Question

please!!!!!!!!!!!!!!!!!
ans

Answer 1

$\boxed{\begin{minipage}{15 em} Consider a general pair of linear equations: \\ \\ a_1x+b_1y+c_1=0 \\ a_2x+b_2y+c_2=0 \\ \\ \text{For infinitely many solutions, the } \\ \text{required condition is:} \\ \\ \\ \boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}} \end{minipage}}$


We can use the same condition even if the equations are in the form:

$a_1x+b_1y=c_1 \\ a_2x+b_2y=c_2$
________________________________________

$\begin{array}{lc}\bold{a)} & \bold{kx-3y=1} \\ \\ & \bold{2x+ky=2} \end{array}$

For Infinite Solutions:

$\frac{k}{2} = \frac{-3}{k} = \frac{1}{2} \\ \\ \\ \implies \frac{k}{2} = \frac{1}{2} \quad AND \quad \frac{-3}{k} = \frac{1}{2} \\ \\ \\ \implies k=1 \quad AND \quad k=-6$

The value of k cannot be both 1 and -6 simultaneously. So there is no value of k for which equations have infinite solutions.


__________


$\begin{array}{lc}\bold{b)} & \bold{2x+ky=1} \\ \\ & \bold{6x+4y=3} \end{array}$

For Infinite Solutions, we have:



$\frac{2}{6} = \frac{k}{4} = \frac{1}{3} \\ \\ \\ \implies \frac{1}{3} = \frac{k}{4} = \frac{1}{3} \\ \\ \\ \implies \frac{k}{4} = \frac{1}{3} \\ \\ \\ \implies \boxed{\bold{k=\frac{4}{3}}}$


Thus, The value of k is $\bold{\frac{4}{3}}$

__________


$\begin{array}{ll}\bold{a)} & \bold{2x+3y=9} \\ \\ & \bold{6x+(k-2)y=(3k-2)} \end{array}$

For Infinite Solutions:

$\frac{2}{6} = \frac{3}{k-2} = \frac{9}{3k-2} \\ \\ \\ \implies \frac{1}{3} = \frac{3}{k-2} = \frac{9}{3k-2} \\ \\ \\ \implies 3 = \frac{k-2}{3} = \frac{3k-2}{9}$

Let us consider the equalities one by one. All equalities must simultaneously give the same solution of k, otherwise, the equations cannot have infinite solutions.


$\rightarrow 3 = \frac{k-2}{3} \\ \\ \implies k-2=9 \\ \\ \implies k=11$



$\rightarrow 3 = \frac{3k-2}{9} \\ \\ \implies 3k-2 = 27 \\ \\ \implies 3k=29 \\ \\ \implies k = \frac{29}{3}$

Here again, the equalities yield contradictory values of k. 

Thus, there is no value of k for which equations have infinite solutions.