Problem 26 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{array}{l} \frac{x+3}{2}-\frac{y-1}{2}+\frac{z+2}{4}=\frac{3}{2} \\ \frac{x-5}{2}+\frac{y+1}{3}-\frac{z}{4}=-\frac{25}{6} \\ \frac{x-3}{4}-\frac{y+1}{2}+\frac{z-3}{2}=-\frac{5}{2} \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -2.5\), \(y = -1.5\), and \(z = -3.5\).

Step 1 - Clearing the fractions

Multiplying each equation by the least common multiple of its denominators to get rid of the fractions. In this case, for equation 1 LCM is 4, for equation 2, LCM is 12 and for equation 3, LCM is 4. Thus new equations will appear as follows: \(2x + 6 - 2y + 2 + z + 2 = 6 \) \[ln]\(6x - 30 + 4y + 4 - 3z = -50 \) \[ln]\(x - 3 - 2y - 2 + 2z - 6 = -10 \)

Step 2 - Simplifying the equations

After simplifying, we have: \(2x - 2y + z = -2 \) \[ln]\(6x + 4y - 3z = -16 \) \[ln]\(x - 2y + 2z = -9 \)

Step 3 - Solve the system of equations

To find \(x\), \(y\), and \(z\), there are multiple methods such as substitution method and Gaussian elimination method. In here, we use the substitution method to eliminate \(y\) and \(z\) and solve for \(x\), by substituting the obtained \(y\) and \(z\) in 2nd and 3rd equations respectively.

Step 4 - Solve for x

By multiplying equation 1 by 3 and adding to equation 2, we get \(4x = -10\). Hence, \(x = -\frac{10}{4} = -2.5\)

Step 5 - Solve for y

Then substitute \(x = -2.5\) in equation 1, \(2(-2.5) - 2y + (-5) = -2 \Rightarrow -5 - 2y = -2 \Rightarrow 2y = -3 \Rightarrow y = -\frac{3}{2} = -1.5\)

Step 6 - Solve for z

Similarly, substitute \(y=-1.5\) and \(x=-2.5\) into equation 3, \(-2.5 - 2(-1.5) + 2z = -9 \Rightarrow -2 + 2z = -9 \Rightarrow 2z = -7 \Rightarrow z = -\frac{7}{2} = -3.5\)

Key Concepts

Substitution Method

The substitution method is one of the fundamental techniques for solving systems of equations. The idea is simple: solve one of the equations for one variable, and then substitute that expression into the other equation(s).

• First, we solve one equation for one of the variables. This can be either x, y, or z, depending on which is easiest. In many cases, we look for equations where a particular variable has a coefficient of one.
• Next, we substitute that expression into the other equations. This reduces the number of variables in those equations, making them easier to manage.
• We continue this process until we solve for all variables.

The key advantage of the substitution method is its simplicity. However, it can become cumbersome when dealing with larger systems of equations or complex expressions. Despite this, substitution remains a foundational step towards building understanding of more complex methods, such as Gaussian elimination.

Gaussian Elimination

Gaussian elimination is a powerful technique designed for solving systems of linear equations. It involves performing operations to transform the system into an upper triangular matrix. Once in this form, back substitution is used to find the solution for each variable.

• The process begins by selecting a pivot element, typically the first element of the first column. Row operations are used to make zeroes in all positions below the pivot.
• Next, the method proceeds column by column, eliminating variables from additional rows by using similar row operations.
• Once the system is in upper triangular form, variables are solved starting from the bottom row and moving upwards, known as back substitution.

Gaussian elimination is particularly useful for larger systems as it is methodical and consistent. It provides a clear path to solutions even for complex systems.

Least Common Multiple (LCM)

The least common multiple (LCM) is the smallest positive integer that is divisible by each of a given set of numbers. When dealing with a system of equations featuring fractions, it helps to clear these fractions by multiplying through by the LCM of the denominators.

• Identify the denominators in the equations that need simplifying.
• Calculate the LCM of these denominators. For example, if the denominators are 2 and 3, the LCM is 6.
• Multiply each equation by the LCM, which results in removing the fractions, simplifying the equations for further manipulation.

Using the LCM is a straightforward way to simplify initial equations, making solving them, whether through substitution or Gaussian elimination, more feasible.

Clearing Fractions

Clearing fractions is a step often used to simplify equations and make them easier to solve. When equations contain fractions, dealing with common denominators adds complexity.

• To begin clearing fractions, multiply each term of the equation by the least common multiple (LCM) of the denominators. This conversion transforms the equation into a form without fractions.
• The resulting equations are generally simpler and more accessible for further operations, like substitution or Gaussian elimination.
• This technique, while basic, makes the entire solving process more streamlined and opens the door to using more advanced solving methods efficiently.

Mastering the process of clearing fractions aids in tackling a wide array of algebraic problems more effectively and is a critical step, ensuring equations are in their simplest form before applying various solving techniques.