Problem 25 - Precalculus

Question

Solve each system of linear equations. $$\begin{aligned} 2 x+5 y &=9 \\ x+2 y-z &=3 \\ -3 x-4 y+7 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are parametrized as \( x = \frac{9 - 5t}{2} \), \( y = t \), and \( z = \frac{3 - t}{2} \) for any real number \( t \).

Step 1: Label Equations

Label the given system of equations for easy reference. Equation 1: \( 2x + 5y = 9 \)Equation 2: \( x + 2y - z = 3 \)Equation 3: \( -3x - 4y + 7z = 1 \)

Step 2: Eliminate z from Equations 2 and 3

To eliminate \( z \), we'll multiply Equation 2 by 7 and Equation 3 by 1, and then add them to eliminate \( z \):\[7(x + 2y - z) = 7(3) \7x + 14y - 7z = 21 \-3x - 4y + 7z = 1 \\_4x + 10y = 22 \\]This gives us a new equation without \( z \):Equation 4: \( 4x + 10y = 22 \)

Step 3: Solve Equations 1 and 4 for x and y

We now have two equations with two variables:Equation 1: \( 2x + 5y = 9 \)Equation 4: \( 4x + 10y = 22 \)Since Equation 4 is just 2 times Equation 1, they do not provide independent information; thus there's infinite number of solutions for \( x \) and \( y \) (they are dependent).We express \( x \) in terms of \( y \) using Equation 1:\( 2x + 5y = 9 \)\( 2x = 9 - 5y \)\( x = \frac{9 - 5y}{2} \) Let \( y = t \), we then have \( x = \frac{9 - 5t}{2} \).

Step 4: Solve for z in terms of t

Substitute \( x = \frac{9 - 5t}{2} \) and \( y = t \) into Equation 2 to find \( z \):\[\frac{9 - 5t}{2} + 2t - z = 3 \\frac{9 - 5t + 4t}{2} - z = 3 \\frac{9 - t}{2} - z = 3 \-z = 3 - \frac{9 - t}{2} \z = \frac{9 - t}{2} - 3 \z = \frac{9 - t - 6}{2} \z = \frac{3 - t}{2} \]Thus, \( x = \frac{9 - 5t}{2} \), \( y = t \), \( z = \frac{3 - t}{2} \), where \( t \) is any real number.

Key Concepts

linear equations

Linear equations are crucial fundamentals in algebra. They represent equations of straight lines on a Cartesian plane. The general form of a linear equation in two variables, \(x\) and \(y\), is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. In algebraic terms:

• The degree of each term is 1, meaning variables are not raised to any power other than one.
• Linear equations can have one, two, or more variables.
• The graph of a linear equation is always a straight line.

In a system of linear equations, we solve for values that satisfy all equations simultaneously. This system can have a unique solution, no solution, or infinitely many solutions. Understanding the characteristics of linear equations makes it easier to initially analyze their potential solutions before deeper calculations begin.

variable elimination

Variable elimination is an effective method used to solve systems of linear equations. The idea is to remove or eliminate one of the variables in order to simplify the problem and more easily find the remaining unknowns. Here’s how it typically works:

• Identify two equations where one variable can be easily eliminated by addition or subtraction.
• Multiply one or both equations to ensure coefficients are aligned for elimination.
• Perform the addition or subtraction to remove one variable, yielding a simplified system.

This helps transform a complicated multi-variable equation into a clearer, simpler one. In our exercise, we eliminated \(z\) by multiplying one equation and then adding to another. This allowed us to focus on just two variables, significantly simplifying the computation. It’s a strategic step to make solving multi-level problems more manageable.

infinite solutions

In certain systems of equations, particularly when dealing with dependent equations, there arises a case of infinite solutions. This scenario occurs when one equation can be derived from another by multiplication or addition, as was seen in our exercise. Here’s what infinite solutions indicate:

• The lines or planes described by the equations coincide, appearing as the same line or plane in a graphical view.
• Each solution to one of the equations also satisfies the others.
• In algebraic terms, one equation is a linear combination of the others, not providing any new information.

When recognizing infinite solutions, we typically express variables in terms of a free parameter. In our solved example, we expressed \(x\), \(y\), and \(z\) as functions of \(t\), an arbitrary value. Recognizing infinite solutions ensures an accurate understanding of the relationship among variables, crucial to solving such algebraic systems.