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Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:47pm On Jun 18, 2013
Ortarico:

Yes you're right, cause with the options it's easier to work out what/how the examiner wants the equation to be.

A little error there Chief Ortarico

=> log25x5 - log1/255
=> log(5)^2(x)5 - log(25)(-1)5
=> (2x)log55 - log(5)2(-1)5
=> (2x)log55 - (-2)log(5)5
Since log55 = 1
=> (2x)log55 - (-2)log(5)5
=> 2x .(1) + 2
=> 2x + 2
=> 2(x + 1)

So the final answer should be => 2(x + 1) which is option B as solved by Omosivie

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Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 7:51pm On Jun 18, 2013
doubleDx:

A little error there Chief Ortarico

=> log25x5 - log1/255
=> log(5)^2(x)5 - log(25)(-1)5
=> (2x)log55 - log(5)2(-1)5
=> (2x)log55 - (-2)log(5)5
Since log55 = 1
=> (2x)log55 - (-2)log(5)5
=> 2x .(1) + 2
=> 2x + 2
=> 2(x + 1)

So the final answer should => 2(x + 1)

Ok thanks general doubleDx

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 8:52am On Jun 19, 2013
To find the sum of positive integers...1+2+3+...+n=n(n+1)/2...
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 2:32pm On Jun 19, 2013
Pls help me solve this: 1/root 2 -1 minus 1/root 2 +1. Thanks
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 4:56pm On Jun 19, 2013
Omosivie,:
Pls help me solve this: 1/root 2 -1 minus 1/root 2 +1. Thanks

NB: _/ is for the root
1/(_/2 - 1) - 1/(_/2 + 1)
The LCM is (_/2 - 1)(_/2 + 1):
=> (_/2 + 1) - (_/2 - 1)/(_/2 - 1)(_/2 + 1)
Clearing the bracket we'll get:
=> _/2 + 1 - _/2 + 1/_/4 + _/2 - _/2 - 1

=> 1 + 1/2 - 1
=> 2. . . . . . . .ans

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 6:36pm On Jun 19, 2013
Ortarico:

NB: _/ is for the root
1/(_/2 - 1) - 1/(_/2 + 1)
The LCM is (_/2 - 1)(_/2 + 1):
=> (_/2 + 1) - (_/2 - 1)/(_/2 - 1)(_/2 + 1)
Clearing the bracket we'll get:
=> _/2 + 1 - _/2 + 1/_/4 + _/2 - _/2 - 1

=> 1 + 1/2 - 1
=> 2. . . . . . . .ans
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 1:28pm On Jun 21, 2013
Pls help me out with this two simultaneous equations:
1, xy=-3; x+y=2
2, x+2y=0; x^2+y^2=20
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 6:54pm On Jun 21, 2013
Omosivie,:
Pls help me out with this two simultaneous equations:
1, xy=-3; x+y=2
2, x+2y=0; x^2+y^2=20
wat r d opti0ns?
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:06pm On Jun 21, 2013
Omosivie,:
Pls help me out with this two simultaneous equations:
1, xy=-3; x+y=2
2, x+2y=0; x^2+y^2=20

Let's go. . . .
1. xy = -3. . . . . .eqn(i)
x + y= 2. . . . . . .eqn(ii)
from eqn(ii):
x= 2 - y. . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(i):
xy= -3
(2 - y) y= -3
2y - y2= -3
Re-arrangng gives:
y2 - 2y - 3= 0
(y - 3)(y + 1)
y= 3 or -1
:. x= 1 or -3
To check:
xy= -3; (-1*3)= -3
x + y= 2; (-1 + 3)= 2

2. x + 2y= 0. . . . . .eqn(i)
x2 + y2= 20. . . . .eqn(ii)
from eqn(i):
x + 2y= 0
x= -2y. . . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(ii):
x2 + y2= 20
(-2y)2 + y2= 20
4y2 + y2= 20
5y2= 20
y2= 4
y= +/- 2
substitute y= 2 in2 eqn(i):
x + 2y= 0
x + 2(2)= 0
x + 4= 0
x= -4
To check:
x + 2y= 0; -4 + 2(2)= 0
x2 + y2= 20; (-4)2 + (2)2= 20

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Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 8:42pm On Jun 21, 2013
Ortarico:

Let's go. . . .
1. xy = -3. . . . . .eqn(i)
x + y= 2. . . . . . .eqn(ii)
from eqn(ii):
x= 2 - y. . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(i):
xy= -3
(2 - y) y= -3
2y - y2= -3
Re-arrangng gives:
y2 - 2y - 3= 0
(y - 3)(y + 1)
y= 3 or -1
:. x= -1 or 3
To check:
xy= -3; (-1*3)= -3
x + y= 2; (-1 + 3)= 2

