Problem 20 - College Algebra and Calculus: An Applied Approach

Question

Solve the system by elimination Then state whether the system is consistent inconsistent. $$\left\\{\begin{array}{l}5 u+6 v=24 \\ 3 u+5 v=18\end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of linear equations is u = 6/7, v = 18/7. The system is consistent.

Step 1: Multiply the equations by suitable numbers

The goal is to have the coefficient of one of the variables identical in both equations, to enable elimination. To achieve this, multiply equation 1 by 3 and equation 2 by 5. Then we have:\[\begin{array}{l}15 u+18 v=72 \ 15 u+25 v=90\end{array}\]

Step 2: Subtract the equations

Now, subtract equation 2 from equation 1 to eliminate u. The result is:\[0u - 7v = -18\]Simplify this to get the value of v:\[v = \frac{18}{7}\]

Step 3: Subtitute v into the original equation

Substitute v = 18/7 into either of the original equations to solve for u. Let's use the first one:\[ 5u + 6(\frac{18}{7}) = 24 \]After simplifying this, we derive u to be:\[ u = \frac{6}{7} \]

Step 4: Check if the system is consistent or inconsistent

Finally, substitute u = 6/7 and v = 18/7 into both equations to check if they hold true. If they do, the system is consistent; if not, it's inconsistent. Here they hold true, ergo the system of equations is consistent.

Key Concepts

Elimination Method in Solving Systems of Equations

The elimination method is a popular technique for solving systems of linear equations. It works by strategically eliminating one of the variables, allowing you to solve a simpler equation. This is done by Adjusting the coefficients of one of the variables to be equal and then subtracting or adding the equations.

For example, consider the system given:

• 5u + 6v = 24
• 3u + 5v = 18

To use elimination, multiply each equation by numbers that make the coefficients of either \( u \) or \( v \) identical. In this case, we multiply the first equation by 3 and the second by 5 to get:

• 15u + 18v = 72
• 15u + 25v = 90

Subtracting these equations results in elimination of the \( u \) variable. This gives us a simpler equation in \( v \), which we can solve directly.

Ultimately, elimination method is reliable for finding solutions without requiring graphical insights, adhering to clear algebraic transformations.

Consistent Systems of Equations

A system of equations is termed as consistent if it has at least one solution. This can indicate one solution, many solutions, but never zero solutions. A consistent system points to intersecting or overlapping lines when plotted graphically, showing a point or multiple points of intersection.

With the solved example, after applying the elimination method, we found specific values for \( u \) and \( v \):

• \( u = \frac{6}{7} \)
• \( v = \frac{18}{7} \)

To verify consistency, these values are substituted back into the original equations. If both equations hold true for these variables, it confirms that the system is consistent.

This means, the unique solution satisfies both equations, demonstrating they intersect at this single point in two-dimensional space.

Algebraic Solutions for Linear Systems

Algebraic solutions involve the process of finding values for variables that satisfy a given set of equations. For linear systems, these involve either substitution, elimination, or matrix methods to arrive at the solution.

In the exercise given, the algebraic solution was found using elimination, resulting in:

• \( u = \frac{6}{7} \)
• \( v = \frac{18}{7} \)

Algebraic solutions offer precise answers through manipulations and operations on the coefficients and constants involved. Such solutions help in understanding relationships between variables and can be verified by substituting the solutions back into original equations.

This approach is especially crucial in fields requiring exact calculations, such as engineering and physics, where accuracy is essential.