Question

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

Answer 1

options are :

a) \[\frac{3}{4}\]

b)\[\frac{4}{3}\]

c) \[-4\]

d) \[\frac{-4}{3}\]

Answer 2

i think there is easier way out....

Answer 3

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
\[c=-\frac{4}{3}\]

Answer 4

^^ exactly

Answer 5

@hartnn: Oh yeah

Answer 6

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

Answer 7

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

Answer 8

Yes.

Answer 9

@ash2326 gd and then

Answer 10

and why would that have an infinite # of solutions?

Answer 11

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

Answer 12

ash2326's solution is mostly correct

Answer 13

this one doesn't satisfy that

Answer 14

ok! then @Zarkon

Answer 15

3x-4y=7 and x+cy=13

\[x-\frac{4}{3}y=\frac{7}{3}\]

\[x+cy=13\]

Answer 16

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

Answer 17

if c=-4/3 then you get

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

ie \[\frac{7}{3}=13\]

Answer 18

thus no solution

Answer 19

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

Answer 20

and then solve it

Answer 21

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 .

Answer 22

@mayankdevnani what r u doing?

Answer 23

@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.

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

@mayankdevnani where are you not getting ?

Answer 28

step by step

Answer 29

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

Answer 30

yes...your way is fine

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

-4/3

Answer 35

ya.. but how??

Answer 36

that will be a prallal line

Answer 37

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

Answer 38

got it @mayankdevnani ?

Answer 39

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

Answer 40

@mayankdevnani clear?

Answer 41

ok... @mathslover

Answer 42

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

Answer 43

neither do they overlap

Answer 44

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

Answer 45

:)

Answer 46

@mathslover is my approach correct?

Answer 47

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

Answer 48

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

Answer 49

yes it is similar to that of @ash2326