Question

2x-3y-1=0 ,kx+5y-7=0 find the value of k for the equation having unique solutions ​

Answer 1

Given :-

2x - 3y - 1 = 0          ...(i)

kx + 5y - 7 = 0        ...(ii)

To find :-

The value of k for which the equation has an unique solution.

Solution :-

We know, for having an unique solution, the following condition must be satisfied :

$\displaystyle{\frac{a_1}{a_2}\neq \frac{b_1}{b_2} }$

In the given equations, we have :

• $a_1=2$, $b_1=-3$,

• $a_2=k$ and $b_2=5$

A/q, the equations have an unique solution. So, the condition must be satisfied.

Now, putting these values :

$\displaystyle{\frac{2}{k}\neq \frac{-3}{5} }\\\\\displaystyle{\implies 10\neq -3k}\\\\\boxed{\displaystyle{\implies k\neq \frac{-10}{3} }}$

∴ So, we can say that, the value of 'k' can be any integer except -10/3.

Answer 2

$\huge{\underline{\sf{Given\: question:-}}}$

$=2x - 3y = 1 \\= kx + 5y = 7$

$\huge{\underline{\sf{Solution:-}}}$

★ To find:-

The value of k for which the following system of equation has unique solution.

$= 2x - 3y - 1 = 0$

$[a_{1} = 2, \: b_{1} = 3, \: c_{1} = - 1]$

$= kx + 5y - 7 = 0$

$[a_{2} = k ,\: b_{2} = 5 ,\: c_{2} = 7]$

$= \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

★ Now,

By comparing, and substituting the values, we get:-

$=\frac{2}{k}\neq \frac{ - 3}{5}$

$=\large{\boxed{\sf{k \neq {\dfrac{ - 10}{3}}}}}$

Therefore, The value of k is all real number except $k \neq \frac{ - 10}{3}$

So, hence proved!