Problem 2 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{array}{rr}x+3 y-z= & -3 \\ 3 x-y+2 z= & 1 \\ 2 x-y+z= & -1\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (-1, 2, 3)\).

Step 1: Understand the system of equations

We are given a system of three linear equations in three variables:1. \(x + 3y - z = -3\)2. \(3x - y + 2z = 1\)3. \(2x - y + z = -1\)Our goal is to find the values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.

Step 2: Eliminate one variable from two equations

Choose any two equations and eliminate one variable. Let's eliminate \(z\). Subtract equation 1 from equation 2:\[\begin{align*}(3x - y + 2z) - (x + 3y - z) =& 1 - (-3)\ 2x - 4y + 3z =& 4\end{align*}\]Now subtract equation 1 from equation 3:\[\begin{align*}(2x - y + z) - (x + 3y - z) =& -1 - (-3)\x - 4y + 2z =& 2\end{align*}\]

Step 3: Solve the two-variable system

Now we have two equations in \(x\) and \(y\):1. \(2x - 4y + 3z = 4\)2. \(x - 4y + 2z = 2\)Subtract the second new equation from the first:\[\begin{align*}(2x - 4y + 3z) - (x - 4y + 2z) =& 4 - 2\x + z =& 2\end{align*}\]

Step 4: Solve for another variable

From the equation \(x + z = 2\), solve for \(z\):\[z = 2 - x\]

Step 5: Substitute back to find solutions

Substitute \(z = 2 - x\) into one of the original equations. Let's use equation 3:\[2x - y + (2 - x) = -1\]Simplifying gives:\[x - y + 2 = -1\]\[x - y = -3\]

Step 6: Solve for remaining variables

Now we have two equations:1. \(x - y = -3\)2. \(z = 2 - x\)Substitute \(x = y - 3\) into the equation for \(z\):\[z = 2 - (y - 3)\]\[z = 5 - y\]Substitute \(y = 2\) into the equation \(x - y = -3\) gives:\[x = -1\]Using \(x\) in equation for \(z\),\[z = 5 - 2 = 3\].

Step 7: Verify the solution

Substitute \(x = -1\), \(y = 2\), \(z = 3\) back into the original equations to ensure they are satisfied. All three check out:1. \(-1 + 3(2) - 3 = -3 \) 2. \(3(-1) - 2 + 2(3) = 1 \)3. \(2(-1) - 2 + 3 = -1 \)Thus, the solution \(x = -1\), \(y = 2\), \(z = 3\) is correct.

Key Concepts

Linear equations

Linear equations are the simplest form of algebraic equations made up of variables and constants. They represent straight lines when graphed on a coordinate plane, hence the name 'linear'. These equations can take various forms, such as:

• Standard Form: \[ ax + by + cz = d \] (as seen in the system of equations above)
• Slope-Intercept Form: \[ y = mx + b \]

In a system of linear equations, we aim to find values for the variables that satisfy all the equations simultaneously. Each equation in the set represents a constraint on the possible values of the variables. Solving such systems is pivotal in understanding intersection points between lines, finding the values that work under different conditions.
In the given exercise, each equation represents a three-dimensional plane. The solution to the system where all three equations meet gives the point that satisfies all three conditions. This point represents the values of the variables \((x, y, z)\). This exercise is a classic case of working with three linear equations involving three variables.

Elimination method

The elimination method is a systematic approach to solve systems of linear equations. It involves strategically adding or subtracting equations to eliminate one or more variables, simplifying the equation set. By canceling out a variable, we're left with fewer variables in the resulting equations, bringing us a step closer to a solution.
In exercises like ours, the elimination method often starts by labeling equations for clarity. Then, one variable is targeted for elimination by creating new equations through operations between the original equations. In this set, the system deliberately set out to eliminate variable \(z\) by subtracting equation 1 from equations 2 and 3.

• Consider \(3x - y + 2z = 1\) minus \(x + 3y - z = -3\), leading to the first new equation.
• Similarly, \(2x - y + z = -1\) minus \(x + 3y - z = -3\) serviced the elimination to deliver another equation.
These subtractions effectively reduce the system from three equations with three variables to two equations with two variables. Once one variable is eliminated, the new system is tackled by moving to the next method, often variable substitution.

Variable substitution

Variable substitution is a technique involving solving one equation for a particular variable, then substituting its value into another equation. This process simplifies equations by expressing a variable in terms of others, effectively reducing the complexity of the system.
In our exercise, after reducing the system to fewer variables using elimination, the next step was to find an expression for \(z\) in terms of \(x\). Given the equation \(x + z = 2\), you solve for \(z\) to get \(z = 2 - x\).
By substituting \(z = 2 - x\) back into one of the original equations, we continued simplifying and were eventually able to substitute across the chain of equations:

• For example, into \(2x - y + (2 - x) = -1\), further leading to \(x - y = -3\).

Through substitution, systems become much simpler, making it easier to isolate each variable. This controlled substitution progressively unveils solutions for each variable, one at a time, leading us towards solving the whole system.

Solving equations

Solving equations is the ultimate goal when working with systems like our example. Once we've used elimination and substitution to simplify the system sufficiently, you reach the stage where you solve directly for variables, aligning with the given constraints.
By the last stages of the step-by-step solution, our task is to explicably solve linear equations like \(x - y = -3\) and \(z = 5 - y\). Here, logical deduction follows:

• Find \(y\) from \(x - y = -3\).By plugging in such expressions back into another simplified formula.
• Derive \(x\) and solve for it yielding \(-1\), then finally apply these derived values in the expression for \(z\) to get \(z = 3\).

By substituting and consistently checking each step's solution with all original equations, we verify our result. Like in our resolution, checking ensures correctness: all original system equations are satisfied by the set values \(x = -1\), \(y = 2\), \(z = 3\). This practical method ensures that once we reach a solution, it fits all boundaries set by the initial system, solving the exercise conclusively.