Problem 9 - Beginning and Intermediate Algebra

Question

Solve the system of equations. $$\begin{aligned} 2 x+y-3 z &=1 \\ x-4 y+z &=6 \\ 4 x-7 y-z &=13 \end{aligned}$$

Step-by-Step Solution

Answer

The solution of the given system of equations is: \( x = 1, y = 3, z = 2 \)

Step 1: Eliminate variable x from equation 2 and 3

Multiply the first equation by -2 and then add it to the second equation. Similarly, multiply the first equation by -4 and then add it to the third equation. The goal is to eliminate variable x from equation 2 and 3. $$\begin{aligned} (-2)\cdot (2x+y-3z) &= (-2)\cdot 1 \\ -4x-2y+6z &= -2 \end{aligned}$$ Now, add the modified first equation to the second equation: $$\begin{aligned} (-4x-2y+6z) + (x-4y+z) &= -2 + 6 \\ -3x-6y+7z &= 4 \end{aligned}$$ Similarly, for the third equation: $$\begin{aligned} (-4)\cdot (2x+y-3z) &= (-4)\cdot 1 \\ -8x-4y+12z &= -4 \end{aligned}$$ Add the modified first equation to the third equation: $$\begin{aligned} (-8x-4y+12z) + (4x-7y-z) &= -4 + 13 \\ -4x-11y+11z &= 9 \end{aligned}$$ Now we have a system of two equations with two variables (y and z): $$\begin{aligned} -3x-6y+7z &= 4 \\ -4x-11y+11z &= 9 \end{aligned}$$

Step 2: Eliminate variable y from the new system of equations

Multiply the first new equation by \( \frac{11}{6} \) and the second new equation by 2. Then, add these two new modified equations to eliminate variable y. $$\begin{aligned} \frac{11}{6}\cdot(-3x-6y+7z) &= \frac{11}{6}\cdot 4 \\ -\frac{33}{2}x-11y+\frac{77}{2}z &= 22 \end{aligned}$$ and $$\begin{aligned} 2\cdot(-4x-11y+11z) &= 2\cdot 9 \\ -8x-22y+22z &= 18 \end{aligned}$$ Now, add these two modified equations: $$\begin{aligned} \left(-\frac{33}{2}x-11y+\frac{77}{2}z\right) + (-8x-22y+22z) &= 22+18 \\ -25x + 33z &= 40 \end{aligned}$$

Step 3: Solve for x in terms of z

Divide both sides of the last equation by -25 to solve for x in terms of z: $$\begin{aligned} x &= -\frac{33}{25}z+\frac{40}{25} \end{aligned}$$

Step 4: Substitute the expression for x into the first new equation, and solve for y

Substitute the expression for x in the new equation 1: $$\begin{aligned} -3\left(-\frac{33}{25}z+\frac{40}{25}\right) -6y + 7z &= 4 \end{aligned}$$ Solve for y by rearranging the equation: $$\begin{aligned} y &= -\frac{43}{30}z+\frac{49}{30} \end{aligned}$$

Step 5: Substitute the expressions for x and y into the original equation 1 and solve for z

Substitute the expressions for x and y into the original equation 1: $$\begin{aligned} 2\left(-\frac{33}{25}z+\frac{40}{25}\right) + \left(-\frac{43}{30}z+\frac{49}{30}\right) - 3z &= 1 \end{aligned}$$ Solve for z: $$\begin{aligned} z &= 2 \end{aligned}$$

Step 6: Find x and y using the expressions for x and y in terms of z

Now that we have z, we can find x and y using the expressions: $$\begin{aligned} x &= -\frac{33}{25}(2)+\frac{40}{25} \\ x &= 1 \end{aligned}$$ and $$\begin{aligned} y &= -\frac{43}{30}(2)+\frac{49}{30} \\ y &= 3 \end{aligned}$$

Step 7: The solution of the system of equations

The solution of the given system of equations is: $$\begin{aligned} x &= 1 \\ y &= 3 \\ z &= 2 \end{aligned}$$

Key Concepts

Linear Equations

Linear equations are a type of equation where each term is either a constant or the product of a constant and a single variable. They are represented in the standard form as \( ax + by + cz = d \), where \( a, b, c, \) and \( d \) are constants, and \( x, y, \) and \( z \) are variables. This specific form allows us to easily handle and manipulate them to find solutions.
Linear equations often appear in systems, like in our original problem with three equations and three variables. Solving such a system requires finding the values of the variables that satisfy all equations simultaneously.
To solve systems of linear equations, we can use various methods like substitution, elimination, or matrix operations. The goal is to simplify the system into a form where the solutions are apparent.

• Linear equations form the basis of linear algebra and have numerous applications in science, engineering, and economics.
• They represent relationships with constant rates of change.

Elimination Method

The elimination method is a powerful technique for solving systems of linear equations. It involves algebraically manipulating the equations to eliminate one or more variables, thereby reducing the number of variables.
This method often involves:

• Multiplying equations by constants to align coefficients.
• Adding or subtracting equations to cancel out a variable.

For the given exercise, we first eliminated the variable \( x \) by transforming the original equations. This was done by aligning the coefficients of \( x \) across the equations and then adding these equations together to remove the \( x \) variable. This reduced the system to one with two variables, \( y \) and \( z \).
The elimination method is especially useful in larger systems, providing a clear pathway to simplify the problem gradually down to a single variable solution.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. In the context of our system of linear equations, this means finding the x, y, and z that satisfy all given equations simultaneously.
The process typically includes:

• Using algebraic strategies like substitution or elimination to reduce complexity.
• Simplifying equations via arithmetic operations to isolate variables.
• Returning to any modified equations to check the solutions.

For the particular steps taken in the solution, once the system was reduced through elimination, each subsequent variable was isolated by substituting back into previous equations. This ensures that the solutions are validated against the original constraints of the equations.
It is essential to approach solving systematically and verify each step to avoid errors in calculation, especially when dealing with fractions or negative coefficients.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions and equations. It is a critical skill in mathematics, particularly when dealing with systems of equations.
For our system, algebraic manipulation was used to:

• Multiply equations to create coefficients that allow elimination.
• Rearrange equations to simplify and isolate variables.
• Perform arithmetic operations like addition or subtraction across equations.

In practice, this involved multiplying the original equations to align terms and then adding or subtracting them to effectively nullify a variable, leading to a simpler equation.
Mastery of algebraic manipulation improves problem-solving efficiency and accuracy, crucial for more complex systems that might involve non-linear terms or larger variable sets.