Question

# For the given pair of linear equations a1x+b1y+c1=0 and a2x+b2y+c2=0, the solution (x,y) is

A

((b1c2-b2c1)/(a1b2-a2b1), (c1a2-c2a1)/(a1b2-a2b1))

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B

((b2c2-b1c1)/(a1b2-a2b1), (c2a2-c1a1)/(a1b2-a2b1))

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C

((b2c1-b1c2)/(a1b2-a2b1), (c1a1-c2a2)/(a1b2-a2b1))

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D

None of these

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Solution

## The correct option is D ((b1c2-b2c1)/(a1b2-a2b1), (c1a2-c2a1)/(a1b2-a2b1)) Let us use Elimination method to solve the given pair of equations Given two equations are a1x+b1y+c1=0 and a2x+b2y+c2=0 Multiply first equation by b2 and second equation by b1, to get b2a1x+b2b1y+b2c1=0 b1a2x+b1b2y+b1c2=0 Subtracting these two equations (b2a1 – b1a2)x + (b2b1 – b1b2)y + (b2c1– b1c2) = 0 i.e., (b2a1 – b1a2)x = b1c2 – b2c1 Therefore x=(b1c2−b2c1)(a1b2−a2b1), Substitute this x in any of the given equation we get y=(c1a2−c2a1)(a1b2−a2b1)

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