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hi

solve 2^x+2y=1 nd 3^2x+y=27 simultaenously pls thnks

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Reply 1
Original post by tariqballe
solve 2^x+2y=1 nd 3^2x+y=27 simultaenously pls thnks


y=27−32xy=27-3^{2x} substitute that into 2x+2y=12^{x}+2y=1 to solve for x and then plug back in for y
Reply 2
Original post by Robbie242
y=27−32xy=27-3^{2x} substitute that into 2x+2y=12^{x}+2y=1 to solve for x and then plug back in for y


Actuall ythrough elimination
Reply 3
Original post by tariqballe
Actuall ythrough elimination


:confused::confused::confused:

What on earth does this mean
Original post by tariqballe
Actuall ythrough elimination


I assume you want it using the other method? It's much easier to substitute in cases like this because otherwise you'd end up taking away numbers with different powers (which is a lot harder to solve).
Reply 5
Original post by k4l397
I assume you want it using the other method? It's much easier to substitute in cases like this because otherwise you'd end up taking away numbers with different powers (which is a lot harder to solve).


Its actually 2^x+2y=2^0 so itz easier eliminating
Reply 6
Original post by tariqballe
Its actually 2^x+2y=2^0 so itz easier eliminating


nonsense


please explain what your issue is - if you feel that you can answer the question using a different one to the one suggested then your question is answered - but the suggestion made by Robbie is the sensible approach
Reply 7
Original post by tariqballe
solve 2^x+2y=1 nd 3^2x+y=27 simultaenously pls thnks


Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=12^{x + 2y} = 1 and 32x+y=273^{2x + y} = 27, which are much simpler to solve?
Original post by davros
Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=12^{x + 2y} = 1 and 32x+y=273^{2x + y} = 27, which are much simpler to solve?


By Jove, I think he's cracked it!
Reply 9
Original post by Mr M
By Jove, I think he's cracked it!


Robbie's equations were too hard so I changed the problem into one I could solve more easily :smile:

Ready for some 'Tough Young Teachers' tonight?
Original post by davros
Ready for some 'Tough Young Teachers' tonight?


I prefer reality TV so I've chosen Benidorm.
Original post by davros
Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=12^{x + 2y} = 1 and 32x+y=273^{2x + y} = 27, which are much simpler to solve?


prsom
Reply 12
Original post by TenOfThem
nonsense


please explain what your issue is - if you feel that you can answer the question using a different one to the one suggested then your question is answered - but the suggestion made by Robbie is the sensible approach


2^x+2y=2^0
x+2y=0----i
then 3^2x+y=3^3
so 2x+y=3----ii
so x+2y=0--*2
2x+y=3 --*1

therefore 2x+4y=0---ii
-
2x+y=3
0+3y=-3
so y=-1 then plug it bck into any equations to find x
Original post by tariqballe
2^x+2y=2^0
x+2y=0----i
then 3^2x+y=3^3
so 2x+y=3----ii
so x+2y=0--*2
2x+y=3 --*1

therefore 2x+4y=0---ii
-
2x+y=3
0+3y=-3
so y=-1 then plug it bck into any equations to find x


So, now you have shown that you were asking a completely different question to the one that you posted

You can do the question so there is no issue

And, either elimination or substitution are very straightforward with the equations that you had



If you want continued help, I suggest that in future

•

Post the correct question

•

Thank people for their help or suggestions

•

Let people know when you have solved the problem

•

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Reply 14
Original post by tariqballe
2^(x+2y)=2^0
x+2y=0----i


Those are brackets. :tongue:
Reply 15
Original post by BabyMaths
Those are brackets. :tongue:


lol.........
Is this FP1???
*adopts a look of fear*


Posted from TSR Mobile
Reply 17
Original post by benwalters1996
Is this FP1???
*adopts a look of fear*


Posted from TSR Mobile


Looks more like GCSE to be honest :smile:
Reply 18
Original post by davros
Looks more like GCSE to be honest :smile:


Why leave it so late? :wink:
At least the word 'solve' is correct in the opening post. That's something.

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