Problem 65 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use any method to solve the system. \(\left\\{\begin{array}{c}x-5 y=21 \\ 6 x+5 y=21\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = 6 \) and \( y = -3 \)

Step 1. Observing Necessary Cancellations

By adding the two equations directly, it is observed that the terms involving 'y' get canceled out, simplifying the problem greatly. The system of equations becomes: \( x + 6x = 21 + 21 \)

Step 2. Simplifying to Find 'x'

Combine like terms on both sides to find the value of 'x'. Thus: \( 7x = 42 \). Solving for 'x' gives: \( x = 42 / 7 \) , hence \( x = 6 \)

Step 3. Substituting 'x' to Find 'y'

Substitute the value of 'x' into the first equation of the system to find the value of 'y'. This gives: \( 6 - 5y = 21 \). Solving for 'y' gives: \( -5y = 21 - 6 \), then \( -5y = 15 \), hence \( y = 15 / -5 \), finally resulting in \( y = -3 \)

Key Concepts

Algebraic Methods

The term 'algebraic methods' in mathematics refers to a set of techniques used to solve equations, where unknown variables are represented by symbols. In the context of systems of equations - like the ones involving 'x' and 'y' in our exercise - algebraic methods are indispensable tools.

These methods include a variety of approaches, such as the substitution method, elimination method, graphical method, and matrix method. Each method has its particular use-case. For instance, the elimination method is especially useful when you can easily cancel out one variable by adding or subtracting the equations. In the original exercise, the elimination method is chosen intuitively by observing that adding the two equations will cancel out the 'y' variables, demonstrating the importance of a strategic approach to algebra.

Substitution Method

The substitution method is an algebraic technique particularly effective for systems of equations where one equation can be easily solved for one variable in terms of another. The basic strategy of the substitution method involves three key steps:

• First, solve one of the equations for one variable.
• Next, substitute the expression obtained into the other equation.
• Finally, solve the resulting equation to find the value of the second variable.

Returning to our exercise, once we find the value of 'x' using the elimination method, we could also apply the substitution method. After determining that \( x = 6 \), we substitute \( x \) into the first equation \( x - 5y = 21 \), simplifying to find the value of 'y'. This approach ensures that students fully understand the flexibility and interchangeability of algebraic methods.

Linear Equations

Linear equations form the foundation for much of algebra and consist of equations that graph as straight lines on the Cartesian plane. Any equation that can be rearranged into the standard form \( Ax + By = C \), where \( A \), \( B \), and \( C \) are constants, is considered a linear equation.

In our exercise, both \( x - 5y = 21 \) and \( 6x + 5y = 21 \) are linear equations. These can be recognized by the fact that 'x' and 'y' are to the first power only - no exponents, square roots, or other operations that would change the linearity of the equation. Understanding the properties of linear equations is key to recognizing patterns and employing the correct methods to solve them efficiently.