Checking Solutions for Both Equations

Introduction

Key Concepts

Understanding Simultaneous Linear Equations

Simultaneous linear equations consist of two or more linear equations with the same set of variables. The solution to these equations is the set of variable values that satisfy all equations simultaneously. In the IB MYP 4-5 curriculum, students are expected to solve such systems using various methods, including substitution, elimination, and graphical analysis.

Methods of Solving Simultaneous Equations

There are primarily three methods to solve simultaneous linear equations:

• Graphical Method: Involves plotting each equation on a graph and identifying the point(s) where the lines intersect.
• Substitution Method: Entails solving one equation for one variable and substituting this expression into the other equation.
• Elimination Method: Involves adding or subtracting equations to eliminate one variable, making it easier to solve for the remaining variable.

Graphical Method

The graphical method provides a visual representation of the solutions. For example, consider the following system of equations:

$$ \begin{align} y &= 2x + 3 \\ y &= -x + 1 \end{align} $$

By plotting both equations on a graph, the intersection point represents the solution:

$$ \begin{align} 2x + 3 &= -x + 1 \\ 3x &= -2 \\ x &= -\frac{2}{3} \\ y &= 2\left(-\frac{2}{3}\right) + 3 = \frac{5}{3} \end{align} $$

Thus, the solution is $\left(-\frac{2}{3}, \frac{5}{3}\right)$.

Substitution Method

The substitution method involves solving one equation for a variable and substituting this expression into the other equation. Consider the system:

$$ \begin{align} x + y &= 5 \\ 2x - y &= 1 \end{align} $$

Solving the first equation for $y$:

$$ y = 5 - x $$

Substituting into the second equation:

$$ 2x - (5 - x) = 1 \\ 3x - 5 = 1 \\ 3x = 6 \\ x = 2 $$>

Then, $y = 5 - 2 = 3$. Thus, the solution is $(2, 3)$.

Elimination Method

The elimination method eliminates one variable by adding or subtracting equations. Take the system:

$$ \begin{align} 3x + 2y &= 16 \\ 2x - 2y &= 4 \end{align} $$>

Adding both equations:

$$ (3x + 2y) + (2x - 2y) = 16 + 4 \\ 5x = 20 \\ x = 4 $$>

Substituting back into the first equation:

$$ 3(4) + 2y = 16 \\ 12 + 2y = 16 \\ 2y = 4 \\ y = 2 $$>

Thus, the solution is $(4, 2)$.

Checking Solutions

After obtaining a solution, it is essential to verify its validity by substituting the values back into the original equations. This step ensures that the solution satisfies all equations in the system.

Using the substitution method example:

$$ \begin{align} x + y &= 5 \\ 2x - y &= 1 \end{align} $$>

Solution: $(2, 3)$

Substituting into the first equation:

$$ 2 + 3 = 5 \quad \checkmark $$>

Substituting into the second equation:

$$ 2(2) - 3 = 4 - 3 = 1 \quad \checkmark $$>

Since both equations are satisfied, the solution is correct.

Importance of Checking Solutions

Checking solutions serves multiple purposes:

• Accuracy: Ensures that no computational errors have been made during the solving process.
• Consistency: Verifies that the solution fits all equations in the system.
• Confidence: Builds students' confidence in their problem-solving abilities.

Common Mistakes to Avoid

When checking solutions, students should be aware of common pitfalls:

• Arithmetic Errors: Mistakes in basic calculations can lead to incorrect solutions.
• Sign Errors: Incorrect handling of positive and negative signs can alter the solution.
• Misapplication of Methods: Choosing an inappropriate method or incorrectly applying it can lead to errors.

Example Problem and Solution Verification

Consider the system:

$$ \begin{align} 4x - y &= 11 \\ 2x + 3y &= 13 \end{align} $$>

Using the elimination method:

Multiply the first equation by 3:

$$ 12x - 3y = 33 $$>

Add to the second equation:

$$ (12x - 3y) + (2x + 3y) = 33 + 13 \\ 14x = 46 \\ x = \frac{46}{14} = \frac{23}{7} $$>

Substitute $x$ into the first equation:

$$ 4\left(\frac{23}{7}\right) - y = 11 \\ \frac{92}{7} - y = 11 \\ y = \frac{92}{7} - 11 = \frac{92}{7} - \frac{77}{7} = \frac{15}{7} $$>

Solution: $\left(\frac{23}{7}, \frac{15}{7}\right)$

Checking in the second equation:

$$ 2\left(\frac{23}{7}\right) + 3\left(\frac{15}{7}\right) = \frac{46}{7} + \frac{45}{7} = \frac{91}{7} = 13 \quad \checkmark $$>

Both equations are satisfied, confirming the solution is correct.

