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Given: 3x-4y=7 and x+cy=13 , for what value of "c" will the two equations not have a solution?

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options are : a) \[\frac{3}{4}\] b)\[\frac{4}{3}\] c) \[-4\] d) \[\frac{-4}{3}\]
i think there is easier way out....
Rewrite first equation \[x-\frac{4}{3}y=7\] for no solution we need to have parallel lines second equation \[x+cy=13\] compare this with the first \[c=-\frac{4}{3}\]

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Other answers:

^^ exactly
@hartnn: Oh yeah
WELL FOLLOW @ash2326 THAT ONE MUCH EASIER..... agreed with @hartnn
no solution is of the form: ax+b=c ax+b=d
@ash2326 gd and then
and why would that have an infinite # of solutions?
because ax+b=c and ax+b=c are satisfied by infinite number of points (x,y) simultaneously.
ash2326's solution is mostly correct
this one doesn't satisfy that
ok! then @Zarkon
3x-4y=7 and x+cy=13 \[x-\frac{4}{3}y=\frac{7}{3}\] \[x+cy=13\]
yaa ... i also agree with @Zarkon don't you @mathslover
if c=-4/3 then you get \[x-\frac{4}{3}y=\frac{7}{3}\] and \[x-\frac{4}{3}y=13\] ie \[\frac{7}{3}=13\]
thus no solution
isn't like this \[3x-4y=7\] \[3x=7-4y\] \[x-y=\frac{7}{12}\] \[x=\frac{7}{12}+y\]
and then solve it
What I did is this : \[\large{7+4y = 3(13-cy)}\] \[\large{7+4y=39-3cy}\] \[\large{32= 4y+3cy}\] We have to find the value of c when the RHS becomes zero which will give no solution. \[\large{4y+3(c)(y)=0}\] \[\large{c = \frac{-4y}{3y}}\] \[\large{c=\frac{-4}{3}}\] hence at c = -4/3 we get no solutions for the variables in the given two equations .
@mayankdevnani what r u doing?
@mayankdevnani you need to concentrate hardly on understanding questions... You are solving for x or y but in the question we have to find the value for c which makes the solutions for the equation as null set.
how did you get it......... @mathslover
btw , @Zarkon sir , was my method acceptable? I know there are easy methods too.. but that is what I did..
ANSWER is correct... but i did'nt understand..... plz tell me once again.....
@mayankdevnani where are you not getting ?
step by step
THAT WAS exactly what I was thinking @mathslover THAT IS ABSOULETLY CORRECT
yes...your way is fine
OK thanks @sauravshakya Let us wait for honorable zarkon sir ..
OK thanks a lot @Zarkon .. that fine made a lot sense sir.. :)
@mathslover how did you get it \[7+4y=1(13-cy)\]
ya.. but how??
that will be a prallal line
3x-4y = 7 x = (7+4y) / 3 and then : x + cy = 13 x = 13-cy 13-cy = (7+4y)/3 (since they are equal to x) 3(13-cy) = 7+4y
just put -4/3 at C and you will have a parallel set of line and parallel set of lines have no solution
3x-4y=7 and 3x-4y=39 are parallel lines how could they have a solution ..they intersect nowhere
neither do they overlap
thnx.... @Zarkon @ghazi @sauravshakya @ash2326 but special thnx to @mathslover
@mathslover is my approach correct?
Isnt your method same as @ash2326 's method........ @ghazi
i didn't see ...i just came here read the question and identified these are parallel lines...i guess yes both have same solution and we don't need to solve anything because already it is visible that it's case of parallel line when c= -4/3 hence no solution
yes it is similar to that of @ash2326

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