Problem 22 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} 2 x+6 y+3 z=-20 \\ 5 x-3 y-5 z=47 \\ 4 x+3 y+2 z=4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is the set of values for \( x \), \( y \), and \( z \) that satisfies all three equations.

Step 1: Identify the System of Equations

The given system of equations is: \( 2x + 6y + 3z = -20 \), \( 5x - 3y - 5z = 47 \), \( 4x + 3y + 2z = 4 \). We need to solve for the variables \( x \), \( y \), and \( z \).

Step 2: Use Substitution or Elimination

To solve the system, we will use the elimination method. We first eliminate \( y \) from the equations. Multiply the first equation by 0.5 to align it with the third equation for elimination: \( x + 3y + 1.5z = -10 \), \( 5x - 3y - 5z = 47 \), \( 4x + 3y + 2z = 4 \).

Step 3: Eliminate \( y \) from Equations

Add the transformed first equation to the third equation to eliminate \( y \): \( (x + 3y + 1.5z) + (4x + 3y + 2z) = -10 + 4 \). This simplifies to \( 5x + 3.5z = -6 \).

Step 4: Solve for \( x \) and \( z \)

Now, use the eliminated equations. We have two equations: \( 5x + 3.5z = -6 \) and the original second equation \( 5x - 3y - 5z = 47 \). Use these to express one variable in terms of the other. Subtract the inflation-equation from the original to isolate variables.

Step 5: Substitute and Simplify

Use the resulting expressions from Step 4 to substitute back into one of the original equations to solve for \( y \). Simplifying leads to values for all three variables.

Key Concepts

Elimination Method

The elimination method is a powerful tool in solving systems of equations. It's like strategically removing one variable at a time to simplify the system. Imagine you have a team working on a project, and you want to focus on specific roles. You systematically delegate tasks to each person until everyone is working effectively. Here, you're eliminating one variable, letting you focus on others more clearly.

• Identify pairs of equations where you can eliminate a variable easily.
• Manipulate equations using addition or subtraction to cancel out a variable when they are added.
• Repeat the process to reduce the system, ideally to a point where you can solve straightforwardly.

In our solution, we paired equations that allowed us to eliminate the variable 'y'. This involved adjusting coefficients so the terms with 'y' cancel when equations are added.

Solving for Variables

Solving for variables in a system of equations is like putting pieces of a puzzle together. Each variable represents a piece, and our goal is to find which piece fits where. When tackling a system of equations:

• Try to isolate one of the variables first; this can be done by either elimination or substitution methods.
• Once one variable is isolated, substitute it back into other equations to find the remaining unknowns.

In solving our problem, eliminating 'y' left us with simpler equations in terms of 'x' and 'z'. We could then solve for one variable and backtrack to find others.

Linear Algebra

Linear algebra is like the language of systems of equations. It's about understanding the relationships between different lines, or, in our case, different equations. A system of linear equations can represent these relationships, often depicted graphically as lines or planes that can intersect.

• A single solution for the system corresponds to lines or planes intersecting at a point.
• No solution means they are parallel and never meet, while infinite solutions occur when equations describe the same line or plane.

In our exercise, we are working towards a single intersection point by using elimination to simplify and solve.

Substitution Method

The substitution method is like a detective solving a mystery, cracking the case one clue at a time. We take an equation and express one variable in terms of another. This reduced form is like the key clue that helps unlock the rest of the puzzle. How it works:

• Solve one equation for one variable in terms of others.
• Substitute this expression into the remaining equations.
• Continue simplifying and substituting until all variables are found.

Though we primarily used the elimination method in our provided solution, substitution became useful when some equations were simplified, and further solving required expressing one variable using another.