Problem 11 - Precalculus Mathematics for Calculus

Question

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

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

Step 1: Solve for the variable in the last equation

The last equation is \(-2z = 1\). To solve for \(z\), divide both sides by \(-2\): \[ z = \frac{1}{-2} = -\frac{1}{2} \]

Step 2: Substitute the value of z into the second equation

The second equation is \(y + 4z = 0\). Substitute \(z = -\frac{1}{2}\) into the equation: \[ y + 4(-\frac{1}{2}) = 0 \]Simplify: \[ y - 2 = 0 \] Add 2 to both sides to isolate \(y\): \[ y = 2 \]

Step 3: Substitute the values of y and z into the first equation

The first equation is \(2x - y + 6z = 5\). Substitute \(y = 2\) and \(z = -\frac{1}{2}\) into the equation: \[ 2x - 2 + 6(-\frac{1}{2}) = 5 \]Simplify the terms: \[ 2x - 2 - 3 = 5 \] Combine like terms: \[ 2x - 5 = 5 \] Add 5 to both sides: \[ 2x = 10 \] Divide both sides by 2 to solve for \(x\): \[ x = 5 \]

Step 4: Conclusion: Combine the solutions

The solution to the system of equations is \(x = 5\), \(y = 2\), and \(z = -\frac{1}{2}\). This means the values satisfy all three equations of the system.

Key Concepts

Triangular system

A triangular system is a type of system of linear equations where the equations are structured in a way that each equation only involves a specific subset of variables, and the system is typically organized in a manner that resembles a triangle when written in standard form. In linear algebra, a triangular system can either be upper or lower triangular.

• Upper Triangular System: In this setup, the first equation involves all the unknowns, the second equation involves all unknowns except the first one, and so on. It forms a triangle matrix with non-zero values above the diagonal.
• Lower Triangular System: Here, each equation contains unknowns up to a certain point, and the system has non-zero values below the diagonal.

Solving a triangular system is straightforward because the arrangement allows one to start from the bottom-most equation and work upwards or vice versa, simplifying the calculations as you go. This characteristic makes the triangular form highly advantageous when using methods like back-substitution.

Equation solving

Equation solving is finding the values of variables that satisfy given mathematical equations. To solve equations systematically, one usually performs operations to isolate the variable of interest.

• Begin by simplifying the equation, combining like terms and constants.
• Use inverse operations to isolate variables (e.g., addition to counter subtraction).
• Check your solution by substituting it back into the original equation.

In the context of solving systems, equation solving involves strategies such as substitution, elimination, and matrix operations. Each method is chosen based on the structure of the system and the convenience of applying these procedures. For the given triangular system, using the method of back-substitution simplifies finding solutions.

System of equations

A system of equations consists of multiple equations solved together because they share common variables. The ultimate objective is to find specific values for these variables that satisfy all the equations simultaneously. Systems of equations can be linear or nonlinear, although linear equations are more commonly addressed in beginner studies.

• Linear System: Comprising straight-line equations, solutions are found where the lines intersect.
• Nonlinear System: Includes at least one nonlinear equation, such as a quadratic or exponential equation.

To solve a system, it can be approached by graphical methods, where intersections are visually identified, or by algebraic methods like substitution, elimination, or matrices for more complex systems. In the specific triangular system given, the ordered approach inherent to the triangular format allows for direct application of back-substitution, translating the interdependencies of the equations into a clear solution pathway.