Problem 2 - Precalculus with Calculus Previews

Question

Solve the given linear system. State whether the system is consistent, with independent or dependent equations, or whether it is inconsistent. $$ \left\\{\begin{array}{l} 2 x-2 y=1 \\ 3 x+5 y=11 \end{array}\right. $$

Step-by-Step Solution

Answer

The system has a unique solution: \( x = \frac{27}{16}, y = \frac{19}{16} \); consistent, independent.

Step 1: Write the System of Equations

We need to solve the following linear system of equations:\[\begin{cases}2x - 2y = 1 \3x + 5y = 11 \end{cases}\]

Step 2: Solve for x in the First Equation

Start by solving the first equation for \(x\) in terms of \(y\):\[2x - 2y = 1\]Add \(2y\) to both sides:\[2x = 2y + 1\]Divide each term by 2:\[x = y + \frac{1}{2}\]

Step 3: Substitute x into the Second Equation

Substitute \(x = y + \frac{1}{2}\) into the second equation:\[3(y + \frac{1}{2}) + 5y = 11\]This simplifies to:\[3y + \frac{3}{2} + 5y = 11\]

Step 4: Simplify and Solve for y

Combine like terms:\[8y + \frac{3}{2} = 11\]Subtract \(\frac{3}{2}\) from both sides:\[8y = 11 - \frac{3}{2}\]Convert 11 to \(\frac{22}{2}\):\[8y = \frac{22}{2} - \frac{3}{2}\]\[8y = \frac{19}{2}\]Divide by 8:\[y = \frac{19}{16}\]

Step 5: Substitute back to find x

Use \(y = \frac{19}{16}\) in \(x = y + \frac{1}{2}\):\[x = \frac{19}{16} + \frac{1}{2}\]Convert \(\frac{1}{2}\) to \(\frac{8}{16}\):\[x = \frac{19}{16} + \frac{8}{16} = \frac{27}{16}\]

Step 6: Analyze the Solution

We have found the solution \(x = \frac{27}{16}\) and \(y = \frac{19}{16}\). Since a unique solution exists, the system is consistent with independent equations.

Key Concepts

Consistent vs Inconsistent Systems

A system of linear equations can either be consistent or inconsistent. Consistent systems are those that have at least one solution, whereas inconsistent systems have none. This classification is crucial in understanding whether or not the equations can be solved simultaneously.

A consistent system can further be divided into:

• Independent: The system has exactly one solution. This occurs when the graphs of the equations intersect at a single point.
• Dependent: The system has infinitely many solutions. This happens when the equations describe the same line, so they overlap completely.

In the given exercise, we find the solution as \( x = \frac{27}{16} \) and \( y = \frac{19}{16} \). Since there is one distinct solution, it's a consistent system with independent equations.

Independent vs Dependent Equations

When working with systems of equations, understanding the difference between independent and dependent equations helps to identify the nature of the solutions.

• Independent Equations: These equations meet at exactly one solution point, indicating each equation contributes novel information to solve the system. The slope and intercept for each equation are different, which means the lines will intersect at one point.
• Dependent Equations: These equations are essentially the same line, meaning one is a multiple or manipulation of the other. They result in infinitely many solutions because they overlap, having the same slope and intercept.

In the exercise example, the solution is unique with precise values for both variables, indicating independent equations that intersect at this point.

Solving Linear Equations

The process of solving linear equations involves finding values for the variables that satisfy all given equations in the system. There are several methods for solving them, with substitution and elimination being the most common.

Substitution Method:1. Solve one equation for one variable in terms of the others.2. Substitute this expression into the other equation(s).3. Solve the simplified equation for the remaining variable.4. Substitute back to find the other variable.

In the exercise, the substitution method was used. First, solve the first equation for\( x \) as \( x = y + \frac{1}{2} \), then substitute this into the second equation to solve for\( y \). This leads to finding specific values for both\( x \) and\( y \). This clear, step-by-step method ensures a correct solution to the linear system.