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# Maths Question watch

1. When 1 is added to the numerator and denominator of the fraction m/n the new fraction is 3/2. When 1 is subtracted from the numerator and denominator of the fraction m²/n² the new fraction is 21/8. Find the possible values of m and n.

I wrote this as two simultaneous equations:

(m+1)/(n+1) = 3/2
2(m+1) = 3(n+1)
2m + 2 = 3n + 3
2m = 3n + 1

(2m)² = (3n + 1)²
4m² = 9n² + 6n + 1
8m² = 18n² + 12n + 2 <---(1)

And the second equation:

(m² - 1)/(n² - 1) = 21/8
21/8 = (m² - 1)/(n² - 1)
21(n² - 1) = 8(m² - 1)
21n² - 21 = 8m² - 8 <---(2)

Substitute (1) into (2):

21n² - 21 = (18n² + 12n + 2) - 8
21n² - 21 = 18n² + 12n - 6
3n² - 12n - 15 = 0
n² - 4n + 5 = 0

And there are no answers for n.
Can someone help me see where I've gone wrong please. Thanks.
2. Try factorising the second equation using the difference between two squares.
3. 3n² - 12n - 15 = 0
n² - 4n + 5 = 0
you've divided the -15 by -3 for some reason

so you'd have:

n² - 4n - 5 = 0

which is
(n-5)(n+1)=0

so n= 5 or n= -1
4. (Original post by SUKBarracuda)
you've divided the -15 by -3 for some reason

so you'd have:

n² - 4n - 5 = 0

which is
(n-5)(n+1)=0

so n= 5 or n= -1
Thanks, that's (one) error, but still not the right answer.
5. (Original post by mik1a)
When 1 is added to the numerator and denominator of the fraction m/n the new fraction is 3/2. When 1 is subtracted from the numerator and denominator of the fraction m²/n² the new fraction is 21/8. Find the possible values of m and n.

I wrote this as two simultaneous equations:

(m+1)/(n+1) = 3/2
2(m+1) = 3(n+1)
2m + 2 = 3n + 3
2m = 3n + 1

(2m)² = (3n + 1)²
4m² = 9n² + 6n + 1
8m² = 18n² + 12n + 2 <---(1)

And the second equation:

(m² - 1)/(n² - 1) = 21/8
21/8 = (m² - 1)/(n² - 1)
21(n² - 1) = 8(m² - 1)
21n² - 21 = 8m² - 8 <---(2)

Substitute (1) into (2):

21n² - 21 = (18n² + 12n + 2) - 8
21n² - 21 = 18n² + 12n - 6
3n² - 12n - 15 = 0
n² - 4n + 5 = 0

And there are no answers for n.
Can someone help me see where I've gone wrong please. Thanks.
i.e 21(n+1)(n-1)=8(m+1)(m-1)
so 21(n+1)(n-1)=8((3n+1)/2+1)((3n+1)/2-1)
so 21(n+1)(n-1)=(12n+12)(3n/2-1/2)
=18n^2+12n-6
so 3n^2-12n-15=0
so 3(n^2-4n-5)=0
so 3(n-5)(n+1)=0
6. n=5 is correct but n cannot be -1
7. (Original post by hornblower)
Try factorising the second equation using the difference between two squares.
How does that help? It just gives a more complicated equation to simplify.
8. (Original post by lgs98jonee)

so 3n^2-12n-15=0
so 3(n^2-4n-5)=0
so 3(n-5)(n+1)=0
The top line is what I got, but that isn't the right answer.
9. (Original post by mik1a)
The top line is what I got, but that isn't the right answer.
when n=5,m=8 and so (8+1)/(5+1)=9/6=3/2
and (8^2-1)/(5^2-1)=63/24=21/8.
10. (Original post by IntegralAnomaly)
when n=5,m=8 and so (8+1)/(5+1)=9/6=3/2
and (8^2-1)/(5^2-1)=63/24=21/8.
But I can't guess then in an exam and prove they are right. Does anyone have a proper working?
11. (Original post by mik1a)
But I can't guess then in an exam and prove they are right. Does anyone have a proper working?
guess? u worked it out earlier! n^2-4n-5=0
12. (Original post by IntegralAnomaly)
guess? u worked it out earlier! n^2-4n-5=0
Oh yeah duh. THanks!

I thought that the two values of n were one value of n and one of m! You can make some really stupid assumptions when you're not concentrating on it!

Thanks everyone! (good start to P2!)
13. (m+1)/(n+1) = 3/2
2(m+1) = 3(n+1)
2m + 2 = 3n + 3
2m = 3n + 1

(2m)² = (3n + 1)²
4m² = 9n² + 6n + 1
8m² = 18n² + 12n + 2 <---(1)

And the second equation:

(m² - 1)/(n² - 1) = 21/8
21/8 = (m² - 1)/(n² - 1)
21(n² - 1) = 8(m² - 1)
21n² - 21 = 8m² - 8 <---(2)

Substitute (1) into (2):

21n² - 21 = (18n² + 12n + 2) - 8
21n² - 21 = 18n² + 12n - 6
3n² - 12n - 15 = 0
n² - 4n - 5 = 0
(n-5)(n+1)

so n=5 or -1

but can't be -1

so n=5

therefore m = 8.

Your working was all right except for that one mistake...

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