Applications of Checking Solutions

Verifying solutions is not only crucial in academic settings but also in real-world applications such as engineering, economics, and computer science. Accurate solutions ensure the reliability of models and systems based on these equations.

Advanced Techniques for Solution Verification

In more complex systems, especially those involving more variables or non-linear equations, automated methods and software tools like MATLAB or graphing calculators can assist in verifying solutions efficiently.

Theoretical Foundations

The ability to check solutions stems from the fundamental principles of algebra and linear systems. Understanding the consistency and independence of equations ensures that solutions are both unique and valid.

Numerical Examples

Let’s explore another example:

$$ \begin{align} 5x + 2y &= 14 \\ 3x - y &= 5 \end{align} $$>

Using the substitution method:

From the second equation:

$$ y = 3x - 5 $$>

Substitute into the first equation:

$$ 5x + 2(3x - 5) = 14 \\ 5x + 6x - 10 = 14 \\ 11x = 24 \\ x = \frac{24}{11} $$>

Then, $y = 3\left(\frac{24}{11}\right) - 5 = \frac{72}{11} - \frac{55}{11} = \frac{17}{11}$

Solution: $\left(\frac{24}{11}, \frac{17}{11}\right)$

Checking in the first equation:

$$ 5\left(\frac{24}{11}\right) + 2\left(\frac{17}{11}\right) = \frac{120}{11} + \frac{34}{11} = \frac{154}{11} = 14 \quad \checkmark $$>

Checking in the second equation:

$$ 3\left(\frac{24}{11}\right) - \frac{17}{11} = \frac{72}{11} - \frac{17}{11} = \frac{55}{11} = 5 \quad \checkmark $$>

The solution satisfies both equations.

Intersection of Graphs and Solutions

Graphically, the solution to a system of linear equations corresponds to the intersection point of their graphs. If the lines intersect at exactly one point, there is a unique solution. If they are parallel, there is no solution, and if they coincide, there are infinitely many solutions.

Determining the Nature of Solutions

The determinant of the coefficient matrix plays a pivotal role in identifying the nature of solutions:

$$ \text{For a system:} \begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} $$>

The determinant ($D$) is calculated as:

$$ D = a_1b_2 - a_2b_1 $$>

Interpretation:

• If $D \neq 0$: The system has a unique solution.
• If $D = 0$ and the system is consistent: There are infinitely many solutions.
• If $D = 0$ and the system is inconsistent: There is no solution.

Example of Determinant Application

Consider the system:

$$ \begin{align} 2x + 3y &= 6 \\ 4x + 6y &= 12 \end{align} $$>

Calculating the determinant:

$$ D = (2)(6) - (4)(3) = 12 - 12 = 0 $$>

Since $D = 0$, we examine consistency. The second equation is a multiple of the first, indicating infinitely many solutions.

Advantages of Checking Solutions

• Ensures Accuracy: Minimizes errors in calculations.
• Builds Understanding: Reinforces conceptual grasp of linear systems.
• Develops Critical Thinking: Encourages thoroughness and attention to detail.

Limitations and Challenges

• Time-Consuming: Especially with complex or large systems.
• Requires Attention to Detail: Small mistakes can lead to incorrect conclusions.
• Not Always Necessary: In some cases, the solution method inherently ensures correctness.

Strategies to Effectively Check Solutions

• Systematic Substitution: Carefully substitute solution values into each equation.
• Use of Technology: Employ calculators or software to verify results.
• Cross-Verification: Solve the system using a different method and compare results.

Comparison Table

Aspect	Advantages	Limitations
Graphical Method	Visual representation aids understanding. Identifies nature of solutions easily.	Less precise for exact solutions. Time-consuming for manual plotting.
Substitution Method	Efficient for systems where one equation is easily solvable. Provides exact solutions.	Can be cumbersome with complex equations. Requires careful algebraic manipulation.
Elimination Method	Systematic and adaptable to any linear system. Efficient for eliminating variables.	May involve fractions, complicating calculations. Requires precise arithmetic.
Checking Solutions	Ensures accuracy and consistency. Reinforces understanding of the system.	Can be time-consuming. Requires additional steps after solving.

Summary and Key Takeaways