Problem 48 - College Algebra with Modeling and Visualization

Question

Solve the system, if possible. $$y+2 z=-5$$ $$3 x \quad-2 z=-6$$ $$-x-4 y \quad=11$$

Step-by-Step Solution

Answer

The solution is \( x = -3 \), \( y = -2 \), \( z = -\frac{3}{2} \).

Step 1: Label the Equations

Let's label the given equations for reference:1. \( y + 2z = -5 \) 2. \( 3x - 2z = -6 \) 3. \( -x - 4y = 11 \)

Step 2: Solve for x from Equation 2

From equation 2, \( 3x - 2z = -6 \), solve for \( x \):\[ 3x = 2z - 6 \]\[ x = \frac{2z - 6}{3} \]

Step 3: Solve for y from Equation 1

From equation 1, \( y + 2z = -5 \), solve for \( y \):\[ y = -5 - 2z \]

Step 4: Substitute y into Equation 3

Substitute \( y = -5 - 2z \) from Step 3 into equation 3:\[ -x - 4(-5-2z) = 11 \]

Step 5: Simplify Equation 3

Simplify the equation:\[ -x + 20 + 8z = 11 \]\[ -x + 8z = -9 \] (after subtracting 20 from both sides)

Step 6: Substitute x from Step 2

Now substitute \( x = \frac{2z - 6}{3} \) from Step 2 into the simplified equation:\[ -\left(\frac{2z - 6}{3}\right) + 8z = -9 \]

Step 7: Solve for z

Multiply the whole equation by 3 to eliminate the fraction:\[ - (2z - 6) + 24z = -27 \]Simplify:\[ -2z + 6 + 24z = -27 \]\[ 22z = -33 \]\[ z = -\frac{33}{22} = -\frac{3}{2} \]

Step 8: Solve for y using z value

Substitute \( z = -\frac{3}{2} \) back into \( y = -5 - 2z \):\[ y = -5 - 2\left(-\frac{3}{2}\right) \]\[ y = -5 + 3 \]\[ y = -2 \]

Step 9: Solve for x using z value

Substitute \( z = -\frac{3}{2} \) back into \( x = \frac{2z - 6}{3} \):\[ x = \frac{2(-\frac{3}{2}) - 6}{3} \]\[ x = \frac{-3 - 6}{3} \]\[ x = \frac{-9}{3} \]\[ x = -3 \]

Step 10: Verify the Solutions

Substitute \( x = -3 \), \( y = -2 \), and \( z = -\frac{3}{2} \) back into the original equations to verify:- For \( y + 2z = -5 \): \( -2 + 2(-\frac{3}{2}) = -5 \) correct.- For \( 3x - 2z = -6 \): \( 3(-3) - 2(-\frac{3}{2}) = -6 \) correct.- For \( -x - 4y = 11 \): \( -(-3) - 4(-2) = 11 \) correct.

Key Concepts

Substitution Method

The substitution method is a technique used in solving systems of equations. It involves isolating one variable in terms of the others and then substituting this expression into the remaining equations. This method is particularly useful when it is straightforward to solve for one variable. In our problem, we begin by solving one equation for a specific variable and then substitute that expression into the other equations.

• Find an equation with one variable isolated or easy to isolate.
• Solve this equation for that variable.
• Substitute the expression into the other equations, simplifying them as you proceed.

This iterative process reduces the system to one that eventually has each variable solved in terms of basic numbers or known quantities. In our exercise, the variable substitution was performed neatly across a few steps, gradually simplifying the system.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. At its core, linear algebra allows us to solve systems of linear equations like the one given.

The equations in linear algebra can be seen both as lines in coordinate geometry, or as row vectors. Here are some points to understand better:

• A linear equation represents a line in a two-dimensional plane.
• When several equations are considered together, they form a system, which can be visualized as multiple lines.
  

In this particular problem, we deal directly with linear equations in three variables, where each equation describes a plane in a three-dimensional space. The solution of this system is the point where all planes intersect.

Equation Solving

Solving equations is a fundamental skill in algebra and by extension, any mathematical application. It involves finding all possible values of the variables that make the equation true.
The process can be broken down into several manageable steps:

• Rearrange the equation to isolate the variable you're solving for.
• Perform any necessary simplifications, such as combining like terms or multiplying through by a common factor to clear fractions
  .

Solving a system, such as the one in this exercise, typically involves manipulating equations to express variables in terms of others and substituting these expressions back. It's a systematic effort to reduce complexity until each variable is independently solved.

Verification of Solutions

Verification is an essential step to ensure that a solution is correct. It involves substituting the determined values back into the original equations to confirm that they satisfy each equation.
It's a straightforward but important step in equation solving:

• Take each solution and plug it back into the original equations.
• Ensure the left-hand side equals the right-hand side for all equations.

For our system, the verification process confirmed that our solution values for \(x = -3\), \(y = -2\), and \(z = -\frac{3}{2}\) satisfied all the original equations correctly. This reassures us that no mistakes were made during the calculation process.