Problem 11 - Algebra A Combined Function

Question

Solve each system. $$ \left\\{\begin{aligned} x &+5 z=0 \\ 5 x+y &=0 \\ y-3 z &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Express x in terms of z

We begin with the first equation: \(x + 5z = 0\). To express \(x\) in terms of \(z\), rearrange the equation to get \(x = -5z\).

Step 2: Substitute for x in the second equation

Substitute \(x = -5z\) into the second equation \(5x + y = 0\). This gives us: \(5(-5z) + y = 0\), which simplifies to \(-25z + y = 0\). Thus, we find that \(y = 25z\).

Step 3: Substitute for y in terms of z into the third equation

With \(y = 25z\), substitute into the third equation \(y - 3z = 0\). We get: \(25z - 3z = 0\), which simplifies to \(22z = 0\).

Step 4: Solve for z

From \(22z = 0\), solving for \(z\) gives us \(z = 0\).

Step 5: Solve for x and y using z = 0

With \(z = 0\), substitute back to find \(x\) and \(y\). Using \(x = -5z\), we have \(x = 0\). And using \(y = 25z\), we have \(y = 0\).

Key Concepts

Linear Equations

Linear equations are the building blocks of algebra and involve expressions set equal to one another that create straight lines when graphed. In the system given, each of the equations is linear because each can be expressed in the form of \(ax + by + cz = d\) where \(a\), \(b\), and \(c\) are constants, and \(x\), \(y\), and \(z\) are variables. These equations represent planes in three-dimensional space:

• In the first equation, \(x + 5z = 0\), the coefficients \(1\) and \(5\) represent the relationship between \(x\) and \(z\) when \(y\) is zero.
• The second equation \(5x + y = 0\) highlights how \(y\) depends on \(x\).
• The third equation, \(y - 3z = 0\), connects \(y\) and \(z\).

Linear equations like these form the cornerstone of solving systems because they can be manipulated algebraically to find specific values for variables. Understanding how these linear relationships intersect is key in solving the system.

Solving Systems

When we talk about solving systems of equations, we're talking about finding values for the variables that satisfy all the equations in the system simultaneously. In the given exercise, we focused on three linear equations with three variables. The approach is systematic:

• First, express one of the variables in terms of another. This simplifies and reduces the number of variables in the initial steps.
• Next, substitute this expression into another equation, creating a system with fewer variables. This substitution process reduces complexity.
• By repeatedly substituting and simplifying, the system eventually devolves into equations with a single variable, which are straightforward to solve.

In our case, expressing \(x\) and then \(y\) in terms of \(z\) allowed us to eventually find that \(z = 0\). With \(z = 0\), it became easy to substitute back and find \(x\) and \(y\). It's crucial to remember that each step must maintain the balance of the equations, allowing us to trace back and find values for each variable correctly.

Variables

Variables in mathematical equations are symbols representing unknown values that we aim to determine through solving equations. In this exercise, the variables \(x\), \(y\), and \(z\) represent unknown quantities whose values need to satisfy all given equations simultaneously. Here's how they work together:

• Each variable interacts with others through coefficients in the equations, representing how a change in one affects the others.
• In finding \(x = -5z\), \(x\) is directly dependent on the value of \(z\).
• The discovery that \(y = 25z\) further illustrates the dependence between \(y\) and \(z\).
• Finally, solving for one variable like \(z = 0\) leads us to calculate the exact values for \(x\) and \(y\) based on the interactions defined by the equations.

Understanding variables and their roles in these equations is essential because they are the placeholders that allow us to interpret, re-arrange, and solve mathematical relationships in systems of equations.