Problem 12 - Algebra A Combined Function

Question

Solve each system. $$ \left\\{\begin{aligned} x-5 y &=0 \\ x &-z=0 \\ -x &+5 z=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Equations

We have three equations in the system: 1) \(x - 5y = 0\), 2) \(x - z = 0\), and 3) \(-x + 5z = 0\).

Step 2: Solve for One Variable

From the second equation \(x - z = 0\), we can express \(x\) in terms of \(z\): \(x = z\).

Step 3: Substitute to Find Another Variable

Use the expression for \(x\) from Step 2 in the third equation \(-x + 5z = 0\): Substituting \(x = z\) gives \(-z + 5z = 0\). This simplifies to \(4z = 0\), therefore \(z = 0\).

Step 4: Substitute Back to Find Remaining Variables

With \(z = 0\), substitute back to find \(x\) using \(x = z\). Thus, \(x = 0\). Now use the first equation \(x - 5y = 0\) with \(x = 0\), giving \(-5y = 0\), which implies \(y = 0\).

Step 5: Conclude the Solutions

The solution to the system of equations is \(x = 0\), \(y = 0\), and \(z = 0\).

Key Concepts

Linear Equations

Linear equations are fundamental expressions used to describe relationships between variables. In our given system, we observe three linear equations. These equations are expressed in terms of unknowns, which need to be solved to find a specific set of values for each variable. Each equation is formulated in the first degree, meaning no exponents or powers higher than one appear for any variable. This property makes linear equations straightforward yet powerful tools for modeling various scenarios in real life.
To better understand, consider the first equation \(x - 5y = 0\). Here, the term \(x\) and the term \(-5y\) are both linear, as they simply involve the variables \(x\) and \(y\). The operation here is straightforward: the sum or difference of each variable's product with its coefficient equals zero. This characteristic allows for easy manipulation in solving systems of equations.

Substitution Method

The substitution method is a crucial technique for solving systems of equations. It involves solving one equation for a variable and then substituting this expression into the other equations. This approach is particularly useful when one equation is easily transformed to express one variable in terms of others.
In the given system, we started using the substitution method from the second equation, \(x - z = 0\). Solving it for \(x\) was straightforward: \(x = z\). This pivotal step simplified our problem significantly, as it allowed us to use the derived expression for \(x\) in other equations to find additional variables with ease.

• Step 1: Simplify one equation to express one variable in terms of another.
• Step 2: Substitute the expression into the remaining equations.
  
• Step 3: Solve these simplified equations to find the values of all unknowns.

Solution of Variables

The ultimate goal in solving systems of linear equations is to determine the values of the unknown variables. In our example, we were tasked with finding the values of \(x\), \(y\), and \(z\).
Using the substitution method, we first identified that \(z = 0\). With \(z\) determined, we easily found \(x = 0\) because \(x = z\). Finally, substituting \(x = 0\) back into the first equation \(x - 5y = 0\), we concluded \(y = 0\) as well. Thus, all variables are solved systematically, giving us the unique solution set:

• \(x = 0\)
• \(y = 0\)
• \(z = 0\)

Step-by-Step Problem Solving

A methodical, step-by-step approach is essential for solving complex systems of equations accurately. Let's recap the process applied in our case. It shows how logical sequencing and using each piece of information leads us to the complete solution.
First, identify all the equations provided in the system and check for one variable that can be easily expressed in terms of another. Next, use this expression to simplify the entire set of equations by substituting into the others. This simplification leads to discovering one unknown at a time.
Once one variable is found, continue by substituting back into former equations to uncover other values. These organized steps allow for managing even the most involved problems with clarity and confidence.
Remember, each problem is different, so while approaches like substitution are powerful, they need to be adapted to suit the structure of each unique exercise.