For what value of $k$, the equation $ 2x-3y=4$ and $3x-ky=5$ meet at exactly one point i.e., point whose abscissa is $-1$?

Given: The equation $ 2x-3y=4$ and $3x-ky=5$ meet at exactly one point i.e., point whose abscissa is $-1$.

To do: To find the value of $k$.

Solution:

Given equation $ 2x-3y=4$ ......  $( i)$

$3x-ky=5$ ...... $( ii)$

As given: $x=-1$, put this value in $( i)$

$2x-3y=4$

$\Rightarrow 2( -1)-3y=4$

$\Rightarrow -2-3y=4$

$\Rightarrow -3y=6$

$\Rightarrow y=\frac{6}{-3}$

$\Rightarrow y=-2$

Now we put $x=-1$ and $y=-2$ in $( ii)$

$3x-yk=5$

$\Rightarrow 3( -1)-k( -2)=5$

$\Rightarrow -3+2k=5$

$\Rightarrow 2k=8$

$\Righhtarrow k=4$

Thus, for $k=4$, the equation $ 2x-3y=4$ and $3x-ky=5$ meet at exactly one point i.e., point whose abscissa is $-1$.
 

Updated on: 10-Oct-2022

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