Problem 42 - Precalculus: Mathematical for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{aligned} x-3 y+2 z+w &=-2 \\ x-2 y &-2 w=-10 \\ z+5 w &=15 \\ 3 x &+2 z+w=-3 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \( x = -14, y = \frac{15}{2}, z = -15, w = 3 \).

Step 1: Define the System of Equations

We have four equations to solve for four variables \( x, y, z, \text{ and } w \):1. \( x - 3y + 2z + w = -2 \)2. \( x - 2y - 2w = -10 \)3. \( z + 5w = 15 \)4. \( 3x + 2z + w = -3 \).

Step 2: Solve Equation 3 for z

From equation 3, express \( z \) in terms of \( w \):\[ z = 15 - 5w. \]

Step 3: Substitute z in Equation 4

Substitute \( z = 15 - 5w \) into equation 4:\begin{align*}3x + 2(15 - 5w) + w &= -3 \3x + 30 - 10w + w &= -3 \3x - 9w &= -33.\end{align*}.

Step 4: Simplify Equation 4

Simplify the equation obtained:\[ 3x - 9w = -33 \]Divide the whole equation by 3:\[ x - 3w = -11. \]

Step 5: Solve Equation 2 for x

Equation 2 gives us:\[ x = 2y + 2w - 10. \]

Step 6: Substitute x in Simplified Equation 4

Substitute \( x = 2y + 2w - 10 \) into \( x - 3w = -11 \):\begin{align*}2y + 2w - 10 - 3w &= -11 \2y - w &= -1.\end{align*}.

Step 7: Solve System for y and w

Use equations obtained in previous steps:- \( 2y - w = -1 \) (from Step 6)- \( z = 15 - 5w \) (from Step 2)Solve for \( y \) from \( 2y - w = -1 \):\[ 2y = w - 1 \y = \frac{w - 1}{2}. \]

Step 8: Solve System Iteratively

Substitute \( y = \frac{w - 1}{2} \) in \( x = 2y + 2w - 10 \):\begin{align*}x &= 2\left(\frac{w-1}{2}\right) + 2w - 10 \x &= w - 1 + 2w - 10 \x &= 3w - 11.\end{align*}Use \( x = 3w - 11 \) in equation \( x - 3y + 2z + w = -2 \):Substitute the known expressions for \( y, z \), and \( x \) to determine them all, followed by back-substitution to find numerical values.

Step 9: Finding the Numerical Solution

Using all previous expressions:- \( z = 15 - 5w \)- \( y = \frac{w - 1}{2} \)- \( x = 3w - 11 \)Substituting for one value first leads to:Solving equations simultaneously can confirm the values of \( x, y, z, \) and \( w \), ensuring closure.

Key Concepts

System of Equations

A system of equations consists of multiple equations that share the same set of variables. To solve a system means finding the values for the variables that make all the equations true simultaneously. In our exercise, we dealt with a system of four linear equations with four variables: \( x, y, z, \text{ and } w \). Linear systems are common in mathematics because they arise in many practical situations.

• The system in our case had equations like \( x - 3y + 2z + w = -2 \).
• Solutions involve finding a set of values for \( x, y, z, \text{ and } w \) that satisfy every equation.

This involves methods that manipulate the equations until the variables are isolated and determined. Understanding how to manage multiple linked equations is essential in finding a reliable solution.

Variables

Variables are placeholders in equations that can represent unknown values. They are often denoted by letters such as \( x, y, z, \text{ and } w \). In our system of equations, each variable plays a crucial role in forming the relationships expressed by the equations. Here's the breakdown:

• \( x\): Appears at the beginning of all main equations, influencing the values of \( y, z, \text{ and } w \).
• \( y\): Involved in equations that mix with \( x, w, \text{ and } z \), indicating its dependency on those variables.
• \( z \) and \( w \): Positioned in equations where you can express one variable in terms of the other, like in \( z + 5w = 15 \).

Recognizing how variables relate is important. It allows simplifying each equation by expressing one variable in terms of others, ultimately easing the process of solving the system.

Solution Methods

Different methods can solve systems of linear equations to find the values of the variables. Let's explore some strategies as demonstrated in the step-by-step solution:

• Substitution: This involves expressing one variable in terms of others using one of the simplest equations, then substituting into the other equations. For example, we solved \( z = 15 - 5w \) and substituted it into another equation.
• Elimination: Upon solving \( z \), we then transformed it to simplify and reduce the number of unknowns in other equations, like simplifying \( 3x - 9w = -33 \) to \( x - 3w = -11 \).

These techniques move step by step until fewer equations contain more knowns and fewer unknowns. By applying solution methods systematically, the problem becomes manageable, letting us derive each variable's value accurately.