Question

15. Find the value of k for which pair of linear equations 3x + 2y = -5 and x - ky =2 has a unique solution
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Answer 1

$\huge\mathfrak\red{ }$

given series is 0.7 + 0.77 + 0.777 + ......... to nth term

therefore ,

S_nS

n

= 7×(0.1 + 0.11 + 0.111 +......to nth term)

=7/9 ( 0.9 + 0.99 + 0.999 + ...... to nth term )

=7/9 ( 9/10 + 99/100 + 999/1000 + ..... to nth term )

= 7/9 { ( 1 - 1/10 ) + ( 1 - 1/100 ) + ( 1 - 1/1000 ) + .... to nth term }

= 7/9 { 1+1+1+....to nth term } - { 1/10 + 1/100 + 1/1000 +.... to nth term}

= 7/9 { n - 1/10 ( 1 - 1/10 ) ^n / 1 - 1/10 }

= 7/9 { n - 1 ( 1 - 1/10 )^n / 9 }

= 7/9 { n - ( 1 - 1/10 )^n / 9 }

\huge\mathfrak\blue{ proved }proved

Answer 2

Answer:

$\bigstar{\bold{k\neq \dfrac{-2}{3} , k \in R}}$

Step-by-step explanation:

$\Large{\underline{\sf{Given:}}}$

A pair of linear equations:

• 3x + 2y = -5
• x - ky = 2

$\Large{\underline{\sf{To\:Find:}}}$

• The value of k so that the equations have a unique solution

$\Large{\underline{\sf{Solution:}}}$

➟ Here we have to find the value of k so that the given pair of equations is consistent and has a unique solution.

➟ If a pair of equations are consistent, we know that,

    $\tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

➟ Here a₁ = 3

              a₂ = 1

              b₁ = 2

              b₂ = -k

➟ Substituting the datas we get,

    $\tt \dfrac{3}{1} \neq \dfrac{2}{-k}$

➟ Cross multiplying we get,

    -3k ≠ 2

       k ≠ -2/3

➟ Hence k can take the value of any real number except -2/3.

    $\boxed{\bold{k\neq \dfrac{-2}{3} , k \in R}}$

$\Large{\underline{\sf{Notes:}}}$

➟ If a pair of linear equations is:

➵ consistent and has a unique solution then,

    $\tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

➵ consistent and has infinite number of solutions,

    $\tt \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

➵ inconsistent and has no solution,

    $\tt \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$