Question

# One equation of a pair of dependent linear equations is 5x+7y=2. The second equation can be

A
10x+14y+1=0
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B
10x14y+4=0
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C
10x14y4=0
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D
10x+14y4=0
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Solution

## The correct option is B 10x+14y−4=0Whentherearetwoconcistentdependentequationsasa1x+b1y+c1=0&a2x+b2y+c2=0thentheequationswillhaveinfinitelymanysolutionswhenthelinesarecoincident.i.ea1a2=b1b2=c1c2.Theequationsarenotconsistentdependentifa1a2=b1b2≠c1c2ora1a2≠b1b2=c1c2.Letusinvestigateeachoption.OptionA⟶Heretheequationsare5x+7y=2&10x+14y=−1.Soa1=5,b1=7,c1=2&a2=10,b2=14&c2=1.∴a1a2=510=12,b1b2=714=12andc1c2=−21=−2.Hereweseethata1a2=b1b2≠c1c2.∴Thelinesarenotcoincident.Sotheyarenotdependent.OptionB⟶Heretheequationsare5x+7y=2&10x−14y=−4.Soa1=5,b1=7,c1=−2&a2=10,b2=−14&c2=4.∴a1a2=510=12,b1b2=7−14=−12andc1c2=−24=−12.Hereweseethata1a2=c1c2≠b1b2.∴Thelinesarenotcoincident.Sotheyarenotdependent.OptionC⟶Heretheequationsare5x+7y=2&10x−14y=4.Soa1=5,b1=7,c1=−2&a2=10,b2=−14&c2=−4.∴a1a2=510=12,b1b2=7−14=−12andc1c2=−2−4=12.Hereweseethatb1b2≠c1c2=a1a2.∴Thelinesarenotcoincident.Sotheyarenotdependent.OptionD⟶Heretheequationsare5x+7y=2&10x+14y=4.Soa1=5,b1=7,c1=−2&a2=10,b2=14&c2=−4.∴a1a2=510=12,b1b2=714=12andc1c2=−2−4=12.Hereweseethata1a2=c1c2=b1b2.∴Thelinesarecoincident.Sotheyaredependent.∴Thesecondequationcanbe10x+14y=4.Anss−OptionD

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