91Ó°ÊÓ

A system of equations consists of two or more equations with the same variables. When solving such a system, the goal is to find values that satisfy all equations simultaneously. In this exercise, we worked with two equations of the form:

• \( \frac{x}{3} + \frac{2y}{3} = 2 \)
• \( \frac{x}{2} - 2y = \frac{5}{2} \)

To solve these, we need to find values for \( x \) and \( y \) that make both equations true. This process often involves using methods like elimination or substitution, especially when the equations involve fractions. By converting to a form without fractions, there’s less complexity, making it easier to apply algebraic techniques and find solutions.

Fractions can make algebraic equations more complex, but they’re very common. A fraction consists of a numerator, the top part, and a denominator, the bottom part. When dealing with equations that have fractions, the first step is often to clear the fractions, making the equation easier to handle. For the given system:

• The first equation had denominators of 3.
• The second equation had a denominator of 2 in one term.

By multiplying every term by these denominators, we eliminated the fractions. For example, multiplying the first equation by 3 transformed \( \frac{x}{3} + \frac{2y}{3} = 2 \) into \( x + 2y = 6 \). This simplification is crucial as it allows you to focus on solving the system using straightforward algebraic methods instead of getting bogged down by fractional terms.

The elimination method involves adding or subtracting equations to remove one of the variables, making it easier to solve for the other. In our exercise, after clearing fractions, we had:

• \( x + 2y = 6 \)
• \( x - 4y = 5 \)

To eliminate \( x \), we subtracted the second equation from the first. Doing this left us with just \( y \), specifically, \( 6y = 1 \). This is a quick and reliable way to solve a system of equations where equations are already nicely aligned, as it drastically reduces the complexity of the system, eventually leading to the solution in fewer steps.

Though we did not use the substitution method heavily in this particular solution, it is an essential technique for solving systems where one of the variables can be easily expressed in terms of the other. Let's see how it works: once you solve one of the equations for one variable, you substitute that expression into the other equation.For example, if we had used substitution here, we could express \( x \) from one of our simplified equations, say \( x = 6 - 2y \), and then replace \( x \) in the other equation, where \( x = 5 + 4y \). This transforms a complex system into a simpler equation with just one variable. While this method is not always the quickest, it provides a clear path when equations are convenient to substitute.