Problem 29 - College Algebra

Question

In Exercises \(29-30,\) solve each system for \((x, y, z)\) in terms of the nonzero constants \(a, b,\) and \(c\) $$ \left\\{\begin{array}{c} a x-b y-2 c z=21 \\ a x+b y+c z=0 \\ 2 a x-b y+c z=14 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system in terms of the nonzero constants \(a\), \(b\), and \(c\) are \(x = -\frac{7}{6a}\), \(y = -\frac{35}{3b}\), \(z = \frac{21}{2c}\)

Step 1: subtract second equation from first

The first approach will be to subtract the second equation from the first one to cancel out the \(y\) term. This results in: \[2cz = 21 \] and after rearranging we will have \(z = \frac{21}{2c}\).

Step 2: substitute \(z\) into the second and third equations

Substituting \(z\) from step 1 into the second and third equations. This will provide an equation that involves only \(x\) and \(y\), making it easier to isolate \(x\). The substitution results in the following equations: \[ax + by = -\frac{21}{2} \] \[2ax - by = 14 - \frac{21}{2} \]

Step 3: solve for \(x\) and substitute \(x\),\(z\) into the first equation

Summing the two previous equations, we get \(3ax = -\frac{7}{2} \), thus \(x = -\frac{7}{6a}\). Substituting \(x\) and \(z\) into the first equation, we get: \[a(-\frac{7}{6a}) - by - 2c(\frac{21}{2c}) = 21 \] Solving for \(y\), we will get \(y = -\frac{35}{3b}\).

Step 4: Recap the solutions

After simplifying the equations for \(x\), \(y\), and \(z\), we obtain the values of these variables in terms of the constants \(a\), \(b\), and \(c\): \[x = -\frac{7}{6a}, y = -\frac{35}{3b}, z = \frac{21}{2c} \]

Key Concepts

Step-by-step solutions

When solving systems of linear equations, a step-by-step approach can make even the trickiest problems manageable. Begin by analyzing each equation and figuring out which variables can be isolated. During computation, it’s critical to address each variable step by step. By taking a systematic approach, we methodically eliminate variables and solve for one variable at a time.

• Start by identifying and eliminating variables systematically.
• Substitute the values obtained into the other equations.
• Simplify at each step to maintain clarity.

This structured method not only enhances understanding but improves accuracy. Each step builds upon the previous one and leads to a full solution when followed carefully.

Algebraic methods

Algebraic methods are essential when dealing with more than one equation. These methods include substitution and elimination, among others. Substitution involves solving one equation for a variable and then substituting this into another equation. On the other hand, elimination involves adding or subtracting equations to remove a variable, thus simplifying the system.

• Substitution helps express one variable in terms of others.
• Elimination is useful to reduce the number of equations and variables.
• Both methods require careful manipulation and precise calculation.

Each technique has its situations where it is most effective, and sometimes a combination of these methods yields the best results.

Variables and constants

In algebra, variables and constants form the backbone of equations. Understanding these concepts is key to solving mathematical systems. Variables like \(x\), \(y\), and \(z\) represent unknown values that we seek to determine.

• Variables are the changing parts of the equation, whose values we solve for.
• Constants like \(a\), \(b\), and \(c\) are fixed values that modify the behavior of the equation.
• Aligning variable terms and identifying constant terms correctly is crucial.

Understanding how variables interact with constants through operations like multiplication or division is vital. They define the relationship between terms within a system of equations.

Solving for unknowns

The ultimate goal in algebraic equations is to solve for unknowns by isolating the variables. This process, known as isolating, involves using algebraic manipulations to find the value of each variable present in the system.

• Begin by simplifying the system to single-variable equations if possible.
• Use inverse operations to undo the effects of constants on the variable.
• Continue until all variables are expressed in terms of known quantities.

In the given problem, this means systematically solving for \(z\), \(x\), and finally, \(y\). It’s important to handle each operation carefully to avoid errors and to double-check solutions by substituting back into the original equations.