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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+14y4=0
Whentherearetwoconcistentdependentequationsasa1x+b1y+c1=0&a2x+b2y+c2=0thentheequationswillhaveinfinitelymanysolutionswhenthelinesarecoincident.i.ea1a2=b1b2=c1c2.Theequationsarenotconsistentdependentifa1a2=b1b2c1c2ora1a2b1b2=c1c2.Letusinvestigateeachoption.OptionAHeretheequationsare5x+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=b1b2c1c2.Thelinesarenotcoincident.Sotheyarenotdependent.OptionBHeretheequationsare5x+7y=2&10x14y=4.Soa1=5,b1=7,c1=2&a2=10,b2=14&c2=4.a1a2=510=12,b1b2=714=12andc1c2=24=12.Hereweseethata1a2=c1c2b1b2.Thelinesarenotcoincident.Sotheyarenotdependent.OptionCHeretheequationsare5x+7y=2&10x14y=4.Soa1=5,b1=7,c1=2&a2=10,b2=14&c2=4.a1a2=510=12,b1b2=714=12andc1c2=24=12.Hereweseethatb1b2c1c2=a1a2.Thelinesarenotcoincident.Sotheyarenotdependent.OptionDHeretheequationsare5x+7y=2&10x+14y=4.Soa1=5,b1=7,c1=2&a2=10,b2=14&c2=4.a1a2=510=12,b1b2=714=12andc1c2=24=12.Hereweseethata1a2=c1c2=b1b2.Thelinesarecoincident.Sotheyaredependent.Thesecondequationcanbe10x+14y=4.AnssOptionD

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