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Kx+3y=k- - 3 12x+ky=k. Find the …

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Kx+3y=k- - 3 12x+ky=k. Find the value of k.

Posted by Ajay Rathor 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

The given equations are
kx + 3y - (k - 3) = 0 ......... (i)
12x + ky - k = 0 ........... (ii)
The system of linear equations is in the form of
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Compare (i) and (ii), we get
a₁= k ,b₁= 3, c₁ = -(k - 3),
a₂=12 ,b₂= k ,c₂ = -k
For a unique solution, we must have
{tex}\frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }{/tex}
{tex}\frac { k } {1 2 } \neq \frac { 3 } { k }{/tex}
{tex}\Rightarrow k ^ { 2 } \neq 36 {/tex}
{tex}\Rightarrow k \neq \pm 6{/tex}
Thus, for all real value of k other than {tex}\pm 6{/tex}, the given system of equations will have a unique solution.

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