Step 3: Assume y = 0

Next, assume \( y = 0 \). Substitute \( y = 0 \) into the second equation:\[ 2(0) - z = 4 \] This simplifies to \( -z = 4 \) or \( z = -4 \). Now substitute \( z = -4 \) into the first equation:\[ x + 2(-4) = 1 \] \[ x - 8 = 1 \] \[ x = 9 \]Check the third equation, \( xyz = 0 \), which is satisfied since \( y = 0 \). Thus, another solution is \((9, 0, -4)\).

Step 4: Assume z = 0

Finally, assume \( z = 0 \). Substitute \( z = 0 \) into the first equation:\[ x + 2(0) = 1 \] This results in \( x = 1 \).Substitute \( z = 0 \) into the second equation: \[ 2y - 0 = 4 \] This gives \( y = 2 \).Check the third equation, \( xyz = 0 \), which is satisfied because \( z = 0 \). Thus, another solution is \((1, 2, 0)\).

Step 5: Verify Solutions

The solutions derived meet the condition set by the third equation. Each case satisfies at least one variable as zero:- \((0, \frac{9}{4}, \frac{1}{2})\)- \((9, 0, -4)\)- \((1, 2, 0)\)Each solution passes all given equations in the system.

Key Concepts

Linear Equations

Linear equations are a fundamental concept in algebra and appear frequently in various mathematical problems. A linear equation in its simplest form looks like this: \( ax + by = c \). It is called "linear" because it represents a straight line when graphed on a coordinate plane. Each equation in a system can have one or more variables.

In the exercise given, the system includes two linear equations:

• \( x + 2z = 1 \)
• \( 2y - z = 4 \)

These equations are linear because each variable (\(x, y, z\)) is raised only to the first power.

Linear equations are solved by finding the values of variables that satisfy all given equations simultaneously. In systems of equations, this involves using techniques such as substitution, elimination, or graphing, which are explored through the steps given in the original solution.

Algebraic Solutions

An algebraic solution involves finding values for the variables through algebraic manipulation rather than direct computation or estimation. This process hinges on understanding the relationships between equations and variables.

In the lingo of systems of equations, finding an algebraic solution means solving for all variables such as \(x\), \(y\), and \(z\) in such a way that every equation in the system holds true. This can often be done by assuming values for one variable, as shown in the original solution for the given exercise.

For instance, steps like assuming \(x = 0\), allow us to simplify and solve for the other variables. These simple yet effective strategies reveal where paths between equations intersect, thereby leading to the solutions for the system.

Variables Manipulation

Variables manipulation is a key process in solving systems of equations efficiently. It involves rearranging equations and expressions to isolate variables and solve for unknowns.

Consider the process of assuming a variable like \(x = 0\) from the system, which simplifies the equations and reveals useful information about \(y\) and \(z\). This is an example of using substitution to reduce the complexity of the system and hone in on solutions.

Also, knowing that \(xyz = 0\) simplifies the system by setting at least one of the variables to zero, aiding in faster resolution. Manipulations such as these are foundational to mastering algebra and offer systematic strategies for approaching complex equations. The ability to rearrange and modify equations is crucial for problem-solving in mathematics.