Author Topic: simultaneous equations! (Read 1014 times) Tweet Share

mangmang712 · « **on:** January 20, 2009, 05:45:38 pm »

Hi!

these questions have probably been asked before, so sorry if it is repetitive!
im finding it difficult to distinguish between unique, infinite and no solutions in simultaneous equations, what is the difference?
for unique solutions is the determinant not equal to 0?

and there is also this:

Find the value of 'm' for which the simultaneous equations have no solution
(m+3)x + my = 12 and (m-1)x + (m-3)y = 7

The question suggests to use a calculator. Can you answer the question without it?
Thankyou!

Flaming_Arrow · « **Reply #1 on:** January 20, 2009, 05:54:59 pm »

no solution is when lines are parallel

$(m+3)x + my = 12$

$(m-1)x + (m-3)y = 7$

rearrange for y

$y= \frac{ 12-x(m+3)}{m}$

$y= \frac{7-x(m-1)}{m-3}$

make gradient equal eachother

$\frac{m+3}{m} = \frac{m-1}{m-3}$

$m^2-9 = m^2-m$

$\implies \boxed{m = 9}$

no solution is when lines are parallel
one solution is when they intersect once
infinite solutions are when lines are the same( same gradient and y intercept)

mangmang712 · « **Reply #2 on:** January 20, 2009, 06:10:26 pm »

Thank you that cleared up everything!
I was stuck on that for a while

mangmang712 · « **Reply #3 on:** January 24, 2009, 09:35:08 am »

Oh I have another question!
Can someone please clarify how $y=a((cb+d)/(ab+1))-c$ becomes $(ad-c)/(ab+1)$
It is part of a literal simultaneous equation
Sorry, I am not too good at these

Damo17 · « **Reply #4 on:** January 24, 2009, 10:45:25 am »

Quote from: mangmang712 on January 24, 2009, 09:35:08 am

Oh I have another question!
Can someone please clarify how $y=a((cb+d)/(ab+1))-c$ becomes $(ad-c)/(ab+1)$
It is part of a literal simultaneous equation
Sorry, I am not too good at these

$y=\frac{a(cb+d)}{ab+1} - c$

$=\frac{acb+ad}{ab+1} - c$

$= \frac{acb+ad}{ab+1} - \frac{c(ab+1)}{ab+1}$

$=\frac{acb+ad}{ab+1} - \frac{cab+c}{ab+1}$

$=\frac{(acb+ad)- (cab+c)}{ab+1}$

$= \frac{ad-c}{ab+1}$

mangmang712 · « **Reply #5 on:** January 24, 2009, 10:56:29 am »

Thanks!

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