O-Level Additional Math

[koguma]

Herbie:

Thanks! Thanks!
Ds tried using y=-xy-2 into eqn 2 and couldn't solve it.

When you make \"y\" the subject, i.e. \"y\" appears on the left hand side of the equation, the right hand side cannot have y.

Hence you cannot have y = -xy-2 since the \"y\" appears on both side of the equation.

[Herbie]

Hi koguma! Many thanks for yr explanation!

[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.