Solving a System of Equations: x - y 6 and x2y2 26

This article delves into solving the system of equations x - y 6

and x2y2 26

. We will explore the methods of algebraic substitution and factoring to find the solutions step-by-step.

Step-by-Step Solution

The given system of equations is:

Equation 1: x - y 6 Equation 2: x2y2 26

Substitution Method

First, solve Equation 1 for y:

y x - 6

Substitute this expression for y into Equation 2:

x2 (x - 6)2 26

Expanding (x - 6)2:

x2 (x2 - 12x 36) 26

Multiplying and combining like terms:

x4 - 12x3 36x2 - 26 0

Rewrite the equation as a quadratic in terms of x2:

2(x2)2 - 12(x2) 10 0

Divide the entire equation by 2:

(x2)2 - 6(x2) 5 0

Factor the quadratic equation:

(x2 - 1)(x2 - 5) 0

Solve for x2 by setting each factor to zero:

x2 - 1 0 rArr; x ±1

x2 - 5 0 rArr; x ±√5

Now find the corresponding y values using y x - 6:

For x 1:

y 1 - 6 -5

So one solution is (1, -5).

For x -1:

y -1 - 6 -7

This does not satisfy the original equation.

For x √5:

y √5 - 6 (does not yield an integer solution)

For x -√5:

y -√5 - 6 (does not yield an integer solution)

Thus, the solutions to the system of equations are:

(1, -5) and (5, -1)

Alternative Methods

Another approach is to directly substitute and solve as follows:

From x - y 6, we get:

x y 6 (Equation 1)

Substitute this expression for x into the second equation:

(y 6)2(y2) 26

Expanding:

y4 12y2 36y2 26

Combine like terms:

y4 12y2 36 - 26 0

y4 12y2 10 0

This equation reveals the same quadratic structure as before, confirming the previous steps.

Thus, the solutions are:

(1, -5) and (5, -1)

Conclusion

The solutions to the system of equations x - y 6 and x2y2 26 are (1, -5) and (5, -1). These solutions can be found using algebraic substitution and factoring, ensuring a thorough understanding of the underlying algebraic principles.

To verify these solutions, you can substitute them back into the original equations to ensure they satisfy both conditions.