Problem 11 - Algebra and Trigonometry

Question

\(7-12=\) Use back-substitution to solve the triangular system. $$ \left\\{\begin{aligned} 2 x-y+6 z &=5 \\ y+4 z &=0 \\\\-2 z &=1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

\(x = 5\), \(y = 2\), \(z = -\frac{1}{2}\).

Step 1 - Solve for z

The equation \(-2z = 1\) is the simplest as it involves only one variable. Solve this equation to find \(z\): \[-2z = 1\ z = \frac{-1}{2}\]

Step 2 - Substitute z into the second equation

Use the value of \(z = \frac{-1}{2}\) in the second equation, \(y + 4z = 0\), and solve for \(y\): \y + 4\left(\frac{-1}{2}\right) = 0\ y - 2 = 0\ y = 2\.

Step 3 - Substitute z and y into the first equation

Substitute \(z = \frac{-1}{2}\) and \(y = 2\) into the first equation, \(2x - y + 6z = 5\), to find \(x\): \2x - 2 + 6\left(\frac{-1}{2}\right) = 5\2x - 2 - 3 = 5\2x - 5 = 5\2x = 10\x = 5\.

Key Concepts

Back-Substitution

Back-substitution is a method commonly used to solve triangular systems of linear equations. It's especially handy when dealing with systems arranged in an upper triangular form. An upper triangular system is characterized by having all zeroes below the diagonal, which makes it easier to solve. The process begins by solving the equation involving the last variable and then proceeds upwards, subbing in known values as it goes:

• Start with the bottom-most equation and solve for its single variable.
• Use the value found to substitute into the next equation up, solving for another variable.
• Continue this process, moving up through the equations, solving as you proceed.

This step-by-step procedure ensures that all the calculations are manageable, leveraging simplicity by solving for one unknown at a time.

Linear Equations

Linear equations are mathematical expressions that form straight lines when plotted on a graph. They follow the standard format of variables raised to the power of one, with coefficients and constants:

• The basic form of a linear equation in two variables is represented as \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
• In the context of this exercise, we dealt with a system of three linear equations in three variables: \( x \), \( y \), and \( z \).

These types of equations are also crucial because they describe how variables relate to each other in a linear fashion. Linear relationships are foundational, appearing in algebra, and real-world applications like economics and physics.

Algebraic Solution

Finding an algebraic solution to a system of equations involves manipulating the equations to find the values of the unknowns. Algebraic methods ensure an exact solution, unlike numerical approaches which tend to approximate. Here's how these solutions generally work:- First, arrange the equations neatly in a system, as shown in the problem.- If possible, eliminate variables using methods like substitution or elimination.For this particular exercise:

• The solution process began by isolating \( z \) from the third equation since it only contained one variable.
• Once \( z \) was known, its value was substituted into the second equation to find \( y \).
• Finally, the known values of \( z \) and \( y \) were substituted back into the first equation to solve for \( x \).

By carefully applying these algebraic techniques, the exact values of each variable in the linear system were efficiently determined.