The suckyness of simultaneous equations - please help! (1 Viewer)

[Desert Dessert]

Hi guys,

I'm sure that y'all probably have better things to do...but would anyone be able to help me with this problem?! I've been struggling for ages...thank you soooooo much!

solve simultaneously:

3^x + 3^y = 10/3

3^(x+y) = 1

thanks again...even a couple of helpful tips would be greatly appreciated!

 

[kami]

Hi guys,

I'm sure that y'all probably have better things to do...but would anyone be able to help me with this problem?! I've been struggling for ages...thank you soooooo much!

solve simultaneously:

3^x + 3^y = 10/3

3^(x+y) = 1

thanks again...even a couple of helpful tips would be greatly appreciated!

3^0 = 1
x + y = 0
x = y

2(3^x) = 10/3
3^x = 5/3

and then you make x the subject and use your calculator to find out what it is.

I'm not 100% sure if that is how you're meant to do it since its been a while for me. But oh well, I tried.

 

[watatank]

What level math is that? That is a bitch of a question, never seen anything like it. Consulted a four unit student for help after I got stuck...

3^0 = 1

so (x+y) = 0

x = -y

3^x + 3^(-x) = 10/3

3^x + 1/3^x = 10/3

(3^x)^2 + 1 = (3^x).10/3 (multiply everything by 3^x)

Let 3^x be "a"

a^2 - (10/3). a + 1 = 0

Use quadratic formula (can't b bothered to do this part) to find that a = 3, 1/3

3^x = 3, 1/3

x = +/- 1

so y = +/- 1


Don't worry about it a whole lot that you couldn't do it, I couldnt do it either without help and I did both two and three unit maths for HSC. Your profile says you don't do ext maths 1 or 2....it's more likely that it will appear in a difficult three unit paper, and the chances of that question (or similar) appearing in two unit is slim to none

 

[Desert Dessert]

No kiddin'. Well, it's 2 unit question - assignment really...my teacher's nuts. Fantastic, but nuts. Thank you sooooooooo much for the help watatank!!! I was seriously stumped (and completely on the wrong track...)

...I will be very happy if i never see this question, or any question like it EVER again.

Thank you!!!

 

[watatank]

Wow you're in the year below me (or were, seeing as I left SHHS =P)!! Which teacher gave you that evil question?

 

[bos1234]

You might not see this in the 2 unit paper.

solve simultaneously:

3^x + 3^y = 10/3.............1

3^(x+y) = 1..................2 a^b x a^ c = a^(b+c)

2.

3^x X 3^y = 1
3^x = 1/3^y

subbing into 1

1/3^y + 3^y = 10/3

multply everything by 3

1/1^y + 9y = 10

if the base is 1. then anything to the power one is one

1^1000 = 1


1/1^y + 9y = 10
1+9y =10
9y=9
3^2y =3^2
2y=2
y=1

sub 1 back into the 1st eqn and get x =-1

 

[dolphy-chan]

hey all :wave: just here asking for help. If possible, explain to me, how I do Simultaneous equations TT__TT''' because I seriously do not get it....actually I do get it.....it's just that when the textbook puts the equations into words, I don't get...sort of.... :worried:

Some get me rather confused. So if some one culd help me, it would be great. Questions that get me confused:

An equal sided rhombus has the equation on one side 13-2y, another has 5x+6 and the last side with an equation has 3(x-y). Can some one explain to me how to do that?

Then there is this one:
When the numerator and the denominator of a certain fraction are increased by 1, the value is then 5/6. However when the when the numerator and the denominator are each decreased by 5 , the value of the fraction is then 3/4. Find the original fraction.

Is it:
x/y + 1 = 5/6
x/y - 5 = 3/4


Please tell me that is right, because this one confuses me alot ^^;;

Next one :

After a bill was debated in parliament, the members of the house of reps. voted in favour of the bill by a majority of 39. If theree were 97 present, find the number who voted in favour of the bill.

Can some one just tell me how to put this into simultaneous equations? Please? Just explain it to me, if possible ^__^llll


If it helps, these questions came from the year 10 mathscape text book by Clive Meyers, Loyd Dawe and Graham Barnsley. Its the 5.2 text book, exercise 3.5.

All you have to do is explain ti to me and I should probably get it.... no need for answers...that's what B.O.B is for when I have finished XD;;;

Gomen Nasai for wasting your time with my pathetic brain <_<''

dolphy-chan
--

Catch me at: http://rainbowdolphin.deviantart.com/

 

[SoulSearcher]

I suppose the first question is making
13 - 2y = 5x + 6 = 3(x-y)
So you have 2 simultaneous equations
13 - 2y = 3(x-y) ... (1)
5x + 6 = 3(x-y) ... (2)
So you want to remove one of the variables, in this case we'll choose y, so from 1,
13 - 2y = 3x - 3y
y = 3x - 13 ... (3)
Substitute this into (2) to find x, then sub the value of x you found into (3) to find y.

Now I'm thinking the second question is in this format:
(x+1)/(y+1) = 5/6 ... (1)
(x-5)/(y-5) = 3/4 ... (2)
So again, get y in terms of x in the first equation, then substitute y into the second equation to find x, then sub the value of x back into the first equation to find y.

The third equation is asking for the number of people who voted in favour of the bill, which we will give x. The other people will be denoted by y. So
x + y = 97, since the total number of everyone who voted is 97
x - y = 39, which is the difference between the number of people who voted for and against the bill. You should be fine from there.

 

[dolphy-chan]

Arigato. That has helped alot *hugs*

I just hope I go okay for my maths scape test that I have in like 2 days XD

 

[SoulSearcher]

You'll be fine :uhhuh:

 

[dolphy-chan]

O bugger....I think I really AM bad at maths =__='''

I got it wrong...

the answer is x= 3 y=-4
where as I got:

x= 3.17
y= 13.83

thanks for helping though ^^;

I'll see if encarta kids thingo helps

 

[dolphy-chan]

I think that was what my maths teacher said too =__=''''