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a= 3
b= 2
d= 5
a=4b + (6/ c-1)- 15d
okey so il show how far i am. could someone please tell me where im goin wrong. many thanks
3= (8) + (6/ C-1) - 75
75+3 = 8+ (6/ C-1)
78 = 8 + (6/ C-1)
78-8 = (6/C-1)
70 = (6/C-1)
C-1 (70) = 6
C (70) = 7
C= (7/70)
C= 0.1
You're right until the fourth line up from the bottom. If (c-1) definitely has brackets round then 6=70(c-1) you need to multiply out to get 6=70c-70 then rearrange. You should get c=76/70 = 1.1 approx
but if i then subsitute 1.1 value for C then the answer should equal a= (3)?
don't sub in 1.1 (the answer is really 1.0857....) as this is the rounded up answer. if you put c=76/70 into the original equation then you will get a=3.
You were told by Willow T that your bracketing is faulty; should be:Quote:
Originally Posted by decoy808
a = 4b + 6 / (c-1) - 15d
Now re-arrange:
6 / (c-1) = a - 4b + 15d
c - 1 = 6 / (a - 4b + 15d)
c = 6 / (a - 4b + 15d) + 1
Hi ineed to solve this exercise. Who can help me?
Let A subset of R (real numbers) A no empty. A is annotated if and only if exist M (positive rel) such that absolute value of x is less than or equal that M, for all x in A