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

- mayankdevnani

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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- katieb

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- mayankdevnani

options are :
a) \[\frac{3}{4}\]
b)\[\frac{4}{3}\]
c) \[-4\]
d) \[\frac{-4}{3}\]

- hartnn

i think there is easier way out....

- ash2326

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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## More answers

- hartnn

^^ exactly

- ParthKohli

@hartnn: Oh yeah

- anonymous

WELL FOLLOW @ash2326 THAT ONE MUCH EASIER..... agreed with @hartnn

- hartnn

no solution is of the form:
ax+b=c
ax+b=d

- ParthKohli

Yes.

- mayankdevnani

@ash2326 gd and then

- Zarkon

and why would that have an infinite # of solutions?

- hartnn

because ax+b=c and ax+b=c are satisfied by infinite number of points (x,y) simultaneously.

- Zarkon

ash2326's solution is mostly correct

- Zarkon

this one doesn't satisfy that

- mayankdevnani

ok! then @Zarkon

- Zarkon

3x-4y=7 and x+cy=13
\[x-\frac{4}{3}y=\frac{7}{3}\]
\[x+cy=13\]

- mayankdevnani

yaa ... i also agree with @Zarkon
don't you @mathslover

- Zarkon

if c=-4/3 then you get
\[x-\frac{4}{3}y=\frac{7}{3}\]
and
\[x-\frac{4}{3}y=13\]
ie \[\frac{7}{3}=13\]

- Zarkon

thus no solution

- mayankdevnani

isn't like this
\[3x-4y=7\]
\[3x=7-4y\]
\[x-y=\frac{7}{12}\]
\[x=\frac{7}{12}+y\]

- mayankdevnani

and then solve it

- mathslover

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 .

- anonymous

@mayankdevnani what r u doing?

- mathslover

@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.

- mayankdevnani

how did you get it......... @mathslover

- mathslover

btw , @Zarkon sir , was my method acceptable? I know there are easy methods too.. but that is what I did..

- mayankdevnani

ANSWER is correct... but i did'nt understand..... plz tell me once again.....

- mathslover

@mayankdevnani where are you not getting ?

- mayankdevnani

step by step

- anonymous

THAT WAS exactly what I was thinking @mathslover THAT IS ABSOULETLY CORRECT

- Zarkon

yes...your way is fine

- mathslover

OK thanks @sauravshakya Let us wait for honorable zarkon sir ..

- mathslover

OK thanks a lot @Zarkon .. that fine made a lot sense sir.. :)

- mayankdevnani

@mathslover how did you get it
\[7+4y=1(13-cy)\]

- ghazi

-4/3

- mayankdevnani

ya.. but how??

- ghazi

that will be a prallal line

- mathslover

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

- mathslover

got it @mayankdevnani ?

- ghazi

just put -4/3 at C and you will have a parallel set of line and parallel set of lines have no solution

- ghazi

@mayankdevnani clear?

- mayankdevnani

ok... @mathslover

- ghazi

3x-4y=7 and 3x-4y=39 are parallel lines how could they have a solution ..they intersect nowhere

- ghazi

neither do they overlap

- mayankdevnani

thnx.... @Zarkon @ghazi @sauravshakya @ash2326
but special thnx to @mathslover

- ghazi

:)

- ghazi

@mathslover is my approach correct?

- anonymous

Isnt your method same as @ash2326 's method........ @ghazi

- 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

- ghazi

yes it is similar to that of @ash2326

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