Question

If 2x+3y=5 and kx-4y=6 have unique solution, then find k​

Answer 1

Answer:

this solution will help you.

Answer 2

Linear Equations

Formula:

We consider two linear equations

$\quad\quad a_{1}x+b_{1}y+c_{1}=0\quad...(1)$

$\quad\quad a_{2}x+b_{2}y+c_{2}=0\quad...(2)$

The two equations $(1)$ and $(2)$ will have unique solution, when

$\quad\quad \frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}$

Solution:

The two given linear equations are

$\quad\quad 2x+3y-5=0$

$\quad\quad kx-4y-6=0$

Comparing these with the above equations, we get

$\quad\quad a_{1}=2,\:b_{1}=3,\:c_{1}=-5$

$\quad\quad a_{2}=k,\:b_{2}=-4,\:c_{2}=-6$

For equation solution,

$\quad \frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}$

$\Rightarrow \frac{2}{k}\neq \frac{3}{-4}$

$\Rightarrow \frac{k}{2}\neq-\frac{4}{3}$

$\Rightarrow k\neq -\frac{8}{3}$

$\therefore$ except for $k=-\frac{8}{3}$, for any other values of real $k$, we will obtain unique solution for the given system of linear equations.

However the solution set of $k$ is

$\quad\quad k=\mathbb{R}-\{-\frac{8}{3}\}$

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