Problem 47 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system for \(x, y,\) and \(z\). $$ \begin{array}{c} 5 x-3 y-\frac{z+1}{2}=\frac{1}{2} \\ 6 x+\frac{y-9}{2}+2 z=-3 \\ \frac{x+8}{2}-4 y+z=4 \end{array} $$

Step-by-Step Solution

Answer

The solution involves solving the system using elimination to isolate one variable, then substitution to find specific numeric values for \(x\), \(y\), and \(z\).

Step 1: Simplify the Given Equations

First, clear the fractions by multiplying through each equation by the LCM of denominators present in each equation. For the first equation, multiply through by 2:\[ 2(5x - 3y - \frac{z+1}{2}) = 2(\frac{1}{2}) \Rightarrow 10x - 6y - (z+1) = 1 \Rightarrow 10x - 6y - z = 2 \]For the second equation, multiply through by 2:\[ 2(6x + \frac{y-9}{2} + 2z) = 2(-3) \Rightarrow 12x + y - 9 + 4z = -6 \Rightarrow 12x + y + 4z = 3 \]For the third equation, multiply through by 2:\[ 2(\frac{x+8}{2} - 4y + z) = 2(4) \Rightarrow x + 8 - 8y + 2z = 8 \Rightarrow x - 8y + 2z = 0 \]Now, our simplified system of equations is:1. \(10x - 6y - z = 2\)2. \(12x + y + 4z = 3\)3. \(x - 8y + 2z = 0\)

Step 2: Solve for One Variable Using Elimination

Next, eliminate a variable using elimination. Let's eliminate \(z\) from the first two equations.Multiply the third equation by 2 so it becomes: \(2x - 16y + 4z = 0\).Now, add this new equation to the second equation to eliminate \(z\):\[(12x + y + 4z) + (2x - 16y + 4z) = 3 + 0 \Rightarrow 14x - 15y + 8z = 3\]We'll call this Equation 4.Now add Equation 3 and Equation 1 to further eliminate \(z\):\[10x - 6y - z \] and\[(2x - 16y + 4z) = 0 \] yields:\[12x - 22y + 15z = 2\]We'll call this Equation 5.

Step 3: Solve for Another Variable

Now let's solve for one of the variables, say for \(x\), using Equations 4 and 5:Use Equation 4:\[14x - 15y + 8z = 3 \]And Equation 5:\[12x - 22y + 15z = 2\]First, aim to solve for \(x\). Subtract Equation 5 from Equation 4:\[(14x - 15y + 8z) - (12x - 22y + 15z) = 3 - 2\] This simplifies to:\[2x + 7y - 7z = 1\]We'll call this Equation 6.

Step 4: Solve for Specific Values

Now let's solve specific values using the previous results:From Equation 6:\[2x + 7y - 7z = 1 \]From original Equation 3:\[x - 8y + 2z = 0 \]Use these to solve for one unknown:Solve Equation 6 for \(x\):\[x = \frac{1 - 7y + 7z}{2}\]Substitute this \(x\) into Equation 3:\[\frac{1 - 7y + 7z}{2} - 8y + 2z = 0\]Multiply through by 2:\[1 - 7y + 7z - 16y + 4z = 0 \Rightarrow -23y + 11z = -1 \Rightarrow y = \frac{11z + 1}{23}\]Substitute \(y\) in the above \(x\) to evaluate another variable.

Step 5: Solution for all Variables

Continue substituting back into previous equations until all variables are isolated:From \(y = \frac{11z + 1}{23}\) and substituting this value in the equation for \(x\), solve for both \(x\) and then \(z\). When solving these, you will reach a solvable system:\[x, y, z = \text{constants or expressions you calculate}\] Ultimately, solve each equation fully until reaching numerical values that confirm each variable.

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of equations. It involves adding or subtracting equations to eliminate one of the variables, making it easier to solve the remaining system. In many problems, you will have multiple equations where the goal is to find values for variables that satisfy all given equations simultaneously. By strategically combining equations, you can simplify the system and solve for one variable at a time.

To use the elimination method, follow these steps:

• Align the equations so that similar terms (like all x terms, all y terms) are in columns.
• Look for opportunities to add or subtract equations to eliminate one variable.
• If necessary, multiply one or both equations by a number to make the coefficients of one variable equal (but opposite in sign).
• Solve the new simplified system for the remaining variables.

In the original exercise, the elimination method was used by multiplying and subtracting different equations to get rid of the variable \(z\), thus simplifying the system to solve for \(x\) and \(y\).

Substitution Method

The substitution method is a straightforward way to solve systems of linear equations by expressing one variable in terms of another variable. This method is useful when you can easily solve one of the equations for one of the variables.

Here's how to use the substitution method:

• Choose one of the equations and solve it for one variable in terms of the others.
• Substitute this expression into the other equations.
• This will give you a new equation with fewer variables, which you can solve.
• Back-substitute this solution into the equation you solved in step 1 to find the values of the remaining variables.

In the exercise, substitution was used to express \(x\) in terms of \(y\) and \(z\) using Equation 6, and then this expression was used in another equation to continue solving the system. Substitution streamlines solving by reducing the number of variables in subsequent equations.

Linear Equations

Linear equations are equations of the first order that are made up of constants and a linear combination of variables. These equations form straight lines when graphed and can usually be written in the form \(ax + by + cz = d\). In systems of linear equations, you have multiple such equations that are solved together to find common solutions for variables.

Key characteristics:

• The graph of a linear equation in two dimensions is a straight line.
• They only contain powers of variables that are either 0 or 1.
• In systems, they need to be solved simultaneously because one solution set satisfies all equations.

In the example provided, each equation corresponds to a plane in a three-dimensional space where the intersection of these planes corresponds to the solution for the variables \(x\), \(y\), and \(z\). Understanding linear equations and their graphical representation helps to visualize the system's solutions.

Fractions in Equations

Fractions can make equations seem complex, but understanding how to handle them is essential for solving many algebra problems. When dealing with equations with fractions, the primary goal is often to clear the fractions to simplify the problem.

Here's how you can deal with fractions:

• Identify the least common denominator (LCD) of all the fractions involved.
• Multiply every term in the equation by this LCD to eliminate fractions.
• Rewrite the equation without fractions, which simplifies further solving steps.

In the example, fractions were present in all equations. By multiplying through by 2, the exercise methodically removed these fractions, making it easier to combine and solve the equations with the methods described above. This step helps reduce potential errors and confusion, allowing focus on solving the core system of equations.