O-Level Additional Math
-
(x+1)/(x-1) is of the form
D = d.q + r where D = x+1, q =1, d = x-1, r =2
whereas (-x-1)/(1-x) is of the form
-D = -d.q -r ==> (-D) = (-d).q + (-r) where (-D) = (-x-1), q = 1, (-d) = (1-x) and -r = -2
Because the divisor is changed to -ve sign, the remainder also has to change to -ve sign so that dividend also changes sign
HTH.SKT:
Can anyone help to clear my DD's doubt for the following:
(x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?
TIA. -
Hi,
Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.
Thanks in advance! -
benorito:
You need to know that product of two numbers = HCF x LCM of those two numbersHi,
Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.
Thanks in advance!
280 * y = 6160 * 40
y = (6160 * 40)/280 = 880 -
iFruit:
Thank you !!
You need to know that product of two numbers = HCF x LCM of those two numbers
280 * y = 6160 * 40
y = (6160 * 40)/280 = 880 -
Hi,
Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.
(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.
TIA. -
SKT:
Using Cramer’s ruleHi,
Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.
(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.
TIA.
x = [30m - 20(m-1)]/ [3m² – (m² -3m +2)] = (10m+20)/ (2m² + 3m -2)
y = [20m -10(m-2)] / (2m² + 3m -2) = (10m+20)/ (2m² + 3m -2)
Recall that equation has
1) infinite solutions when (10m+20) = 0 and (2m²+ 3m -2)=0
2) No solutions when (10m+20) # 0 and (2m² + 3m -2)=0
(2m² + 3m -2)= (2m-1)(m+2) =0----> m=-2 or m=1/2
When m=-2, 10m+20 = 0,
when m=1/2, 10m+20 = 25
So infinite solutions when m=-2
No solutions when m=1/2 -
SKT:
Below is an ELEMENTARY approachHi,
Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.
(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.
TIA.
For linear equation : y = mx +c
(a) 2 lines will have no unique solution if they are the same line i.e. both lines have the same gradient, m and the same y-intercept, c
mx + (m-1)y = 10 that is: y = -mx/(m-1) +10/(m-1)
(m-2)x + 3my = 20 that is: y = -(m-2)/(3m) + 20/(3m)
If the gradients are the same :
-mx/(m-1) = -(m-2)/(3m)
3m² = m²-2m-m+2
2m²+3m-2 = 0
(2m-1)(m+2) = 0
m = 1/2 or -2
If the y-intercepts are the same :
10/(m-1) = 20/(3m)
30m = 20m - 20
m = -2
Therefore, there is no unique solution if m = -2
(b) 2 lines will have no solution if they are parallel i.e. same gradient but different y-intercept.
Therefore, there is no solution if m = 1/2 -
Hi,
However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?
TIA -
SKT:
That is right.Hi,
However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?
TIA
No Unique solution means, either no solution or infinite solutions.
when m=-2, there are infinite solutions (when it's the same line)
so both 1/2 and -2 are answers for (a). -
SKT:
NO UNIQUE solution means these 2 lines are collinear (2 lines overlapping each other, so every point on the lines is a solution or can say there are infinite number of solutions)Hi,
However, the answer provided for (a) is 1/2, -2, wonder what's the meaning of \"no unique solution\". Are the two lines parallel or collinear or both?
TIA
So answer for (a) is -2 ONLY
NO solution means these 2 lines are parallel but NOT collinear (these 2 lines will never meet which is why there is no solution)
So answer for (b) is 1/2
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