Problem 10 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{array}{r}2 x-y+z=0 \\ x-y-2 z=0 \\ 2 x-3 y-z=0\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = 0\), \(y = 0\), \(z = 0\).

Step 1: Identify the System of Equations

We have a system of three equations with three variables: \(x\), \(y\), and \(z\). The equations are:1. \(2x - y + z = 0\)2. \(x - y - 2z = 0\)3. \(2x - 3y - z = 0\).

Step 2: Express Variables in Terms of Each Other

From equation (2), solve for \(x\): \[x = y + 2z\]Substitute \(x = y + 2z\) in equations (1) and (3).

Step 3: Substitute into Equation (1)

Substitute \(x = y + 2z\) into equation (1): \[2(y + 2z) - y + z = 0\]Simplify:\[2y + 4z - y + z = 0\]\[y + 5z = 0\] (Equation 4)

Step 4: Substitute into Equation (3)

Substitute \(x = y + 2z\) into equation (3):\[2(y + 2z) - 3y - z = 0\]Simplify:\[2y + 4z - 3y - z = 0\]\[ -y + 3z = 0\] (Equation 5)

Step 5: Solve for Variables

From equation (4), \(y + 5z = 0\), solve for \(y\):\[y = -5z\]From equation (5), \(-y + 3z = 0\), replace \(y\) with \(-5z\):\[ -(-5z) + 3z = 0\]\[5z + 3z = 0\]Solve for \(z\):\[8z = 0\]\[z = 0\]Substitute \(z = 0\) back to find \(y\) and \(x\):\(y = -5(0) = 0\)\(x = y + 2z = 0 + 2(0) = 0\).

Step 6: Verify Solution

Verify by substituting \(x = 0\), \(y = 0\), \(z = 0\) into the original equations:1. \(2(0) - 0 + 0 = 0\) is correct.2. \(0 - 0 - 2(0) = 0\) is correct.3. \(2(0) - 3(0) - 0 = 0\) is correct.All equations are satisfied.

Key Concepts

Understanding Linear Equations

A linear equation is a mathematical sentence that describes a straight line when plotted on a graph. It's an equation where each term is a constant or the product of a constant and a single variable. Linear equations are of the form:

• \(ax + by + cz + \ldots = d\)

where \(a\), \(b\), \(c\), and \(d\) are constants and \(x\), \(y\), and \(z\) are variables.
In our problem, this type of linear equation helps to find the value of these variables.
Solving a system of linear equations like the ones given means finding a point where all the lines intersect each other if they have a solution.
Understanding the graphical interpretation could be beneficial, such as visualizing these as planes in three-dimensional space which intersect at a common point.

Substitution Method Demystified

The substitution method is a common technique used to solve systems of equations. This approach is straightforward, yet sometimes overlooked in its simplicity.
Let's break it down:

• Step 1: Solve one equation for one variable. In our problem, equation (2) is rearranged to solve for \(x\) in terms of the other variables: \(x = y + 2z\).
• Step 2: Substitute this expression into the other equations. Replace \(x = y + 2z\) in equations (1) and (3) to create new equations involving only \(y\) and \(z\).
• Step 3: Solve the resulting equations for the remaining variables. This will often mean additional substitutions until all variables are expressed in terms of simple constants.

Each substitution simplifies the system and reduces the number of variables, making the equations easier to solve.
This method is particularly effective when one of the equations offers a simple rearrangement to isolate a variable.

Decoding Variables in Equations

Variables are symbols used to represent numbers in mathematical expressions and equations. In algebra, they serve as a placeholder for unknown values we aim to calculate.
Our exercise uses three variables: \(x\), \(y\), and \(z\), which are defined by the given linear equations.
Understanding how to manipulate these variables is crucial:

• Identify: Recognize each variable and attempt to express it in terms of the others whenever possible.
• Substitution: Use methods like substitution to simplify the relationships between variables.
• Solving: Each substituted equation eventually becomes simpler, allowing for straightforward solutions for each variable.

In conclusion, the ultimate goal of working with variables in equations is to establish their numerical values, providing the solution to the system of equations.