Problem 40

Question

Find two systems of equations that have the ordered triple as a solution. (There are many correct answers.) $$\left(0,2, \frac{1}{2}\right)$$

Step-by-Step Solution

Verified
Answer
The first system of equations are \[x = 0\], \[y = 2\] and \[z = \frac{1}{2}\]. The second system of equations are \[x = 0\], \[2y = 4\] and \[2z = 1\].
1Step 1: Creating the first system
For the first system of equations, we could create equations as simple as \[x = 0\], \[y = 2\] and \[z = \frac{1}{2}\], as this will directly yield the given ordered triple upon solving.
2Step 2: Creating the second system
For the second system, we can select equations which are not as direct as the first one but yield the given ordered triple. One such system of equations could be \[x = 0\] (reusing this equation for simplicity), \[2y = 4] and \[2z = 1].

Key Concepts

Ordered Triple as a SolutionSolving Linear SystemsAlgebraic Equations
Ordered Triple as a Solution
Understanding an ordered triple as a solution is fundamental in the study of three-dimensional coordinate systems. In three dimensions, an ordered triple \( (x, y, z) \) represents a specific point in space. To determine if an ordered triple is a solution to a system of equations, each element of the ordered triple must satisfy all equations in the system simultaneously.

For example, consider the ordered triple \( (0, 2, \frac{1}{2}) \). To confirm it as a solution, we substitute \( x \) with 0, \( y \) with 2, and \( z \) with \(\frac{1}{2}\). If the substitution leads to true statements in each equation of the system, then \( (0, 2, \frac{1}{2}) \) is indeed the solution. This concept is pivotal in solving real-world problems involving multiple variables, such as physics simulations or economic modeling. When creating a system of equations, we defined that system such that the ordered triple will always satisfy it, ensuring a clear understanding of how each component in the triple correlates with the variables in the equations.
Solving Linear Systems
Solving linear systems is a cornerstone of algebra, with applications spanning engineering, economics, and science. A linear system consists of two or more linear equations with a common set of variables. The goal of solving such systems is to find the values of these variables that satisfy all equations at once.

There are various methods for solving linear systems, including graphing, substitution, elimination, and matrix operations. The choice of method often depends on the complexity of the system and the number of variables involved. In the case of simple systems, direct substitution—as shown in the creation of the first example system—works seamlessly. For more complex systems, such as systems with variables that include coefficients other than one, manipulation is required to simplify the equations. This often involves multiplying both sides of the equations to create common terms that facilitate the elimination of variables and eventually uncover the solution.
Algebraic Equations
Algebraic equations are expressions that set two algebraic terms equal to each other and often contain one or more variables. The most basic form of an algebraic equation is a linear equation, which is an equation of the first degree. These equations typically take the form \( Ax + By + Cz = D \) in three-dimensional space, where \( A \), \( B \), and \( C \) are coefficients, \( x \), \( y \), and \( z \) are the variables, and \( D \) is a constant.

To solve algebraic equations, we aim to isolate the variable by performing operations that maintain the balance of the equation. This often involves adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. When dealing with multiple algebraic equations in a system, the solutions are the points where the equations intersect. The difficulty can increase with the addition of terms and variables, but the foundational principles remain consistent. Whether you are working with linear, quadratic, or higher-degree equations, the goal is to manipulate and reduce the equations to find a common solution set.