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Q.

Find the value of k for which the system of equations 2x+3y=7    and   8x+ k+4 y28=0    and   has infinitely many solutions.


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a

k = -15

b

k = 8

c

k = 14

d

k = 3 

answer is B.

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Detailed Solution

Given equations are 2x+3y=7    and   8x+ k+4 y28=0 .
 We know that a given pair of linear equations of the form, a 1 x+ b 1 y+ c 1 =0 and  a 2 x+ b 2 y+ c 2 =0 must satisfy a 1 a 2 = b 1 b 2 = c 1 c 2 to say they are having infinitely many solutions.
Let the equations be,
2x+3y7=0 ……………….(1)
8x+ k+4 y28=0 …………….(2)
Compare equation (1), with a 1 x+b,y+ c 1 =0 We get, a 1 =2,  b 1 =3,  c 1 =7. Now, compare equation (2), 8x+ k+4 y28=0 with a 2 x+ b 2 y+ c 2 =0. We get, a 2 =8,  b 2 =k+4,  c 2 =28.
Substitute the above values in the condition a 1 a 2 = b 1 b 2 = c 1 c 2 .
2 8 = 3 k+4 = 7 28 ……………(3)
Consider the 1st and 2nd terms,
2 8 = 3 k+4 2(k+4)=8×3 2k+8=24 2k=16 k=8
Put the value of k=8 in equation (3),
2 8 = 3 8+4 = 7 28 1 4 = 1 4 = 1 4 Hence, for k=8 the given linear equation has infinitely many solutions.
The correct option is (2).
 
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