2. x + 2y= 0. . . . . .eqn(i)
x2 + y2= 20. . . . .eqn(ii)
from eqn(i):
x + 2y= 0
x= -2y. . . . . . .eqn(iii)
substitute eqn(iii) in2 eqn(ii):
x2 + y2= 20
(-2y)2 + y2= 20
4y2 + y2= 20
5y2= 20
y2= 4
y= 2
substitute y= 2 in2 eqn(i):
x + 2y= 0
x + 2(2)= 0
x + 4= 0
x= -4
To check:
x + 2y= 0; -4 + 2(2)= 0
x2 + y2= 20; (-4)2 + (2)2= 20
Thanks, but the answer is not the same with my textbook's answer. According to my textbook, the answer for number one is -3 or 1 and the answer for number two is (4,2) or (-4,2). I don't understand dis textbook and i'm totally fed up.
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 8:45pm On Jun 21, 2013
Leebliss13: wat r d opti0ns?
No options
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:50pm On Jun 21, 2013
Omosivie,:
Thanks, but the answer is not the same with my textbook's answer. According to my textbook, the answer for number one is -3 or 1 and the answer for number two is (4,2) or (-4,2). I don't understand dis textbook and i'm totally fed up.

But that's it now.
At no 1 x is (-3 , 1) then y is (-1 , 3) and at no 2 x is -4, then y is 2. To check that you saw was just to confirm. That's what I got sister

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Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 9:17pm On Jun 21, 2013
Ortarico:

But that's it now.
At no 1 x is (-3 , 1) then y is (-1 , 3) and at no 2 x is -4, then y is 2. To check that you saw was just to confirm. That's what I got sister
Thanks, I definately need glasses
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 9:42pm On Jun 21, 2013
Omosivie,:
Thanks, I definately need glasses

Thanks too, it happens to everyone.
Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 4:25pm On Jun 22, 2013
Remember this bro,the square root of a natural number is + or -...so,from where you got y^2=4...therefore y=+ or - 2...therefore y=+2 or -2...from the first equation x+2y=0...substitute y=-2...therefore x+2(-2)=0...x-4=0...x=4...substitute y=2...x+2y=0...x+2(2)=0...x+4=0...x=-4...therefore,when y=-2,x=4 and when y=2,x=-4
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 4:51pm On Jun 22, 2013
Calculusf(x):
Remember this bro,the square root of a natural number is + or -...so,from where you got y^2=4...therefore y=+ or - 2...therefore y=+2 or -2...from the first equation x+2y=0...substitute y=-2...therefore x+2(-2)=0...x-4=0...x=4...substitute y=2...x+2y=0...x+2(2)=0...x+4=0...x=-4...therefore,when y=-2,x=4 and when y=2,x=-4

All right, but if you look at it well, you'll see that I typed +/- 2, I know the rule boss.

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 1:46pm On Jun 24, 2013
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them
1, m^3 + mn^2-nm^2-n^3/m^4-n^4
2, -12a^2x/-8a^2b^2
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 2:39pm On Jun 24, 2013
Omosivie,:
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them
1, m^3 + mn^2-nm^2-n^3/m^4-n^4
2, -12a^2x/-8a^2b^2

1. m3 + mn2 - nm2 - n3/m4 - n4
m3 + n3 + mn2 - nm2/m4 - n4
(m - n)3 + mn(n - m)/(m - n)4
mn(n - m)/(m - n) . . . . .ans
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 2:40pm On Jun 24, 2013
2. Please re-type, there's a mistake or I don't get you well.
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 3:09pm On Jun 24, 2013
Ortarico: 2. Please re-type, there's a mistake or I don't get you well.
-12a^2 x / -8a^2 b^2
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 3:17pm On Jun 24, 2013
Ortarico:

1. m3 + mn2 - nm2 - n3/m4 - n4
m3 + n3 + mn2 - nm2/m4 - n4
(m - n)3 + mn(n - m)/(m - n)4
mn(n - m)/(m - n) . . . . .ans

The answer @ the back of my textbook is 1/m+n
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 6:47pm On Jun 24, 2013
Omosivie,:
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them
1, m^3 + mn^2-nm^2-n^3/m^4-n^4


{m3 + mn2 - nm2 - n3}/(m4 - n4)

Simplifying m4 - n4 yields => (m2^2 - n2^2 => (m2 - n2)(m2 + n2)...difference of 2 squares.

