Problem 99 - A Graphical Approach to College Algebra

Question

Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Solve the given system for \(x\) and \(y\) by using the equations relating \(t\) to \(x\) and \(u\) to \(y\).

Step-by-Step Solution

Answer

The solution is \( x = 5 \) and \( y = 1 \).

Step 1: Substitute Variables

Start by substituting \( t = \frac{1}{x} \) and \( u = \frac{1}{y} \) into the original equations. This gives the system: 1. \( 5t + 15u = 16 \) 2. \( 5t + 4u = 5 \).

Step 2: Solve for One Variable

Subtract equation 2 from equation 1 to eliminate \( t \): \((5t + 15u) - (5t + 4u) = 16 - 5\) This simplifies to: \( 11u = 11 \). Therefore, solve for \( u \): \( u = 1 \).

Step 3: Solve for the Other Variable

Now, substitute \( u = 1 \) back into the second equation: \( 5t + 4(1) = 5 \) This simplifies to: \( 5t + 4 = 5 \) Solve for \( t \): \( 5t = 1 \) \( t = \frac{1}{5} \).

Step 4: Solve for Original Variables

Now we have expressions for \( t \) and \( u \): \( t = \frac{1}{5} \) and \( u = 1 \). Use these to find \( x \) and \( y \): - Since \( t = \frac{1}{x} \), then \( x = \frac{1}{t} = 5 \) - Since \( u = \frac{1}{y} \), then \( y = \frac{1}{u} = 1 \).

Step 5: Verify the Solution

Substitute \( x = 5 \) and \( y = 1 \) back into the original equations to check the solutions: 1. \( \frac{5}{5} + \frac{15}{1} = 1 + 15 = 16 \), which satisfies the first equation. 2. \( \frac{5}{5} + \frac{4}{1} = 1 + 4 = 5 \), which satisfies the second equation.

Key Concepts

Substitution Method

The Substitution Method is a popular technique in solving systems of equations. It involves solving one equation for one variable, and then substituting this expression into another equation. This method is used when one of the equations can be easily solved for a single variable.
For example, if you have equations with variables like in our original problem, you can substitute simpler expressions like fractions or single-variable expressions to make calculations more straightforward.
After substitution, the system becomes simpler, often linear, allowing straightforward solving of one variable. Once this variable is found, you substitute it back into one of the original equations to find the other variable.

Nonlinear Systems

Nonlinear Systems refer to equations where variables are not raised to the first power. These systems can include variables in denominators, squares, or any other non-linear operations.
Our original system is nonlinear because the variables appear in the denominators: \( \frac{5}{x} + \frac{15}{y} = 16 \) and \( \frac{5}{x} + \frac{4}{y} = 5 \).
To work with such systems, we need special strategies, like the substitution method, which can transform these equations into a simpler form. It's important to convert the system into a format that is manageable, often turning it into a pseudo-linear system temporarily for solving.

Variable Transformation

Variable Transformation is a technique used to simplify complex equations by introducing new variables. In our problem, this meant setting \( t = \frac{1}{x} \) and \( u = \frac{1}{y} \).
This transformation turns a nonlinear equation into a linear one, making it easier to solve. For instance, \( \frac{5}{x} \) becomes \( 5t \) and \( \frac{15}{y} \) becomes \( 15u \).
By transforming variables, we essentially rewrite the problem in a more familiar form. Once solved, we transform back to solve for the original variables, ensuring that we understand the complete context of the transformed equations.