never thought id be one of those asking for help but algebras not my strong point.....
A= 3n+1
B= 2(n+2)
C= n-7
D= 2n+1
find the value of n if the expressions B and C are equal.
which 2 expressions could never be equal for any value of n?
is there actually a formula to help work these out, or is it just a case of trial and error with numbers? ive already worked some out for him such as the value of n if A and B are equal is 3, but that was just guesswork. just cant get the others 🙁
if theres a formula at least i can show him to help him work others out.
cheers
think ive just worked out second one is A and D.
For the B & C part...
If B = C
Then
(n-7) = 2(n + 2)
You should be able to rearrange and get the result quite easily.
B= 2(n+2)
C= n-7
2n + 4 = n - 7
n = -11
1)n=-11 surely just algebra based on 2(n+2)=n-7
second part B and D, cos that says 2n+4=2n+1
or 4=1 it can't be!!
A & D are equal when n = 0.
For the second part, you just need to go through each combination, do the equation and work out the value of n.
There is a shortcut, but just giving you that straight off won't let your kid learn anything from this...
B = C
2(n+2) = n-7
multiply out the brackets: 2n + 4 = n - 7
collect terms on each side (subtract one n and 4 from BOTH sides):
2n + 4 - n - 4 = n - 7 - n - 4
n = -11
now you can solve for A, B , C and D by substitution.
The key is to collect like terms all the n's on one side, all the constants on the other. To do this, you must do the same operation on both sides of the equation.
Hope that helps
tis indeed -11 thanks, but i still struggle to understand how you get it from that the formula. told you i was crap at algebra 🙂
EDIT: just seen the above so gonna try and get my head round it ta 🙂
cheers
jon1973 has it.
the checkback is:
if n=-11,
2(n)+4=(n)-7
2(-11)+4=(-11)-7
-22+4=-11-7
-18=-18
And the other part is trial and error, but its quite obvious looking at it that A and D can't possibly be equal if n is the same in both.
which 2 expressions could never be equal for any value of n?
I think that just means trying the same approach you used for B = C and the one that can't be solved is the answer.
So A = B
3n+1 = 2(n+2)
3n + 1 = 2n + 4
n = 3 So it's not that one.
motiv.. A and D are equal for n = 0... 😀
It's all Greek to me.
Rearrange and substitute. That's all you need to do.
motiv.. A and D are equal for n = 0..
Does the question say they are only interested in non-zero solutions?
well, the first one is solved by putting B and C equal to each other.
So we say 2(n+2)= n-7
which expands to 2n+4=n-7
we need to keep both sides equal, so what ever we do to one side we must do to the other. So we fiddle with them until we get to a formula which gives us n=...
in the first place we subtract n from both sides.
which gives n+4=-7
now we subtract 4 from both sides
which gives n= -11
for the second one we look for a case in which it looks like we end up with something a bit strange. For me it was one where we start with 2 ns on each side and an uneven number of other bits. I'm not sure i can operationalise that. I guess the stndard way would be to check each combination
ooh youre right!! so er.....which 2 will it be?
B and D
any reason why mate? any formula to prove it?
thanks
ooh youre right!! so er.....which 2 will it be?
It's B and D
If we assume B equals D then we know that:
B= 2(n+2) and D= 2n+1
Therefore:
2(n+2) = 2n +1
Therefore:
2n + 4 = 2n + 1
Take 2n from each side and we are left with:
4= 1
Which is nonsense so our initial assumption of B equaling D is false.
well, try putting them equal to each other then follow my method of doing the same to each side
wahey, youre waaaaay cleverer than me, but yep i get it!!
i can understand it when its explained, its just trying to get there myself. my lad understands it far better than me.
thanks a lot doods 🙂
aw crap. back to first year of my degree to relearn everything. doh!
motiv, you can join me at the back of the class with our dunce caps on 🙂