Rearranging the numerator & factorising yields =>
=> mn2 - nm2 + m3 - n3
=> mn(n - m) - (n3 - m3)
Now, (n3 - m3) can be rewritten as => (n - m)3 + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)3 + 3mn(n - m)}

Putting the numerator & denominator together yields:

=>> [ mn((n - m) - {(n - m})3 + 3mn((n - m))]/(m2 - n2)(m2 + n2)

Factorising (n - m) from the numerator above yields=>

=>> [(n - m)][mn - {(n - m)2 + 3mn}]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 - 2mn + m2 + 3mn)]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 + m2 + mn)]/(m2 - n2)(m2 + n2)

=>> (n - m)(-n2 - m2)/(m2 - n2)(m2 + n2)

=>> (n - m)(-1)(n2 + m2)/(m2 - n2)(m2 + n2)
=>> (m - n)(n2 + m2)/(m2 - n2)(m2 + n2)


=>> (m - n)/(m2 - n2)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

Hence,

{m3 + mn2 - nm2 - n3}/(m4 - n4) = 1/(m + n)

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 7:12pm On Jun 24, 2013
doubleDx:

{m3 + mn2 - nm2 - n3}/(m4 - n4)

Simplifying m4 - n4 yields => (m2^2 - n2^2 => (m2 - n2)(m2 + n2)...difference of 2 squares.

Rearranging the numerator & factorising yields =>
=> mn2 - nm2 + m3 - n3
=> mn(n - m) - (n3 - m3)
Now, (n3 - m3) can be rewritten as => (n - m)3 + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)3 + 3mn(n - m)}

Putting the numerator & denominator together yields:

=>> [mn[b](n - m)[/b] - {(n - m})3 + 3mn((n - m))]/(m2 - n2)(m2 + n2)

Factorising (n - m) from the above yields=>

=>> [(n - m)][mn - {(n - m)2 + 3mn}]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 - 2mn + m2 + 3mn)]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 + m2 + mn)]/(m2 - n2)(m2 + n2)

=>> (n - m)(-n2 - m2)/(m2 - n2)(m2 + n2)

=>> (n - m)(-1)(n2 + m2)/(m2 - n2)(m2 + n2)
=>> (m - n)(n2 + m2)/(m2 - n2)(m2 + n2)


=>> (m - n)/(m2 - n2)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

Hence,

{m3 + mn2 - nm2 - n3}/(m^4 - n^4) = 1/(m + n)
Wow!!!! This is so complex(one of the reasons why I hate maths). Could this be further maths?? I just pray I'm not given this question in my upcoming exam cos it's too long, difficult and complex. Thanks for solving it.
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 7:14pm On Jun 24, 2013
@ Ortarico, Calucus(fx) and DoubleDx. Pls help me out with the number 2 question
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:33pm On Jun 24, 2013
Omosivie,:
Wow!!!! This is so complex(one of the reasons why I hate maths). Could this be further maths?? I just pray I'm not given this question in my upcoming exam cos it's too long, difficult and complex. Thanks for solving it.

It's not Further Maths lol, the solution is not as long as it appears, though my explanation is, so you could understand better. It's just a matter of understanding two expressions and using them at the right time =>

1. That A2 - B2 = (A - B)(A + B) - difference of two squares
2. That A3 - B3 = (A - B)3 - 3AB(A - B) - difference of two cubes

So that A4 - B4 = (P2 - Q2) => (P - Q)(P + Q), such that =>> P = A2 and Q = B2.

Understanding these expressions and knowing when to apply/fix them in when you factorize makes the question weak and easy to solve.

I hope you catch my drift... so, you see? It's easy, don't be discouraged god sister! grin

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:47pm On Jun 24, 2013
Omosivie,:
-12a^2 x / -8a^2 b^2

Solution

-12a2 x /-8a2 b2

=> This is easy seeing as -4a2 is a factor of the numerator and denomitor, cancelling out & reducing the expression to =>

For the numerator => -12a2/-4a2 = 3x
For the denomitor => -8a2 b2/-4a2 = 2b2

Putting the simplified numerator and denomitor together yields =>

=> 3x/2b2
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:01pm On Jun 24, 2013
doubleDx:

{m3 + mn2 - nm2 - n3}/(m4 - n4)

Simplifying m4 - n4 yields => (m2^2 - n2^2 => (m2 - n2)(m2 + n2)...difference of 2 squares.

