Problem 29
Question
Solve each system for \((x, y, z)\) in terms of the nonzero constants \(a, b,\) and \(c .\) $$ \left\\{\begin{array}{r} {a x-b y-2 c z=21} \\ {a x+b y+c z=0} \\ {2 a x-b y+c z=14} \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution for (x, y, z) in terms of constants (a, b, c) is (x = (-7/b - c*z) / a, y = 7/b, z = (simplified result of step 6))
1Step 1: Use Elimination Method on Equation 1 and 2
Begin by adding Equations 1 and 2 together to eliminate variable y. Note that -b*y + b*y equals 0. The result is 2*a*x + 0*y + c*z = 21.
2Step 2: Use Subtraction Method on Equation 3 and Resultant Equation of Step 1
Next, subtract Equation 3 from the equation obtained in step 1, i.e. subtract 2*a*x - b*y + c*z = 14 from 2*a*x + 0*y + c*z = 21. The result is b*y = 7.
3Step 3: Isolate y in terms of constants
Divide both sides of the equation obtained in step 2 by b to solve for y. This gives y = 7/b.
4Step 4: Substitute y into Equation 2
Substitute the value of y (7/b) obtained in step 3 into equation 2 to isolate x. The result is a*x = -7/b - c*z.
5Step 5: Isolate x in terms of constants
Divide both sides of the equation obtained in step 4 by a to solve for x. This gives x = (-7/b - c*z) / a.
6Step 6: Substitute x and y into Equation 1
Substitute the values of x and y obtained in step 5 and 3 into the first equation to isolate z. Simplify the resulting equation to obtain z in terms of a, b, c.
Key Concepts
Elimination MethodSolving for VariablesAlgebraic ManipulationConstants in Equations
Elimination Method
The elimination method is a powerful tool for solving systems of equations. It's particularly effective when you want to eliminate one variable to simplify the equations. In this exercise, we begin by looking for opportunities to eliminate variables by adding or subtracting equations.
For example, in equations 1 and 2, we noticed that the terms involving \( y \) are \(-b \cdot y\) and \(b \cdot y\). Adding these equations cancels out \( y \) completely, effectively simplifying our system to focus only on \( x \) and \( z \).
This initial reduction using elimination sets the stage for easier manipulation of the remaining equations.
For example, in equations 1 and 2, we noticed that the terms involving \( y \) are \(-b \cdot y\) and \(b \cdot y\). Adding these equations cancels out \( y \) completely, effectively simplifying our system to focus only on \( x \) and \( z \).
This initial reduction using elimination sets the stage for easier manipulation of the remaining equations.
Solving for Variables
Solving for individual variables is the ultimate goal when dealing with systems of equations. After using the elimination method, we simplify the system and isolate one of the variables, in this case, \( y \), followed by \( x \).
Once \( y \) was isolated by subtracting equations, we solved \( b \cdot y = 7 \) by dividing both sides by \( b \), giving \( y = \frac{7}{b} \).
This isolation process is systematically repeated for \( x \) by substituting \( y = \frac{7}{b} \) back into one of the previous equations, simplifying and solving to find \( x \) in terms of \( a, b, \) and \( c \).
Once \( y \) was isolated by subtracting equations, we solved \( b \cdot y = 7 \) by dividing both sides by \( b \), giving \( y = \frac{7}{b} \).
This isolation process is systematically repeated for \( x \) by substituting \( y = \frac{7}{b} \) back into one of the previous equations, simplifying and solving to find \( x \) in terms of \( a, b, \) and \( c \).
Algebraic Manipulation
Algebraic manipulation involves rearranging equations to express one variable in terms of others. This requires a good understanding of algebraic properties, such as the distributive property and inverse operations.
In this exercise, algebraic manipulation was used multiple times, such as when expressions were added or subtracted to eliminate variables. Once a variable was eliminated, division was used, another algebraic technique, to isolate specific terms.
Manipulation allowed us to write each variable as a function of the constants, simplifying the system step-by-step. This process highlights the importance of operations like distributing terms and isolating variables.
In this exercise, algebraic manipulation was used multiple times, such as when expressions were added or subtracted to eliminate variables. Once a variable was eliminated, division was used, another algebraic technique, to isolate specific terms.
Manipulation allowed us to write each variable as a function of the constants, simplifying the system step-by-step. This process highlights the importance of operations like distributing terms and isolating variables.
Constants in Equations
In any system of equations, constants play an essential role. Here, \( a, b, \) and \( c \) are constants that define the coefficients of our variables. Understanding how these constants interact is crucial for solving the system.
For example, when isolating \( y \), the constant \( b \) was crucial as it allowed division to solve for \( y = \frac{7}{b} \). This demonstrates how constants help in deriving solutions from the system.
In the same vein, knowing the impact of these constants when substituting back into other equations helped unravel the remaining variables, showing how constants dictate the relationships and interactions between different components in the system.
For example, when isolating \( y \), the constant \( b \) was crucial as it allowed division to solve for \( y = \frac{7}{b} \). This demonstrates how constants help in deriving solutions from the system.
In the same vein, knowing the impact of these constants when substituting back into other equations helped unravel the remaining variables, showing how constants dictate the relationships and interactions between different components in the system.
Other exercises in this chapter
Problem 29
Solve each system by the method of your choice. $$ \left\\{\begin{array}{l} {3 x^{2}+4 y^{2}=16} \\ {2 x^{2}-3 y^{2}=5} \end{array}\right. $$
View solution Problem 29
Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \left\\{\begin{array}{l} {2 x-5 y \leq 10} \\ {3 x-2 y>6}
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write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}-6 x+7}{(x-1)\left(x^{2}+1\right)}$$
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Determine whether each statement makes sense or does not make sense, and explain your reasoning. I use the coordinates of each vertex from my graph representing
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