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Algebra

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by krusta80 » Sun Mar 04, 2012 7:25 pm
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
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by neerajkumar1_1 » Sun Mar 04, 2012 7:28 pm
try solving it algebraically..
krusta80 wrote:
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
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by krusta80 » Sun Mar 04, 2012 8:20 pm
krusta80 wrote:
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
2mn/(m-n) = 2*[x/(x-y)]*[x/(x+y)]/[x/(x-y)-x/(x+y)] = [2*x^2/(x^2-y^2)]/[(x^2+xy-x^2+xy)/(x^2-y^2)] = 2*x^2/(2xy) = x/y :)
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by neerajkumar1_1 » Sun Mar 04, 2012 8:48 pm
lol..
prob try the other way arnd..
krusta80 wrote:
krusta80 wrote:
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
2mn/(m-n) = 2*[x/(x-y)]*[x/(x+y)]/[x/(x-y)-x/(x+y)] = [2*x^2/(x^2-y^2)]/[(x^2+xy-x^2+xy)/(x^2-y^2)] = 2*x^2/(2xy) = x/y :)
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by krusta80 » Sun Mar 04, 2012 9:00 pm
neerajkumar1_1 wrote:lol..
prob try the other way arnd..
krusta80 wrote:
krusta80 wrote:
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
2mn/(m-n) = 2*[x/(x-y)]*[x/(x+y)]/[x/(x-y)-x/(x+y)] = [2*x^2/(x^2-y^2)]/[(x^2+xy-x^2+xy)/(x^2-y^2)] = 2*x^2/(2xy) = x/y :)
I must admit that I tried doing so the other way...I ended up with a formula containing n and m, but it did not seem to match choice D. Obviously there must be a way to do it, but I think you may be missing the point anyway.

This is a classic question where using the easiest method to solve it saves tons of time.
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by neerajkumar1_1 » Sun Mar 04, 2012 10:37 pm
I agree that the one should pick the easiest way.
but it depends on what u prefer easy..
for me algebra is very easy..
and while picking numbers, u dont always get a certain answer..

anyways.. no issues with your picking numbers soln..

I am just wondering why we are getting straight forward algebra wrong..

there has to be a way to solve it right.. and there must be something we are doing wrong..
I proceeded by equating the x's.. may be we cant do that..

experts can probably help..
krusta80 wrote:
neerajkumar1_1 wrote:lol..
prob try the other way arnd..
krusta80 wrote:
krusta80 wrote:
neerajkumar1_1 wrote:If |x|<>|y|, xy<>0, x/(x+y) = n, and x/(x-y) = m, then x/y= ?

a) 3mn/2
b) 3m/(2n)
c) n(m+2)/2
d) 2mn/(m-n)
e) (n^2 - m^2)/(nm)
This one is probably best solved using substitution.

Let's try x = 2 and y = 3...

n = 2/5
m = -2
x/y = 2/3

a) 3mn/2 = -6/5 NOPE
b) 3m/(2n) = -15/2 NOPE
c) n(m+2)/2 = 0 NOPE
d) 2mn/(m-n) = (-8/5)/(-2-2/5) = (-8/5)/(-12/5) = 2/3 YUP
e) (n^2 - m^2)/(nm) = (4/25 - 4)/(-4/5) = (-96/25)/(-4/5) = 24/5 NOPE

D it is.
2mn/(m-n) = 2*[x/(x-y)]*[x/(x+y)]/[x/(x-y)-x/(x+y)] = [2*x^2/(x^2-y^2)]/[(x^2+xy-x^2+xy)/(x^2-y^2)] = 2*x^2/(2xy) = x/y :)
I must admit that I tried doing so the other way...I ended up with a formula containing n and m, but it did not seem to match choice D. Obviously there must be a way to do it, but I think you may be missing the point anyway.

This is a classic question where using the easiest method to solve it saves tons of time.
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by krusta80 » Sun Mar 04, 2012 11:57 pm
Equation 1
x/(x+y) = n
x = nx+ny
(n-1)x + ny = 0
x/y = n/(1-n)

Equation 2
x/(x-y) = m
x = mx - my
(m-1)x - my = 0
x/y = m/(m-1)

x/y = (x/y + x/y)/2 = [n/(1-n) + m/(m-1)]/2 = [(m-1)n + (1-n)m]/[(1-n)*(m-1)] = (m-n)/[(1-n)*(m-1)]

From here, I am unable to get it to convert to choice D. Maybe some sleep will help!
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by beatthegmat.garry » Mon Mar 05, 2012 4:29 am
@Krusto, you almost got it right:
Equation 1
x/(x+y) = n
x = nx+ny
(n-1)x + ny = 0
y/x = (1-n)/n

Equation 2
x/(x-y) = m
x = mx - my
(m-1)x - my = 0
y/x =(m-1)/m

y/x+y/x = [(1-n)/n + (m-1)/m] = [(1-n)m + (m-1)n]/[nm] = (m-mn+mn-n)/mn= (m-n)/mn

hence, (y/x)*2=(m-n)/mn;
(y/x)=(m-n)/2mn;
x/y=2mn/(m-n) which is D.
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by krusta80 » Mon Mar 05, 2012 4:33 am
beatthegmat.garry wrote:@Krusto, you almost got it right:
Equation 1
x/(x+y) = n
x = nx+ny
(n-1)x + ny = 0
y/x = (1-n)/n

Equation 2
x/(x-y) = m
x = mx - my
(m-1)x - my = 0
y/x =(m-1)/m

y/x+y/x = [(1-n)/n + (m-1)/m] = [(1-n)m + (m-1)n]/[nm] = (m-mn+mn-n)/mn= (m-n)/mn

hence, (y/x)*2=(m-n)/mn;
(y/x)=(m-n)/2mn;
x/y=2mn/(m-n) which is D.
Queue the chorus!! Thank you, sir. Excellent idea of calculating y/x instead and then flipping. :)
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by neerajkumar1_1 » Mon Mar 05, 2012 4:35 am
nice work...
well i was equating the x's from eqn 1 and 2
which i guess was wrong..
beatthegmat.garry wrote:@Krusto, you almost got it right:
Equation 1
x/(x+y) = n
x = nx+ny
(n-1)x + ny = 0
y/x = (1-n)/n

Equation 2
x/(x-y) = m
x = mx - my
(m-1)x - my = 0
y/x =(m-1)/m

y/x+y/x = [(1-n)/n + (m-1)/m] = [(1-n)m + (m-1)n]/[nm] = (m-mn+mn-n)/mn= (m-n)/mn

hence, (y/x)*2=(m-n)/mn;
(y/x)=(m-n)/2mn;
x/y=2mn/(m-n) which is D.
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