Problem 44 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a system of linear equations that has the given solution. (There are many correct answers.) $$(-5,-2,1)$$

Step-by-Step Solution

Answer

One possible system of linear equations with the solution (-5, -2, 1) could be: \n1. \(4x - 3y + 2z = -12\)\n2. \(x + 2y - 3z = -12\)

Step 1: Understand what a linear equation is

A linear equation is an equation that can be written in the form \(Ax + By + Cz = D\), where A, B, C, D are constants, and x, y, z are variables. So to solve this problem, we would create linear equations that satisfy the given solution (-5, -2, 1)

Step 2: Create the first linear equation

The first equation may be chosen as follows: \(4x - 3y + 2z = 11\). If we substitute x, y, and z with -5, -2, and 1 respectively, we get a correct equality: \(4*(-5) - 3*(-2) + 2*1 = -20 + 6 + 2 = -12 = -12\)

Step 3: Create the second linear equation

The second equation can be chosen as: \(x + 2y - 3z = -7\). If we substitute x, y, and z with -5, -2, and 1 respectively, we get a correct equality: \(-5 + 2*(-2) - 3*1 = -5 - 4 - 3 = -12 = -12\)

Step 4: Finalizing the system of linear equations

Therefore, a system of linear equations that have the solution (-5, -2, 1) can be: \n1. \(4x - 3y + 2z = -12\) \n2. \(x + 2y - 3z = -12\)

Key Concepts

Variables and Constants

When studying a system of linear equations, grasping the role of variables and constants is vital. **Variables** are symbols used to represent numbers whose exact values are not immediately known. In linear equations, we often use symbols like \(x\), \(y\), and \(z\). These variables take on values from the ordered solution set provided, such as \((-5, -2, 1)\) in our exercise.
On the other hand, **constants** are the fixed numbers present in the equations. They determine the equation's specific form and are usually represented by the symbols \(A\), \(B\), \(C\), and \(D\).
Understanding the difference between variables and constants helps in forming and manipulating equations to achieve the desired solution.

Linear Equation Form

Linear equations are expressed in a standard format that makes solving them methodical. Typically, these equations look like \(Ax + By + Cz = D\), where \(A\), \(B\), \(C\), and \(D\) are constants, and \(x\), \(y\), and \(z\) are variables. The equation must be linear, meaning that each variable is to the power of one.
To create a custom linear equation that fits a given solution, we insert the solution values into the equation as the variables. If the balance of the equation holds true, it means the solution is valid for that equation.
In our example, equations like \(4x - 3y + 2z = -12\) and \(x + 2y - 3z = -12\) are set up to satisfy the constants by plugging in \((-5, -2, 1)\) as values for \(x\), \(y\), and \(z\).

Solution of Equations

Finding the solution to a system of linear equations means determining the ordered set of values that satisfy all equations simultaneously. For the given values, \((-5, -2, 1)\), we must ensure each equation in the system equates correctly when these values replace the variables.
Start by substituting these values into the equations one by one. For instance, in the exercise, inserting \(-5\), \(-2\), and \(1\) into \(4x - 3y + 2z = -12\) verifies it as a solution because the equation remains balanced as shown: \(4(-5) - 3(-2) + 2(1) = -12\).
The true solution is confirmed if this verification works for all equations in the system. Thus, solving involves strategic substitution followed by validation to ensure each equation maintains equality.