Problem 52 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{aligned} -4 x-y+36 z &=24 \\ x-2 y+9 z &=3 \\ -2 x+y+6 z &=6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = -5,\ y = -4,\ z = 0\).

Step 1: Write the System of Equations

Let's rewrite the given system of linear equations for clarity: \[ \begin{align*} -4x - y + 36z &= 24, \ x - 2y + 9z &= 3, \ -2x + y + 6z &= 6. \end{align*} \] Our goal is to solve this system to find the values of \(x\), \(y\), and \(z\).

Step 2: Use Equation (2) to Express \(x\)

From equation (2), we can express \(x\) in terms of \(y\) and \(z\): \(x = 2y - 9z + 3\).

Step 3: Substitute \(x\) in Equations (1) and (3)

Substitute \(x = 2y - 9z + 3\) into equations (1) and (3): \[ \begin{align*} -4(2y - 9z + 3) - y + 36z &= 24, \ -2(2y - 9z + 3) + y + 6z &= 6. \end{align*} \]

Step 4: Simplify Equations (1) and (3)

Simplify the substituted equations:1. \[-8y + 36z - 12 - y + 36z = 24,\] which simplifies to \(-9y + 72z = 36\).2. \[-4y + 18z + y + 6z = 6,\] which simplifies to \(-3y + 24z = 12\).

Step 5: Solve the Simplified System for \(y\) and \(z\)

Now solve the system of two equations:1. \(-9y + 72z = 36\).2. \(-3y + 24z = 12\). Divide the second equation by 3 to simplify it to \(-y + 8z = 4\).

Step 6: Elimination Method to Solve for \(z\)

Use elimination to solve for \(z\). Multiply the simplified second equation by 9: \[-9(-y + 8z) = -9(4)\] which gives \[9y - 72z = -36\].Now add this to the first equation: \(-9y + 72z = 36\). The terms cancel, confirming that the two are linear combinations, so all solutions are valid.

Step 7: Substitute \(z\) into \(-y + 8z = 4\)

Assume \(z = 0\), which gives \(-y = 4\), thus \(y = -4\).

Step 8: Substitute Values of \(y\) and \(z\) to Solve for \(x\)

Substitute \(y = -4\), \(z = 0\) into equation for \(x\): \(x = 2(-4) - 9(0) + 3 = -8 + 3 = -5\).

Key Concepts

Linear Equations

Linear equations form the foundation of solving linear systems. An equation is termed "linear" when it is composed of variables where each term is either a constant or the product of a constant and a single variable. In simpler words, no variables are raised to any power other than one.
In our exercise, we have three linear equations:

• \(-4x - y + 36z = 24\)
• \(x - 2y + 9z = 3\)
• \(-2x + y + 6z = 6\)

Each equation is a linear equation involving the variables \(x\), \(y\), and \(z\). Together, these equations form a system of linear equations. The aim is to find values for \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.
To do this, we need problem-solving techniques such as the elimination and substitution methods, which help isolate and solve for each variable.

Elimination Method

The elimination method is a popular technique used in solving systems of linear equations. The central idea is to eliminate one variable by manipulating the equations so that it cancels out, allowing you to focus on finding the values of the remaining variables.
In our exercise, the elimination method was used to solve the simplified system for \(y\) and \(z\):

• \(-9y + 72z = 36\)
• \(-3y + 24z = 12\)

To eliminate \(y\), the second equation was divided by 3 to make the coefficients of \(y\) comparable:

• \(-y + 8z = 4\)

To completely eliminate \(y\), both equations were structured so that adding them would cancel \(y\) out. Multiplying the simplified second equation by 9 gave us a new equation:

• \(9y - 72z = -36\)

The first equation could then simply be added to this new equation:

• \(-9y + 72z = 36\)

Once the variables \(y\) get cancelled, you can easily find a consistent value for \(z\). This step often allows verification of system consistency or any special conditions like infinite solutions.

Substitution Method

The substitution method involves solving one equation for one variable and then substituting this expression into the other equations. This progressively simplifies the system.
In our problem, this technique was applied starting with the second equation:

• \(x - 2y + 9z = 3\)

Here, we solved for \(x\):

• \(x = 2y - 9z + 3\)

The next step involved substituting this expression for \(x\) into the first and third equations to help eliminate the variable \(x\). This reduced the system into two simpler equations involving only \(y\) and \(z\):

• \(-9y + 72z = 36\)
• \(-3y + 24z = 12\)

This reduction strategy via substitution makes the remaining steps more manageable, eventually leading to a point where substitution back into any derived equation provides values for the excluded variable, demonstrating the versatility of this method in simplifying complex systems.