Solving Simultaneous Equations by Elimination Method
This article will guide you through solving a system of linear equations using the elimination method. We'll work with the specific example:
(a + 2b)x + (2a - b)y = 2 (a - 2b)x + (2a + b)y = 3
Understanding the Elimination Method
The elimination method aims to solve for one variable by eliminating the other. This is achieved by manipulating the equations to create coefficients with opposite signs for one variable, allowing them to cancel out when the equations are added together.
Steps to Solve the System
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Choose a Variable to Eliminate: In this case, we'll choose to eliminate x.
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Multiply Equations to Match Coefficients: To make the coefficients of x opposites, we'll multiply the first equation by (a - 2b) and the second equation by -(a + 2b):
(a - 2b)[(a + 2b)x + (2a - b)y = 2] -(a + 2b)[(a - 2b)x + (2a + b)y = 3]
This gives us:
(a^2 - 4b^2)x + (2a^2 - 5ab + 2b^2)y = 2(a - 2b) -(a^2 - 4b^2)x - (2a^2 + 5ab + 2b^2)y = -3(a + 2b)
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Add the Equations Together: Notice that the x terms have opposite coefficients, allowing them to cancel out. Adding the equations results in:
-10aby = 2(a - 2b) - 3(a + 2b)
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Solve for y: Simplify the equation and solve for y:
-10aby = -a - 8b y = (a + 8b) / (10ab)
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Substitute y into One of the Original Equations: Substitute the value of y back into either of the original equations. Let's use the first equation:
(a + 2b)x + (2a - b)[(a + 8b) / (10ab)] = 2
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Solve for x: Simplify and solve for x:
(a + 2b)x + (2a^2 + 15ab - 8b^2) / (10ab) = 2 (a + 2b)x = 2 - (2a^2 + 15ab - 8b^2) / (10ab) (a + 2b)x = (20ab - 2a^2 - 15ab + 8b^2) / (10ab) (a + 2b)x = (5ab - 2a^2 + 8b^2) / (10ab) x = (5ab - 2a^2 + 8b^2) / (10ab(a + 2b))
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Solution: Therefore, the solution to the system of equations is:
x = (5ab - 2a^2 + 8b^2) / (10ab(a + 2b)) y = (a + 8b) / (10ab)
Conclusion
By following the steps of the elimination method, we successfully solved for the values of x and y in the system of equations. The elimination method provides a structured approach to simplify equations and solve for unknown variables.