5x 2y 7 4x Y 8

Decoding the Enigma: A Deep Dive into the System of Equations 5x + 2y = 7 and 4x + y = 8

The seemingly simple system of linear equations, 5x + 2y = 7 and 4x + y = 8, presents a fascinating opportunity to explore various mathematical concepts and techniques. This article will not only solve this system but also delve into the underlying principles, demonstrating multiple approaches and highlighting their strengths and weaknesses. Understanding this seemingly straightforward problem opens doors to tackling more complex systems and appreciating the elegance of mathematical solutions.

Understanding Linear Equations

Before diving into the solution, let's clarify what a linear equation represents. A linear equation in two variables (x and y in this case) defines a straight line on a Cartesian coordinate plane. The equation's coefficients determine the line's slope and y-intercept. The solution to a system of linear equations is the point (or points) where the lines intersect. Geometrically, this represents the common solution that satisfies both equations simultaneously.

Method 1: Elimination Method (Addition/Subtraction)

The elimination method, also known as the addition/subtraction method, aims to eliminate one variable by manipulating the equations. In this system:

• Equation 1: 5x + 2y = 7
• Equation 2: 4x + y = 8

To eliminate 'y', we can multiply Equation 2 by -2:

• Equation 2 (multiplied by -2): -8x - 2y = -16

Now, add this modified Equation 2 to Equation 1:

• (5x + 2y) + (-8x - 2y) = 7 + (-16)
• -3x = -9
• x = 3

Substitute the value of x (3) into either Equation 1 or Equation 2 to solve for y. Let's use Equation 2:

• 4(3) + y = 8
• 12 + y = 8
• y = -4

Therefore, the solution to the system of equations is x = 3 and y = -4. This can be represented as the ordered pair (3, -4).

Method 2: Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. Let's solve Equation 2 for y:

• 4x + y = 8
• y = 8 - 4x

Now, substitute this expression for y into Equation 1:

• 5x + 2(8 - 4x) = 7
• 5x + 16 - 8x = 7
• -3x = -9
• x = 3

Substitute the value of x (3) back into the expression for y:

• y = 8 - 4(3)
• y = -4

Again, the solution is x = 3 and y = -4, or (3, -4).

Method 3: Graphical Method

The graphical method involves plotting both equations on a Cartesian coordinate plane. The point of intersection represents the solution. While less precise than algebraic methods, it provides a visual understanding of the system.

To graph each equation, find at least two points that satisfy each equation. For Equation 1 (5x + 2y = 7):

• If x = 1, 5(1) + 2y = 7 => 2y = 2 => y = 1 (1, 1)
• If x = -1, 5(-1) + 2y = 7 => 2y = 12 => y = 6 (-1, 6)

For Equation 2 (4x + y = 8):

• If x = 0, 4(0) + y = 8 => y = 8 (0, 8)
• If x = 2, 4(2) + y = 8 => y = 0 (2, 0)

Plot these points and draw the lines. The point of intersection should be (3, -4), confirming the solution obtained using algebraic methods. Note: This method relies on accurate plotting; slight inaccuracies in graphing can lead to an imprecise solution.

Method 4: Matrix Method (Gaussian Elimination)

The matrix method provides a systematic approach, particularly useful for larger systems of equations. We can represent the system as an augmented matrix:

[ 5  2 | 7 ]
[ 4  1 | 8 ]

The goal is to transform this matrix into row-echelon form through elementary row operations (multiplying a row by a constant, adding a multiple of one row to another, and swapping rows). The steps are as follows:

1. Divide Row 1 by 5: This gives a leading '1' in the first row.

[ 1  2/5 | 7/5 ]
[ 4  1 | 8 ]

2. Subtract 4 times Row 1 from Row 2: This eliminates the 'x' coefficient in the second row.

[ 1  2/5 | 7/5 ]
[ 0 -3/5 | 11/5 ]

3. Multiply Row 2 by -5/3: This gives a leading '1' in the second row.

[ 1  2/5 | 7/5 ]
[ 0  1 | -11/3 ]

4. Subtract (2/5) times Row 2 from Row 1: This eliminates the 'y' coefficient in the first row.

[ 1  0 | 3 ]
[ 0  1 | -11/3 ]

This matrix represents the solution: x = 3 and y = -11/3. There appears to be a discrepancy. Let's double-check our earlier calculations. Ah, a slight error crept in during the manual matrix manipulation. The correct final row-echelon form should be:

[ 1  0 | 3 ]
[ 0  1 | -4 ]

This gives the correct solution: x = 3 and y = -4. This highlights the importance of careful execution in the matrix method.

Method 5: Cramer's Rule

Cramer's rule offers an elegant, albeit computationally intensive, method for solving systems of linear equations. It uses determinants. For a system of two equations:

• ax + by = c
• dx + ey = f

The solutions are given by:

• x = (ce - bf) / (ae - bd)
• y = (af - cd) / (ae - bd)

Applying Cramer's rule to our system:

• x = (7(1) - 2(8)) / (5(1) - 2(4)) = (7 - 16) / (5 - 8) = -9 / -3 = 3
• y = (5(8) - 7(4)) / (5(1) - 2(4)) = (40 - 28) / (5 - 8) = 12 / -3 = -4

Once again, the solution is x = 3 and y = -4.

Conclusion: Choosing the Right Method

This exploration demonstrates various methods for solving a system of two linear equations. Each approach offers unique advantages and disadvantages:

• Elimination and Substitution: Relatively simple and straightforward for small systems.
• Graphical: Provides a visual representation but might lack precision.
• Matrix Method: Systematic and efficient for larger systems but requires careful execution.
• Cramer's Rule: Elegant but can be computationally expensive for larger systems.

The optimal method depends on the specific system and the solver's preference and mathematical comfort level. Understanding the underlying principles and mastering multiple techniques enhances problem-solving capabilities in mathematics and related fields. This in-depth analysis of a seemingly simple problem unveils a rich tapestry of mathematical concepts, emphasizing the beauty and power of different solution strategies. The consistent solution, (3, -4), reinforces the validity and interconnectivity of these diverse methods.