Does r,s lie on the line

Stat 1). You have 2 unknowns which you cannot solve for
(3r + 2 -s) (4r + 9 - s ) = 0
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one
=> 4r + 9 - s = 0 : Two variables. Cannot solve for either one

Stat 2) Same. You have 2 unknowns which you cannot solve for
(4r - 6 - s) (3r + 2 - s ) = 0
=> 4r - 6 - s = 0 : Two variables. Cannot solve for either one
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one

Combined
You can eliminate one of the variables ( r or s ) and find the answer to see if it contains the point.

jcan-gmatter wrote:Stat 1). You have 2 unknowns which you cannot solve for
(3r + 2 -s) (4r + 9 - s ) = 0
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one
=> 4r + 9 - s = 0 : Two variables. Cannot solve for either one

Stat 2) Same. You have 2 unknowns which you cannot solve for
(4r - 6 - s) (3r + 2 - s ) = 0
=> 4r - 6 - s = 0 : Two variables. Cannot solve for either one
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one

Combined
You can eliminate one of the variables ( r or s ) and find the answer to see if it contains the point.

Hmm...
In the first case I have two equations
3r + 2 - s = 0
4r + 9 - s = 0
Can I solve for r ans s.

Similarly for case 2.
So shouldnt answer be D.
your comments.

where have I gone wrong..

gmatrant wrote:

jcan-gmatter wrote:Stat 1). You have 2 unknowns which you cannot solve for
(3r + 2 -s) (4r + 9 - s ) = 0
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one
=> 4r + 9 - s = 0 : Two variables. Cannot solve for either one

Stat 2) Same. You have 2 unknowns which you cannot solve for
(4r - 6 - s) (3r + 2 - s ) = 0
=> 4r - 6 - s = 0 : Two variables. Cannot solve for either one
=> 3r + 2 - s = 0 : Two variables. Cannot solve for either one

Combined
You can eliminate one of the variables ( r or s ) and find the answer to see if it contains the point.

Hmm...
In the first case I have two equations
3r + 2 - s = 0
4r + 9 - s = 0
Can I solve for r ans s.

Similarly for case 2.
So shouldnt answer be D.
your comments.

You cannot solve 3r + 2 - s = 0, 4r + 9 - s = 0 simultaneously to get a value for r and s .... Remember to solve two linear equations simultaneously the variables need to satisfy both the equation eg. 2x+3y=5 and 3x+2y=5 over here both the equations are true

Whereas in the above equation we say 3r+2-s = 0 or 4r+9-s = 0 that is both of them are not true..that is only one of them is equal to 0 ..

Regards.

if y=3x+2 contains (r,s) that means s=3r+2
1. (3r+2-s)*(4r+9-s)=0 then 3r+2-s=0 or 4r+9-s=0 insuff.
2. (4r-6-s)*(3r+2-s)=0 then 4r-6-s=0 or 3r+2-s=0 insuff
but combined 1+2 we get that 3r+2-s=0 so 3r+2=s sufficient

i think the answer should be D cos combining both i gt the same answers as workin for each of the them.

what am i doin wrong?

lawalx ..i will try to be put it in a simple way and as gabriel has put it earlier ..let me know if it makes sense

Let (3r + 2 -s) = a and (4r + 9 - s ) = b (4r - 6 - s) = c

From 1

a * b = 0 so this means either a is 0 or b is 0 ..but we cant say which one

From 2

c * a = 0 so this means either a is 0 or b is 0 ..but we cant say which one


Combining a = 0 ..which gives the same equation as the stem

Hope this helps