x%2B3y-2%3D0-(a2%2B1)x%2B(a-2)y-5%3D0.jpeg "(3a+1)x+3y-2=0 (a2+1)x+(a-2)y-5=0")
Solving a System of Linear Equations
This article will discuss how to solve a system of linear equations, specifically the system given by:
(3a+1)x + 3y - 2 = 0
(a^2 + 1)x + (a-2)y - 5 = 0
Understanding the Problem
We have two equations with two unknowns, x and y. Our goal is to find the values of x and y that satisfy both equations simultaneously.
Methods of Solving
There are several methods to solve systems of linear equations, including:
- Substitution Method: This involves solving one equation for one variable and substituting it into the other equation.
- Elimination Method: This involves manipulating the equations to eliminate one variable and then solving for the remaining variable.
Solving the System
Let's use the elimination method to solve the system:
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Multiply the first equation by (a-2) and the second equation by -3:
- (3a+1)(a-2)x + 3(a-2)y - 2(a-2) = 0
- -3(a^2+1)x - 3(a-2)y + 15 = 0
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Add the two equations together:
- [(3a+1)(a-2) - 3(a^2+1)]x + [3(a-2) - 3(a-2)]y - 2(a-2) + 15 = 0
-
Simplify the equation:
- (a^2 - 5a - 1)x + 13 - 2a = 0
-
Solve for x:
- x = (2a - 13) / (a^2 - 5a - 1)
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Substitute the value of x back into either of the original equations to solve for y:
- Let's use the first equation:
- (3a+1)[(2a - 13) / (a^2 - 5a - 1)] + 3y - 2 = 0
- Simplify and solve for y.
Solution
The solution to the system of equations is:
- x = (2a - 13) / (a^2 - 5a - 1)
- y = [2(a^2 - 5a - 1) - (3a+1)(2a - 13)] / [3(a^2 - 5a - 1)]
Important Note
The solution for x and y depends on the value of a. There are cases where the denominator of these expressions could be zero. In those cases, the system might have no solution or infinitely many solutions.