Author Topic: This is METH-ness! (H/w help trend). (Read 13093 times) Tweet Share

Watery · « **Reply #135 on:** February 10, 2010, 10:21:07 pm »

A question from Essential Maths (cas). Help would be deeply appreciated

Consider the simultaneous equation
mx + 2y = 8
4x - (2-m)y = 2m

a) Find the value of m for which there are:
ii) no solutions ii) indefinitely many solutions

b) Solve the equation in terms of m, for suitable values of m

I have done section a but I don't know how to do (b), especially I dont know what they mean by 'for suitable values of m'
And when I solved it simultanously like normal I checked the answer and it wasn't correct.

the.watchman · « **Reply #136 on:** February 10, 2010, 10:28:58 pm »

So, solving simultaneously:

$4mx+8y = 32$ (1) (from the first eqn)

$4mx-m(2-m)y = 2m^2$ (2) (from the second eqn)

1-2: $8y+m(2-m)y = 32-2m^2$

$y=\frac{2(m+4)}{m+2}$

Then solve for x

Is this what you got?

Biology · « **Reply #137 on:** February 10, 2010, 10:30:02 pm »

That's correct, thanks a lot

the.watchman · « **Reply #138 on:** February 10, 2010, 10:31:25 pm »

Quote from: Biology on February 10, 2010, 10:30:02 pm

That's correct, thanks a lot

No prob

Watery · « **Reply #139 on:** February 10, 2010, 10:39:35 pm »

lol biology you're stuck on the same question?

And thanks watchman!!!

the.watchman · « **Reply #140 on:** February 10, 2010, 10:40:40 pm »

No, it's just m: $m(4x-(2-m)y) = m(2m)$

So, $4mx-m(2-m)y = 2m^2$

EDIT: This post just became USELESS!!!

Biology · « **Reply #141 on:** February 10, 2010, 10:41:21 pm »

^
Yep I was wrong on that part. Doh

Watery · « **Reply #142 on:** February 10, 2010, 10:58:43 pm »

Here's another simultaneous equation question which I have problem with.

Solve for x and y:
$(a+b)x+cy = bc$
$(b+c)y+ax = -ab$

The answer is supposed to be

$x=c$

$y=-a$

I got these mumbo jumbo answers which is far from the solution ><
Help please.

brightsky · « **Reply #143 on:** February 11, 2010, 12:08:38 am »

$(a+b)x + cy = bc$ (1)

$(b+c)y + ax = -ab$ (2)

From (1):

$(a+b)x = bc - cy$

$\implies x = \frac{bc - cy}{a+b}$

Substitute that into equation (2):

$(b+c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab$

Multiply everything by $a+b$ :

$(a+b)(b+c)y + a(bc-cy) = -ab(a+b)$

$(a+b)(b+c)y + abc - acy = -ab (a+b)$

$(a+b)(b+c)y - acy = -ab(a+b+c)$

$y[(a+b)(b+c) - ac] = -ab(a+b+c)$

$y = \frac{-ab(a+b+c)}{(a+b)(b+c) - ac}$

$\implies \frac{-a(ba + b^2 + bc)}{ab + ac + b^2 + bc - ac}$

$\implies \frac{-a(ba+b^2 + bc)}{ab + b^2 + bc}$

$\implies -a$

Substitute this into the equation to get $x$

Watery · « **Reply #144 on:** February 11, 2010, 12:10:26 am »

Omg you're a legend, you just saved me from possible scold and detention lol
:smitten: Thank you so much!

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