Problem 28 - Precalculus: Mathematical for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 3. $$\left\\{\begin{array}{rr} u-30 v= & -5 \\ -3 u+80 v= & 5 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is (25, 1).

Step 1: Write the System in Matrix Form

The given system of linear equations is: \[ \begin{array}{r} u - 30v = -5 \ -3u + 80v = 5 \end{array} \] This can be written in matrix form as \( A \mathbf{x} = \mathbf{b} \), where \( A = \begin{bmatrix} 1 & -30 \ -3 & 80 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} u \ v \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} -5 \ 5 \end{bmatrix} \).

Step 2: Solve for One Variable

Start by solving the first equation for \( u \): \[ u = 30v - 5 \] Substitute this expression into the second equation: \[ -3(30v - 5) + 80v = 5 \] Expand and simplify: \[ -90v + 15 + 80v = 5 \] Combine like terms: \[ -10v + 15 = 5 \]

Step 3: Solve for the Other Variable

Let's solve for \( v \) by isolating it in the simplified equation from Step 2:\[ -10v + 15 = 5 \]Subtract 15 from both sides:\[ -10v = -10 \]Divide both sides by -10:\[ v = 1 \]

Step 4: Plug Back to Find the Other Variable

Now that \( v = 1 \), substitute back to find \( u \) using the expression from Step 2, \( u = 30v - 5 \):\[ u = 30(1) - 5 = 30 - 5 = 25 \]

Step 5: Write the Solution as an Ordered Pair

With \( u = 25 \) and \( v = 1 \), the solution to the system in ordered-pair form is \((u, v) = (25, 1)\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that involve two or more variables. The aim is to find a set of values for the variables that satisfy all the equations simultaneously. These equations can represent anything from equations of lines, planes, or even more complex structures. In the given exercise, we have two linear equations involving variables \(u\) and \(v\): - \(u - 30v = -5\) - \(-3u + 80v = 5\) To solve a system of equations, several methods can be used, including substitution, elimination, and graphing. The solution can be a single set of values, no solution, or infinitely many solutions. This exercise guides through transforming the equations into a manageable form for solving.

Matrix Form

Putting a system of equations into matrix form allows for a structured way of handling and solving the equations. This involves using matrices to represent the system compactly and efficiently. Here's how the given system is represented in matrix form: - Coefficient matrix \( A = \begin{bmatrix} 1 & -30 \ -3 & 80 \end{bmatrix}\) - Variable matrix \( \mathbf{x} = \begin{bmatrix} u \ v \end{bmatrix}\) - Constant terms matrix \( \mathbf{b} = \begin{bmatrix} -5 \ 5 \end{bmatrix}\) Thus, the matrix equation takes the form \( A \mathbf{x} = \mathbf{b} \). This compact form allows us to apply linear algebra techniques for finding solutions, making the process systematic and generally applicable to systems of any size.

Ordered Pair Solution

The solution of a system of equations involving two variables is commonly represented as an ordered pair. It indicates the values for each variable that satisfy all the equations in the system. For our exercise, after solving the system, we found the ordered pair \((u, v) = (25, 1)\). - The first element of the pair corresponds to \(u\). - The second element corresponds to \(v\). Representing solutions in ordered pairs is not just a matter of convention. It also provides a clear and concise way to demonstrate the exact values that make each equation true. In practical scenarios, this method of expressing solutions can be used in graph interpretations, where the ordered pair represents the intersecting point of two lines in a coordinate plane.

Variable Substitution

Variable substitution is a powerful method for solving systems of equations. It involves expressing one variable in terms of another, which can simplify the system and make it easier to solve. In this exercise, we start by isolating \(u\) in the first equation: - \(u = 30v - 5\) This expression is used to substitute \(u\) in the second equation, turning a two-variable equation into a single-variable one: - \(-3(30v - 5) + 80v = 5\) This step is crucial as it reduces complexity, allowing us to solve directly for \(v\). Once \(v\) is determined, substitute back to find \(u\). The substitution process is iterative and helps to break down the initial problem into smaller, tractable parts. It is especially useful in systems with many equations and variables.