Problem 54 - Algebra and Trigonometry

Question

Solve the system of linear equations. $$ \left\\{\begin{aligned} y-z+2 w =0 \\ 3 x+2 y +w=0 \\ 2 x +4 w=12 \\\\-2 x &-2 z+5 w= 6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution in terms of parameter \( w \) is: \( x = 6 - 2w \), \( y = \frac{5w - 18}{2} \), \( z = \frac{9w - 18}{2} \), and \( w \) is arbitrary.

Step 1: Simplify the Equations

Start by simplifying the third equation, which is linear in the variables \(x\) and \(w\): \[ 2x + 4w = 12. \]Divide the whole equation by 2 to simplify:\[ x + 2w = 6. \]This simplified equation will help us solve the system more easily.

Step 2: Substitute for x

From the equation \( x + 2w = 6 \), solve for \( x \):\[ x = 6 - 2w. \] Now, substitute this expression for \( x \) into the second equation \( 3x + 2y + w = 0 \):\[ 3(6 - 2w) + 2y + w = 0. \]

Step 3: Simplify After Substitution

Expand and simplify the substitution in the equation from Step 2:\[ 18 - 6w + 2y + w = 0. \]Combine like terms:\[ 18 - 5w + 2y = 0. \]Rearrange to solve for \( 2y \):\[ 2y = 5w - 18. \]

Step 4: Solve for y

From the equation \( 2y = 5w - 18 \), solve for \( y \):\[ y = \frac{5w - 18}{2}. \] Now we have expressions for \( x \) and \( y \) in terms of \( w \).

Step 5: Substitute into Another Equation

Use the first equation \( y - z + 2w = 0 \) and substitute \( y = \frac{5w - 18}{2} \):\[ \frac{5w - 18}{2} - z + 2w = 0. \]Multiply through by 2 to clear the fraction:\[ 5w - 18 - 2z + 4w = 0. \]

Step 6: Solve for z

Combine like terms and solve for \( z \):\[ 9w - 18 = 2z \]\[ 2z = 9w - 18 \]\[ z = \frac{9w - 18}{2}. \] Now we have \( z \) in terms of \( w \).

Step 7: Verify Solution Using Last Equation

With expressions for \( x, y, \) and \( z \) in terms of \( w \), check against the last equation \(-2x - 2z + 5w = 6 \). Substitute the expressions:\( x = 6 - 2w \) and \( z = \frac{9w - 18}{2} \):\[ -2(6 - 2w) - 2\left(\frac{9w - 18}{2}\right) + 5w = 6. \]Simplify and verify the equality holds.After simplification, all terms cancel and verify the solution for \( w \) as arbitrary.

Key Concepts

Substitution Method

The substitution method is a common technique used to solve systems of linear equations. It is particularly useful when dealing with equations that have multiple variables. The main idea is to solve one of the equations for one variable in terms of the other variables. Then, substitute this expression into the remaining equations, simplifying the system.

Here's how the substitution method was applied in the exercise:

• First, the equation \( x + 2w = 6 \) was simplified and solved for \( x \). This gave \( x = 6 - 2w \).
• This expression for \( x \) was substituted into another equation, \( 3x + 2y + w = 0 \), allowing us to eliminate \( x \) from that equation.
• Through substitution and simplification, each variable was expressed in terms of a single variable \( w \). This continued until all variables were expressed using \( w \).

By systematically substituting and solving for each variable, the substitution method simplifies the process of finding solutions to complex systems.

Linear Algebra

Linear algebra is a branch of mathematics that focuses on vector spaces and linear mappings between those spaces. It is the study of lines, planes, and subspaces, and the relationships between them. In systems of linear equations, linear algebra provides tools and techniques to find solutions efficiently.

For this exercise, the system of linear equations has four equations and four unknowns: \( x, y, z, \) and \( w \). Such systems can be represented by matrix equations, though manual methods like substitution can also be applied when the system isn't too complex.

• Understanding the relationships between equations is key. Each equation can be seen as a plane in four-dimensional space, and the solution corresponds to the intersection point of all these planes.
• Linear algebraic concepts help in comprehending how substitution simplifies equations by reducing dimensions one step at a time.
• The work shown fits into the broader context of linear mappings and vector spaces, where each step is akin to transforming the system into more manageable chunks.

In essence, linear algebra offers a powerful perspective on solving systems of linear equations through concepts that apply broadly across mathematic disciplines.

Equation Simplification

Equation simplification is a foundational step in solving systems of equations. It involves manipulating an equation to make it easier to work with while maintaining its equality.

In the given problem, simplification played a crucial role:

• The equation \( 2x + 4w = 12 \) was first simplified by dividing by 2, resulting in \( x + 2w = 6 \). This straightforward change made the equation easier to handle.
• Simplification was also used after substituting expressions. For example, once a substitution was made, terms like \( 18 - 6w + 2y + w \) were combined and rearranged to simplify to \( 2y = 5w - 18 \).
• Eliminating fractions and reducing expressions to find neat solutions like \( z = \frac{9w - 18}{2} \) made verifying the solutions less complex.

By continually simplifying equations, we reduce computation and error, creating a clear path to the solution. It underscores the importance of simplicity in solving mathematical problems efficiently.