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Question

Solve the two equations using method of elimination.

a1x+b1y+c1=0
a2x+b2y+c2=0


A

x = (b1c2b2c1)(a1b2a2b1), y = (c1a2c2a1)(a1b2a2b1)

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B

x = (b2c2b1c1)(a1b2a2b1), y = (c2a2c1a1)(a1b2a2b1)

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C

x = (b2c1b1c2)(a1b2a2b1), y = (c1a1c2a2)(a1b2a2b1)

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D

x = (b1c2+b2c1)(a1b2+a2b1), y = (c1a2+c2a1)(a1b2+a2b1)

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Solution

The correct option is A

x = (b1c2b2c1)(a1b2a2b1), y = (c1a2c2a1)(a1b2a2b1)


Given two equations are
a1x+b1y+c1=0
and a2x+b2y+c2=0

On multiplying the first equation by b2 and second equation by b1, we get

b2a1x + b2b1y + b2c1 = 0

b1a2x + b1b2y + b1c2 = 0

On subtracting these two equations, we get

(b2a1b1a2)x+(b2b1b1b2)y+(b2c1b1c2)=0

i.e., (b2a1b1a2)x = b1c2b2c1

Therefore x=(b1c2b2c1)(a1b2a2b1) ,

On substituting the value of x in equation (1), we get

a1(b1c2b2c1a1b2a2b1)+b1y+c1=0

b1y=c1a1(b1c2b2c1a1b2a2b1)

b1y=c1(a1b2a2b1)a1(b1c2b2c1)a1b2a2b1

y=b1×(c1a2c2a1)b1×(a1b2a2b1)

y=(c1a2c2a1)(a1b2a2b1)


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