Question

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

• A. $10x+14y+1=0$
• B. $10x-14y+4=0$
• C. $10x-14y-4=0$
• D. $10x+14y-4=0$

Answer 1

Answer: (D)

$10x+14y-4=0$

$Whentherearetwoconcistentdependentequationsas{a}_{1}x+b_{1}y+c_1=0\&a_{2}x+b_{2}y+c_2=0thentheequationswillhaveinfinitelymanysolutionswhenthelinesarecoincident.i.e\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.Theequationsarenotconsistentdependentif\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}or\frac{a_1}{a_2}\ne \frac{b_1}{b_2}=\frac{c_1}{c_2}.Letusinvestigateeachoption.OptionA⟶Heretheequationsare5x+7y=2\&10x+14y=-1.So{a}_1=5,b_1=7,c_1=2\&a_2=10,b_2=14\&c_2=1.\therefore \frac{a_1}{a_2}=\frac{5}{10}=\frac{1}{2},\frac{b_1}{b_2}=\frac{7}{14}=\frac{1}{2}and\frac{c_1}{c_2}=\frac{-2}{1}=-2.Hereweseethat\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}.\therefore Thelinesarenotcoincident.Sotheyarenotdependent.OptionB⟶Heretheequationsare5x+7y=2\&10x-14y=-4.So{a}_1=5,b_1=7,c_1=-2\&a_2=10,b_2=-14\&c_2=4.\therefore \frac{a_1}{a_2}=\frac{5}{10}=\frac{1}{2},\frac{b_1}{b_2}=\frac{7}{-14}=-\frac{1}{2}and\frac{c_1}{c_2}=\frac{-2}{4}=\frac{-1}{2}.Hereweseethat\frac{a_1}{a_2}=\frac{c_1}{c_2}\ne \frac{b_1}{b_2}.\therefore Thelinesarenotcoincident.Sotheyarenotdependent.OptionC⟶Heretheequationsare5x+7y=2\&10x-14y=4.So{a}_1=5,b_1=7,c_1=-2\&a_2=10,b_2=-14\&c_2=-4.\therefore \frac{a_1}{a_2}=\frac{5}{10}=\frac{1}{2},\frac{b_1}{b_2}=\frac{7}{-14}=-\frac{1}{2}and\frac{c_1}{c_2}=\frac{-2}{-4}=\frac{1}{2}.Hereweseethat\frac{b_1}{b_2}\ne \frac{c_1}{c_2}=\frac{a_1}{a_2}.\therefore Thelinesarenotcoincident.Sotheyarenotdependent.OptionD⟶Heretheequationsare5x+7y=2\&10x+14y=4.So{a}_1=5,b_1=7,c_1=-2\&a_2=10,b_2=14\&c_2=-4.\therefore \frac{a_1}{a_2}=\frac{5}{10}=\frac{1}{2},\frac{b_1}{b_2}=\frac{7}{14}=\frac{1}{2}and\frac{c_1}{c_2}=\frac{-2}{-4}=\frac{1}{2}.Hereweseethat\frac{a_1}{a_2}=\frac{c_1}{c_2}=\frac{b_1}{b_2}.\therefore Thelinesarecoincident.Sotheyaredependent.\therefore Thesecondequationcanbe10x+14y=4.Anss-OptionD$