Problem 98 - 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 system in Exercise 97 for \(t\) and \(u\).

Step-by-Step Solution

Answer

The solution is \( t = \frac{1}{5} \) and \( u = 1 \).

Step 1: Substitute Variables

Let's start by making the substitution as mentioned. Let \( t = \frac{1}{x} \) and \( u = \frac{1}{y} \). Substitute these into the original system of equations. The first equation becomes \( 5t + 15u = 16 \). The second equation becomes \( 5t + 4u = 5 \). Now, we have a new system of linear equations:\[\begin{align*} 5t + 15u &= 16 \ 5t + 4u &= 5 \end{align*}\]

Step 2: Eliminate Variable t

To eliminate \( t \), subtract the second equation from the first equation:\[ (5t + 15u) - (5t + 4u) = 16 - 5 \]This simplifies to:\[ 11u = 11 \]

Step 3: Solve for Variable u

Solve the equation for \( u \):\[ u = \frac{11}{11} = 1 \]

Step 4: Substitute u into One of the Equations

Substitute \( u = 1 \) back into the second equation:\[ 5t + 4(1) = 5 \]This simplifies to:\[ 5t + 4 = 5 \]

Step 5: Solve for Variable t

Rearrange the equation to solve for \( t \):\[ 5t = 5 - 4 \]\[ 5t = 1 \]\[ t = \frac{1}{5} \]

Step 6: Present the Solution

Thus, the solutions for \( t \) and \( u \) are \( t = \frac{1}{5} \) and \( u = 1 \).

Key Concepts

Solving Non-Linear Systems

Solving non-linear systems may sound intimidating, but with the right approach, it becomes manageable. In general, a non-linear system consists of two or more equations which involve non-linear relationships between variables. Non-linear refers to equations where variables are not just added or multiplied by constants, but are, for example, in the denominator or within roots.

The exercise provided is a non-linear system because the variables appear in the denominators. When faced with such problems, the key is transforming the non-linear system to a linear-friendly form using clever techniques like substitution. The aim is to break down the original structure into a more familiar linear one, making it easier to solve using standard linear techniques, such as elimination or substitution. By performing the substitution, you temporarily simplify the equations so they no longer contain variables in non-linear forms.

In our exercise, by substituting new variables for the reciprocals, the non-linear system becomes linear, allowing us to use traditional linear methods to find a solution. Ultimately, solving these transformed linear systems unlocks the solution to the original non-linear setup.

Variable Substitution

Variable substitution is a pivotal strategy when dealing with complicated systems of equations, especially non-linear ones. The concept revolves around swapping complex expressions with simpler variables, easing the process of solving the equations.

In the given exercise, variables were present in the denominators. Tackling this directly could be difficult and non-intuitive. Hence, the substitution method was adopted: replacing \( \frac{1}{x} \) with \( t \) and \( \frac{1}{y} \) with \( u \). This shifts the problem into linear territory, simplifying it significantly.

Key benefits of variable substitution include:

• Transformation of complex or non-linear systems into linear forms.
• Reduction of computational effort by simplifying expressions.
• A more straightforward approach to identify solutions.

This substitution methodology can often reveal hidden relationships within the equations and is a powerful tool in any mathematician's arsenal.

System of Equations

A system of equations is a set of two or more equations with the same variables. These can be either linear or non-linear. The goal is to find an intersection point, which means a set of values for the variables that simultaneously satisfy all equations.

For this particular exercise, the system started off as non-linear due to variable positions in denominators. A substitution transformed it into a linear system, allowing us to proceed with more conventional solving methods. The transformed system consisted of:

• \( 5t + 15u = 16 \)
• \( 5t + 4u = 5 \)

By performing variable elimination, where you cancel out one of the variables by combining the equations, solutions for each new variable can be found. In this exercise, once \( u \) was determined, it was substituted back to find \( t \).

The importance of understanding systems of equations lies in their ability to solve real-world problems where multiple conditions must be satisfied simultaneously. Understanding systems, and converting them to more workable forms if needed, is crucial in fields ranging from economics to physics.