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Given: 3x4y=7 and x+cy=13 , for what value of "c" will the two equations not have a solution?
 one year ago
 one year ago
Given: 3x4y=7 and x+cy=13 , for what value of "c" will the two equations not have a solution?
 one year ago
 one year ago

This Question is Closed

mayankdevnaniBest ResponseYou've already chosen the best response.0
options are : a) \[\frac{3}{4}\] b)\[\frac{4}{3}\] c) \[4\] d) \[\frac{4}{3}\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
i think there is easier way out....
 one year ago

ash2326Best ResponseYou've already chosen the best response.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 \[c=\frac{4}{3}\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
WELL FOLLOW @ash2326 THAT ONE MUCH EASIER..... agreed with @hartnn
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
no solution is of the form: ax+b=c ax+b=d
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
@ash2326 gd and then
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
and why would that have an infinite # of solutions?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
because ax+b=c and ax+b=c are satisfied by infinite number of points (x,y) simultaneously.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
ash2326's solution is mostly correct
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
this one doesn't satisfy that
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
ok! then @Zarkon
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
3x4y=7 and x+cy=13 \[x\frac{4}{3}y=\frac{7}{3}\] \[x+cy=13\]
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
yaa ... i also agree with @Zarkon don't you @mathslover
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
if c=4/3 then you get \[x\frac{4}{3}y=\frac{7}{3}\] and \[x\frac{4}{3}y=13\] ie \[\frac{7}{3}=13\]
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
isn't like this \[3x4y=7\] \[3x=74y\] \[xy=\frac{7}{12}\] \[x=\frac{7}{12}+y\]
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
and then solve it
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
What I did is this : \[\large{7+4y = 3(13cy)}\] \[\large{7+4y=393cy}\] \[\large{32= 4y+3cy}\] 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 .
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
@mayankdevnani what r u doing?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
@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.
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
how did you get it......... @mathslover
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
btw , @Zarkon sir , was my method acceptable? I know there are easy methods too.. but that is what I did..
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
ANSWER is correct... but i did'nt understand..... plz tell me once again.....
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
@mayankdevnani where are you not getting ?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
THAT WAS exactly what I was thinking @mathslover THAT IS ABSOULETLY CORRECT
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
OK thanks @sauravshakya Let us wait for honorable zarkon sir ..
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
OK thanks a lot @Zarkon .. that fine made a lot sense sir.. :)
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
@mathslover how did you get it \[7+4y=1(13cy)\]
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
that will be a prallal line
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
3x4y = 7 x = (7+4y) / 3 and then : x + cy = 13 x = 13cy 13cy = (7+4y)/3 (since they are equal to x) 3(13cy) = 7+4y
 one year ago

mathsloverBest ResponseYou've already chosen the best response.5
got it @mayankdevnani ?
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
just put 4/3 at C and you will have a parallel set of line and parallel set of lines have no solution
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
ok... @mathslover
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
3x4y=7 and 3x4y=39 are parallel lines how could they have a solution ..they intersect nowhere
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
thnx.... @Zarkon @ghazi @sauravshakya @ash2326 but special thnx to @mathslover
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
@mathslover is my approach correct?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Isnt your method same as @ash2326 's method........ @ghazi
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
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
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
yes it is similar to that of @ash2326
 one year ago
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