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The point of intersection of lines 2x+3y+4=0 and 6x−y+12=0 is

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a

1,0

b

−2,0

c

−3,0

d

−3,−1

answer is B.

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Detailed Solution

Eliminate y 2x+3y+4+36x−y+12=02x+3y+4+18x−3y+36=020x+40=0x=−2Substitute x=-2 in the equation 2x+3y+4=0 2−2+3y+4=03y=0y=0Therefore, the point of intersection is −2,0

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The point of intersection of lines 2x+3y+4=0 and 6x−y+12=0 is