TYS 'O' LVL maths question
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hey guys, i have doubts about tis question
its 2002 p1 simultaneous equation
21c)
x = 2y + 11
4x +3y = 0
i noe i have to rearrange the equation and do but i still cant get the answer
the answer is x=3 , y= -4
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4(2y+11) +3y = 0
8y + 44 + 3y = 0
11y = -44
y = -4
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can explain
i dont get it -
Since x = 2y + 11,
we just substitute x = 2y + 11 into the second equation 4x + 3y = 0
That's how you get the next part
4(2y+11) +3y = 0
8y + 44 + 3y = 0
11y = -44
y = -4
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OHH !!!
WOW U SO SMART.
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:)
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Originally posted by gunner77:
hey guys, i have doubts about tis question
its 2002 p1 simultaneous equation
21c)
x = 2y + 11
4x +3y = 0
i noe i have to rearrange the equation and do but i still cant get the answer
the answer is x=3 , y= -4
I'm surprised this was found in O levels at all. This is linear simultaneous equations, taught at sec2 level?
Ok first, label your equations.
x = 2y + 11---------- (1)
4x +3y = 0----------- (2)
There are two ways to do it. I think the way you've learnt for now is the elimination method. Even though the substitution method is easier, I'll teach this first.
From 1,
x= 2y + 11
4x = 8y + 44 -------(3)
(2)-(3): 4x+3y - 4x = 0 - (8y + 44)
3y = -8y - 44
11y = -44
y= - 4.
Sub y=-4 back into (2)
4x - 12 = 0
4x = 12
x = 3
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kk
wait ah
i see if i got anymore doubts.
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DAMMIT WHY DID I LOSEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE. WHY DID HE HAVE TO UNDERSTAND YOUR POOOOOOOOOOOOOOOOOST.
I suppose I'll skip the substitution method since you taught him already. -.-
P.S. Is minor swearing allowed in the homework forum?
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haha okay.. got to slp soon
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Originally posted by MrSean:
DAMMIT WHY DID I LOSEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE. WHY DID HE HAVE TO UNDERSTAND YOUR POOOOOOOOOOOOOOOOOST.
I suppose I'll skip the substitution method since you taught him already. -.-
P.S. Is minor swearing allowed in the homework forum?
nope.. pls pay 50 cents to me. thanks
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Originally posted by MrSean:
I'm surprised this was found in O levels at all. This is linear simultaneous equations, taught at sec2 level?
Ok first, label your equations.
x = 2y + 11---------- (1)
4x +3y = 0----------- (2)
There are two ways to do it. I think the way you've learnt for now is the elimination method. Even though the substitution method is easier, I'll teach this first.
From 1,
x= 2y + 11
4x = 8y + 44 -------(3)
(2)-(3): 4x+3y - 4x = 0 - (8y + 44)
3y = -8y - 44
11y = -44
y= - 4.
Sub y=-4 back into (2)
4x - 12 = 0
4x = 12
x = 3
bro i used tis method and i tink becoz of its long step i miscalculated, tats why cannot get the ans.
after tinuviel07 use the elimination method, its much clearer and shorter.
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guys one more question
but i cannot find my cube and square in my keyboard leh
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just put x^3 or x3 for cubes lor
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6b ) T x T^3 / sq root of T = T^n
find the value of N
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T^4 / T^1/2 = T^n
For e.g. if the base is the same like
4^5 / 4^2,
you can change it to 4^ (5-2) = 4^3,
so for division, you subtract the power,
if it's 4^5 x 4^2, you change it to 4^(5+2) = 4^7,
so for multiplication, you add the powers together.
T ^ (4-1/2) = T^n
T^ (3.5) = T^n
n = 3.5
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WAH u correct leh
can explain..
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i do like tis..
T^4 = T^2 x T^n
4 = 2 + n
n = 2
why i wrong ah?
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square root of t is T to the power of 1/2
so is T^ 1/2,
you bring it over, it is
T^4 = T^1/2 x T^n
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sq root bring over is not sq meh?
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it's like
1/T^2 =5
you bring T^2, does it become T^1/2?
it becomes
1 = 5 x T^2,
the divide becomes multiply. doesn't affect the root
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ohhh i get it
thanks tinuviel07!
maths pro
thanks for the help
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sure no prob.. see ya
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Originally posted by gunner77:
sq root bring over is not sq meh?
MUST... WIN....Well gunner you could square it if you want.
(T^4/T^0.5)^2 = (T^n)^2
T^8/T^1 = T^2n
T^7 = T^2n
7=2n
n = 3.5
Crap lost again.
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see ya
tmr i post more