O-Level Additional Math

[danlim]

Hi hope someone can help to solve this maths problem


Given that x2-y2=42
X+y=14
Calculate the value of

1) X-Y
2) x2+y2

X2 is actually x square and Y2 is y square

[Skyed]

danlim:

Hi hope someone can help to solve this maths problem


Given that x2-y2=42
X+y=14
Calculate the value of

1) X-Y
2) x2+y2

X2 is actually x square and Y2 is y square

(X2-y2) = (x-y)(x+y)
14(x-y) = 42
Therefore x-y = 3

(X-y)2 = x2 - 2xy + y2
(X+y)2 = x2 + 2xy + y2

Add both of the above together, you get 2x2 + 2y2 = 9 + 196 therefore X2 + y2 = 102.5

[danlim]

Skyed:

danlim:

Hi hope someone can help to solve this maths problem


Given that x2-y2=42
X+y=14
Calculate the value of

1) X-Y
2) x2+y2

X2 is actually x square and Y2 is y square

(X2-y2) = (x-y)(x+y)
14(x-y) = 42
Therefore x-y = 3


(X-y)2 = x2 - 2xy + y2
(X+y)2 = x2 + 2xy + y2

Add both of the above together, you get 2x2 + 2y2 = 9 + 196 therefore X2 + y2 = 102.5

Another question
When I use x=8,y=6
X+y=14
But x-y=2, why not 3?

[koguma]

danlim:

danlim:

Hi hope someone can help to solve this maths problem


Given that x2-y2=42
X+y=14
Calculate the value of

1) X-Y
2) x2+y2

X2 is actually x square and Y2 is y square

Another question
When I use x=8,y=6
X+y=14
But x-y=2, why not 3?

Hi, Just my thought.

I am not sure where you get x=8 and y=6, but I assume that you randomly use 2 numbers to check your answer.

In your question, there are 2 equations,
(1) x2-y2=42
(2) X+y=14

If you put the x=8 and y=6 into the 1st equation x2-y2=42, it does not add up to \"42\".

Although the values fit into the 2nd equation X+y=14, but since it does not fit into equation 1, the 2 values are not correct.

The value for x and y should fit into both equations, and should then fit into the 3rd equation (x-y=3).

From the ans provided by Skyed, you will get x = 8.5 and y = 5.5 .
You substitute this set into all 3 equations and you will get the correct ans.

[nounou]

deleted

[nounou]

Hi, need help on another Q. Kindly help. Thank you.


Evaluate 2012^2 - 2011^2 + 2010^2 - 2009^2 + ...
+ 4^2 - 3^2 + 2^2 - 1^2

:? :?:

:thankyou:

[koguma]

deleted

[koguma]

nounou:

Hi, need help on another Q. Kindly help. Thank you.


Evaluate (2012^2 - 2011^2) + (2010^2 - 2009^2) + ...
+ (4^2 - 3^2) + (2^2 - 1^2)

:? :?:

:thankyou:

Use this formulae : a^2 − b^2 = (a + b)(a − b)

2012^2 - 2011^2
= (2012+2011)(2012-2011)
= 2012+2011

2010^2 - 2009^2
= (2010+2009)(2010-2009)
= 2010+2009

4^2 - 3^2
= (4+3)(4-3)
= 4+3

2^2 - 1^2
= (2+1)(2-1)
= (2+1)

so you need to add 1+2+3+4 .... +2009+2010+2011+2012 to get the ans.

hopefully someone else can give a shorter working answer.

[nounou]

deleted

[nounou]

Hi, thank you very much for your help. I understand your workings.


Q: how to find 1+2+3+4 .... +2009+2010+2011+2012 ???? :?:

Please help. :thankyou: :lovesite:

koguma:

nounou:

Hi, need help on another Q. Kindly help. Thank you.

Evaluate (2012^2 - 2011^2) + (2010^2 - 2009^2) + ...
+ (4^2 - 3^2) + (2^2 - 1^2)

:? :?:

:thankyou:

Use this formulae : a^2 − b^2 = (a + b)(a − b)

2012^2 - 2011^2
= (2012+2011)(2012-2011)
= 2012+2011

2010^2 - 2009^2
= (2010+2009)(2010-2009)
= 2010+2009

4^2 - 3^2
= (4+3)(4-3)
= 4+3

2^2 - 1^2
= (2+1)(2-1)
= (2+1)

so you need to add 1+2+3+4 .... +2009+2010+2011+2012 to get the ans.

hopefully someone else can give a shorter working answer.