Rearranging the numerator & factorising yields =>
=> mn2 - nm2 + m3 - n3
=> mn(n - m) - (n3 - m3)
Now, (n3 - m3) can be rewritten as => (n - m)3 + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)3 + 3mn(n - m)}

Putting the numerator & denominator together yields:

=>> [ mn((n - m) - {(n - m})3 + 3mn((n - m))]/(m2 - n2)(m2 + n2)

***** Factorising (n - m) from the numerator above yields=>

=>> [(n - m)][mn - {(n - m)2 + 3mn}]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 - 2mn + m2 + 3mn)]/(m2 - n2)(m2 + n2)

=>> [n - m][mn - {(n2 + m2 + mn)]/(m2 - n2)(m2 + n2)

=>> (n - m)(-n2 - m2)/(m2 - n2)(m2 + n2)

=>> (n - m)(-1)(n2 + m2)/(m2 - n2)(m2 + n2)
=>> (m - n)(n2 + m2)/(m2 - n2)(m2 + n2) *****


=>> (m - n)/(m2 - n2)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

Hence,

{m3 + mn2 - nm2 - n3}/(m4 - n4) = 1/(m + n)

Respect to you @ general doubleDx. I supposed to have used the law of cubic and quadratic expression but for my hurry I missed it.
Please help explain more at the five starred phase of your work, thanks general!

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 9:12pm On Jun 24, 2013
Ortarico:

Respect to you @ general doubleDx. I supposed to have used the law of cubic and quadratic expression but for my hurry I missed it.
Please help explain more at the five starred phase of your work, thanks general!

[ mn (n - m) - {(n - m)3 + 3mn (n - m)}]/(m2 - n2)(m2 + n2)

If you look at the numerator of the above expression carefully, you will realize that (n - m) is a common factor/coefficient @ mn (n - m) , (n - m)3 and 3mn (n - m)

So that if (n - m) is replaced with say Q, the numerator would be => [ mn Q - {(Q)3 + 3mn(Q)}] which reduces the numerator to =>

mnQ - {Q3 + 3mnQ}
mnQ - Q3 - 3mnQ

Now, if Q is factorized from the above, it becomes

=> Q(mn - Q2 - 3mn)
=> Q(-2mn - Q2)
=> -Q(2mn + Q2)

Remember that Q = (n - m), so that -Q = (m - n), putting back the values of Q & -Q into the numerator's expression, yields =>

(m - n)(2mn + (n - m)2)
(m - n)(2mn + n2+ m2 - 2mn)
(m - n)(n2 + m2)

Putting the simplified numerator back into the expression yields =>

=>> (m - n)(n2 + m2)/(m2 - n2)(m2 + n2)

=>> (m - n)/(m2 - n2)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

I hope this is clearer...

2 Likes

Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 9:41pm On Jun 24, 2013
doubleDx:

[ mn (n - m) - {(n - m)3 + 3mn (n - m)}]/(m2 - n2)(m2 + n2)

If you look at the numerator of the above expression carefully, you will realize that (n - m) is a common factor/coefficient @ mn (n - m) , (n - m)3 and 3mn (n - m)

So that if (n - m) is replaced with say Q, the numerator would be => [ mn Q - {(Q)3 + 3mn(Q)}] which reduces the numerator to =>

mnQ - {Q3 + 3mnQ}
mnQ - Q3 - 3mnQ

Now, if Q is factorized from the above, it becomes

=> Q(mn - Q2 - 3mn)
=> Q(-2mn - Q2)
=> -Q(2mn + Q2)

Remember that Q = (n - m), so that -Q = (m - n), putting back the values of Q & -Q into the numerator's expression, yields =>

(m - n)(2mn + (n - m)2)
(m - n)(2mn + n2+ m2 - 2mn)
(m - n)(n2 + m2)

Putting the simplified numerator back into the expression yields =>

=>> (m - n)(n2 + m2)/(m2 - n2)(m2 + n2)

=>> (m - n)/(m2 - n2)

=>> (m - n)/(m - n)(m + n)

=>> 1/(m + n)

I hope this is clearer...

Oh yes, thanks boss, one love!

1 Like

Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 9:45pm On Jun 24, 2013
You have really done a great job@general doubledx...
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 10:56pm On Jun 24, 2013
^

Thanks bruv!
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 9:31am On Jun 25, 2013
doubleDx:

It's not Further Maths lol, the solution is not as long as it appears, though my explanation is, so you could understand better. It's just a matter of understanding two expressions and using them at the right time =>

1. That A2 - B2 = (A - B)(A + B) - difference of two squares
2. That A3 - B3 = (A - B)3 - 3AB(A - B) - difference of two cubes

So that A4 - B4 = (P2 - Q2) => (P - Q)(P + Q), such that =>> P = A2 and Q = B2.

Understanding these expressions and knowing when to apply/fix them in when you factorize makes the question weak and easy to solve.

I hope you catch my drift... so, you see? It's easy, don't be discouraged god sister! grin
Thanks, I'll cram these algebraic identities/expressions.

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