Question

the no. of triplets of a,b,c for which the system of equations
ax+by=2a-b & (c+1)x+cy=10-a+3b has infinetely many solutions and ,
x=1 and y=3 is one of the solutions is,
(A) exactly one
(B)exactly two

(C)exactly three
(D)infinitly many.

pLz explain.

Answer 1

can u explain it

Answer 2

u can't xplain it

Answer 3

i think ur typing speed is very slow so u hav 2 work on that.

Answer 4

So we know that \[a+3b=2-3\implies a=-3b-1\]and\[4c=10+6b \implies c=\frac{5+3b}{2}\]Seems if we pick a value for b, we can get values for a and c that make it work for point (1,3). This indicates infinitely many triples (a,b,c) that can solve for that point.

If the system is dependent, \[ka=c+1, kb=c,~and~k(2a-b)=10-3a+b\]

Answer 5

After a little substitution,\[k(-3b-1)=\frac{3b+7}{2},~kb=\frac{5+3b}{2},~k(-7b-2)=10+10b+3\]which means we need to solve\[k=\frac{3b+7}{2(-3b-1)}=\frac{5+3b}{2b}=\frac{13+10b}{-7b-2}\]

Answer 6

After a little more work,\[6b^2+14b=-18b^2-36b-10~and~20b^2+26b=-21b^2-41b-10\]simplify a little to get\[17b^2-17b-20=0\]This has two roots, so we expect two solutions.

This is sort of circuitous, and there was enough algebra involved that I can in no way guarantee my calculations, but these posts should give you enough structure to investigate the problem.

Answer 7

thanks

Answer 8

After looking a little more, I think the last polynomial is\[17b^2-17b=0\]which would indicate a single solution (we've had b in the denominator, so it can't be zero). Like I said before, there were quite a few tedious calculations, and I'm not noted for my flawless arithmetic.