Problem 25 - Algebra 2

Question

Solve each system of equations. \(6 x+2 y+4 z=2\) \(3 x+4 y-8 z=-3\) \(-3 x-6 y+12 z=5\)

Step-by-Step Solution

Answer

\(x = \frac{1}{3}, y = -\frac{1}{2}, z = \frac{1}{4}\).

Step 1: Write the system of equations

We begin by writing down the system of linear equations that we need to solve:1. \(6x + 2y + 4z = 2\)2. \(3x + 4y - 8z = -3\)3. \(-3x - 6y + 12z = 5\)

Step 2: Eliminate a variable using the first two equations

To simplify, we'll eliminate one variable by adding or subtracting equations. Multiply the second equation by 2 to get \(6x + 8y - 16z = -6\). Now subtract equation 1 from this new equation:\((6x + 8y - 16z) - (6x + 2y + 4z) = -6 - 2\)This results in:\(6y - 20z = -8\).

Step 3: Eliminate the same variable using the other equations

Multiply equation 3 by 2 to make the coefficients of \(x\) matching:\((-3x - 6y + 12z) \times 2 = -6x - 12y + 24z = 10\)Add this to equation 1:\((6x + 2y + 4z) + (-6x - 12y + 24z) = 2 + 10\)This results in:\(-10y + 28z = 12\).

Step 4: Solve the new system for two variables

We now have two simplified equations:1. \(6y - 20z = -8\)2. \(-10y + 28z = 12\)Let's solve for one variable. Multiply the first equation by 5 and the second by 3:\(30y - 100z = -40\)\(-30y + 84z = 36\)Add these equations:\(-16z = -4\)Divide by -16:\(z = \frac{1}{4}\).

Step 5: Substitute back to find another variable

Substitute \(z = \frac{1}{4}\) into the first reduced equation:\(6y - 20\left(\frac{1}{4}\right) = -8\)\(6y - 5 = -8\)Add 5 to both sides:\(6y = -3\)Divide by 6:\(y = -\frac{1}{2}\).

Step 6: Solve for the remaining variable

Substitute both \(y = -\frac{1}{2}\) and \(z = \frac{1}{4}\) back into the original first equation:\(6x + 2\left(-\frac{1}{2}\right) + 4\left(\frac{1}{4}\right) = 2\)This simplifies to:\(6x - 1 + 1 = 2\)\(6x = 2\)Divide by 6:\(x = \frac{1}{3}\).

Step 7: Solution for the system

We have found the values of \(x\), \(y\), and \(z\) which solve the system:\(x = \frac{1}{3}\), \(y = -\frac{1}{2}\), \(z = \frac{1}{4}\).

Key Concepts

Linear Algebra

Linear algebra is a fascinating field of mathematics that deals with vectors, matrices, and systems of linear equations. It forms the fundamental backbone of many mathematical computations and is essential for various applications in science and engineering. A system of equations, like the one we are solving here, involves finding the values of variables that satisfy multiple linear equations simultaneously. These equations typically look like an algebraic expression with terms that can be a constant or a product of a constant and a variable.
Linear algebra provides the tools and methods, such as matrix operations, to approach and solve these systems, making it a crucial skill in fields ranging from computer science and physics to economics. Imagine solving puzzles where each piece is an equation, and you must find the values of variables to complete the picture. This is essentially what you're doing when solving systems of linear equations.

Elimination Method

The elimination method is a technique used to solve systems of linear equations. It involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the remaining ones. Think of it as simplifying the puzzle by breaking it down into smaller, more manageable parts.
In practice, you might align the equations so that one variable has the same coefficient in each equation. Then, you can subtract (or add) the equations to cancel that variable out. For example, in the exercise given, we simplify the system by eliminating a variable, which lets us focus on solving for the others.
This method is particularly useful when a system has more than two equations, as it helps filter out the complexity step by step until you are left with one variable to solve for.

Substitution Method

The substitution method is another strategic approach to solve systems of linear equations, where you express one variable in terms of another and then substitute this expression into the other equation. This method is like setting a stage where one actor (a variable) takes on the role defined by others, simplifying the equation and reducing the number of unknowns.
In simpler terms, if you isolate one of the variables in one equation, you can replace that variable in another equation with its equivalent expression. This reduces the system to a point where you can easily solve for one of the variables, and subsequently, the others too. In our exercise, substitution let us decay the complex system into a single-variable equation, making the solution straightforward to calculate.

Variables and Equations

Variables and equations are the core components of any system of equations. An equation is a mathematical statement that asserts the equality of two expressions, often containing one or more variables. Variables act like placeholders in these equations, representing unknown values we need to find.
Understanding how to manipulate and balance equations is crucial in solving for variables. By carefully applying algebraic operations like addition, subtraction, multiplication, and division, variables can reveal their value, allowing us to decipher the solution to the system.
For example, our system involves three variables: \(x\), \(y\), and \(z\). Each step towards solving equates to untangling these variables from one another by comparing and adjusting different equations. Mastering this art of interplay between variables and equations helps solve many real-world problems efficiently.