Problem 81 - Intermediate Algebra

Question

Solve the system for real solutions: \(\left\\{\begin{array}{l}\frac{1}{x}+\frac{3}{y}=4 \\\ \frac{2}{x}-\frac{1}{y}=7\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{7}{25}\), \(y = 7\).

Step 1: Express equations in terms of fractions

The given system of equations is \(\frac{1}{x} + \frac{3}{y} = 4\) and \(\frac{2}{x} - \frac{1}{y} = 7\). We begin by recognizing the fractional terms: \(\frac{1}{x}\) and \(\frac{1}{y}\). Let's set \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). By doing this substitution, the equations become: \(u + 3v = 4\) and \(2u - v = 7\). This reformulation helps us work with linear equations rather than fractions.

Step 2: Solve the linear system for u and v

We now have the system: \(u + 3v = 4\) and \(2u - v = 7\). First, solve the first equation for \(u\): \(u = 4 - 3v\). Substitute \(u\) from the first equation into the second equation: \((2(4 - 3v)) - v = 7\). Simplify: \(8 - 6v - v = 7\), which gives \(-7v = -1\), hence \(v = \frac{1}{7}\). Substitute \(v = \frac{1}{7}\) back into \(u = 4 - 3v\), which results in \(u = 4 - 3\left(\frac{1}{7}\right) = \frac{25}{7}\).

Step 3: Convert u and v back to x and y

Recall that \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). So, \(\frac{1}{x} = \frac{25}{7}\) which implies \(x = \frac{7}{25}\). Similarly, \(\frac{1}{y} = \frac{1}{7}\) implies \(y = 7\).

Key Concepts

Substitution Method

The substitution method is a strategy for solving systems of equations, where one equation is solved for one variable and then substituted into the other equation. This method transforms the original system of equations into a simpler form, making it easier to find the solution.

In our problem, we begin by identifying that both equations include fractional terms: \(\frac{1}{x}\) and \(\frac{1}{y}\). By introducing new variables, \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\), we simplify the equations to avoid working directly with fractions. This substitution allows us to rewrite the original equations into:

• \(u + 3v = 4\)
• \(2u - v = 7\)

Now, we can solve the system of equations using traditional algebraic methods.Once \(u\) and \(v\) are found, they are converted back into \(x\) and \(y\) using their definitions. This highlights the flexibility and power of the substitution method, which converts complex systems into more manageable forms.

Linear Equations

Linear equations are algebraic expressions that represent straight lines when graphed. They take the form of \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Solving a system of linear equations is a common mathematical task that involves finding values for the unknown variables that satisfy all equations in the system.

In this exercise, after substitution, our equations \(u + 3v = 4\) and \(2u - v = 7\) form a simple system of linear equations. We can select one of these equations, solve for one variable, and then replace this variable in the other equation, which is a classic approach to solving linear systems.

This method works smoothly because linear equations are predictable and have clear, straightforward solutions. Once we find the values for \(u\) and \(v\) here, substituting them back to find \(x\) and \(y\) is simple and logical.

Fractional Equations

Fractional equations include terms where the variables are in the denominator, such as \(\frac{1}{x}\). These equations can appear complex at first glance due to the fraction format. However, simplifying fractional equations can often be accomplished by suitable substitutions that convert them into linear equations.

In our specific example, each term in the system is a fractional expression. By setting \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\), we transform these fractional equations into linear ones, making the solution process more straightforward and less error-prone.

This method streamlines the calculation by avoiding direct manipulation of fractions, thus maximizing clarity and ease of solving the system. Converting back to the original variables after finding \(u\) and \(v\) ensures comprehensive understanding, providing a neat solution to the equations that initially seemed challenging due to their fractional form.