Question

For what value of K the pair of linear equations 3x - 2y - 7 = 0 and 6 x+ ky +11=0 has unique solution​

Answer 1

Answer:

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\displaystyle \: \longmapsto \: \:⟼ FORMULA TO BE IMPLEMENTED :

A pair of Straight Lines

\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0a

1

x+b

1

y+c

1

=0anda

2

x+b

2

y+c

2

=0

have Unique Solution if \displaystyle \: \: \frac{a_1}{a_2} \neq \frac{b_1}{b_2}

a

2

a

1



=

b

2

b

1

\displaystyle \: \longmapsto \: \:⟼ CALCULATION :

Given pair of linear equations

3x - 2y – 7 = 0 \: \: and \: \: 6x + ky + 11 = 03x−2y–7=0and6x+ky+11=0

Comparing with

\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0a

1

x+b

1

y+c

1

=0anda

2

x+b

2

y+c

2

=0

We get

\displaystyle \: a_1 = 3 \: , \: b_1 = - 2 \: , c_1= - 7 \: and \: \: a_2 = 6 \: , \: b_2 = k \: , \: \: c_2=11a

1

=3,b

1

=−2,c

1

=−7anda

2

=6,b

2

=k,c

2

=11

So the given pair of Straight Lines have Unique Solution if

\displaystyle \: \: \frac{3}{6} \neq \frac{ - 2}{k}

6

3



=

k

−2

\implies \: \: \displaystyle \: 3k \: \ne \: - 12⟹3k



=−12

\implies \: \: \displaystyle \: k \: \ne \: - 4⟹k



=−4

\displaystyle \: \longmapsto \: \:⟼ RESULT :

Hence the given Pair of Straight Lines have Unique Solution for all Real Numbers Except - 4