Question

For what value of k, the system of equations  x + 2y = 3,  5x + ky + 7 = 0  Have (i) a unique solution, (ii) no solution?  Also, show that there is no value of k for which the given system of equation has infinitely namely solutions

Answer 1

$$\begin{aligned} &x+2 y=3 \\ &\Rightarrow x+2 y-3=0 \ldots \text { (i) } \end{aligned}$$
And, $5 x+k y+7=0 \ldots$ (ii)
These equations are of the following form:
$$a 1 x+b 1 y+c 1=0, a 2 x+b 2 y+c 2=0$$
where, $a 1=1, b 1=2, c 1=-3$ and $a 2=5, b 2=k, c 2=7$
(i) For a unique solution, we must have:
$$\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \text { i.e., } \frac{1}{5} \neq \frac{2}{k} \Rightarrow \mathrm{k} \neq 10$$
Thus for all real values of $k$ other than 10 , the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
$$\begin{aligned} &\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ &\Rightarrow \frac{1}{5} \neq \frac{2}{k} \neq \frac{-3}{7} \\ &\Rightarrow \frac{1}{5} \neq \frac{2}{k} \text { and } \frac{2}{k} \neq \frac{-3}{7} \\ &\Rightarrow \mathrm{k}=10, \mathrm{k} \neq \frac{14}{-3} \end{aligned}$$
Hence, the required value of $k$ is 10 .
There is no value of $k$ for which the given system of equations has an infinite number of solutions.

Answer 2

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