Problem 80 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solutions are \( x = 3 \) and \( y = 3 \).

Step 1: Set up the Equations

We have two equations: \( \frac{1}{x} + \frac{2}{y} = 1 \) and \( \frac{2}{x} - \frac{1}{y} = \frac{1}{3} \). These are already arranged in the standard form for solving simultaneous equations.

Step 2: Introduce Substitutions

Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Thus, the system becomes: \( u + 2v = 1 \) and \( 2u - v = \frac{1}{3} \). These equations are linear in terms of \( u \) and \( v \).

Step 3: Solve the System of Equations for u and v

Rewrite the first equation as \( u = 1 - 2v \) and substitute into the second equation: \[ 2(1 - 2v) - v = \frac{1}{3} \]. Simplifying yields: \( 2 - 4v - v = \frac{1}{3} \). Thus, \( 2 - 5v = \frac{1}{3} \).

Step 4: Solve for v

Subtract 2 from both sides: \( -5v = \frac{1}{3} - 2 \). This simplifies to \( -5v = \frac{1}{3} - \frac{6}{3} = -\frac{5}{3} \). Dividing both sides by -5 gives \( v = \frac{1}{3} \).

Step 5: Solve for u

Substitute \( v = \frac{1}{3} \) back into \( u = 1 - 2v \). This gives \( u = 1 - 2\left(\frac{1}{3}\right) = \frac{1}{3} \).

Step 6: Solve for x and y

Recall that \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Thus, \( \frac{1}{x} = \frac{1}{3} \) and \( \frac{1}{y} = \frac{1}{3} \), giving \( x = 3 \) and \( y = 3 \).

Key Concepts

Understanding Linear Equations

Linear equations are fundamental components in algebra that represent relationships between variables. A linear equation in two variables is typically written in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables. These equations graph as straight lines in a coordinate plane.

• When we solve systems of linear equations, our goal is to find the set of values for \( x \) and \( y \) that makes both equations true simultaneously.
• These values are called the "solutions" of the system.
• The intersection point of the lines in the graph corresponds to these solutions.

In the provided exercise, we transformed nonlinear equations by introducing substitutions, resulting in a system of linear equations in terms of new variables \( u \) and \( v \). This transformation is crucial for simplifying and solving complex equations.

The Substitution Method

The substitution method is a powerful technique used to solve systems of equations. It involves solving one equation for one variable and plugging that expression into the other equation.

• Begin by solving one of the equations for one variable, in terms of the other.
• Next, substitute this expression into the second equation.
• This substitution reduces the system to a single equation with one variable, making it easier to solve.

In our example, we solved for \( u \) in terms of \( v \) from the first equation \( u = 1 - 2v \), and then substituted this into the second equation. This allowed us to find \( v \) directly, and subsequently find \( u \). This technique simplifies the process of dealing with simultaneous equations, particularly when they initially appear in a complex form.

Finding Real Solutions

Real solutions of a system are the values that satisfy all given equations within the real number system. They represent measurable, tangible aspects often sought in practical applications.

• Real solutions must meet the conditions set by the equations, without involving imaginary or complex numbers.
• For example, in our solution, once we arrived at \( u = \frac{1}{3} \) and \( v = \frac{1}{3} \), we substituted back to find real values for \( x \) and \( y \) as 3 and 3 respectively.
• In practical terms, this means \( (x, y) = (3, 3) \) is the point where the lines intersect on a coordinate plane.

Thus, any solutions where both \( x \) and \( y \) are real numbers, and which satisfy all original equations, are considered real solutions of that system.