Question

Find x and y in terms of a and b for this system.
ax+by=0
a^2(x)+b^2(y)=1

also, a does not equal 0, b does not equal 0, and a does not equal b

Answer 1

Is that second equation:\[a ^{2x} + b ^{y} = 0\]because that is how it is written.

Answer 2

@Studentc14 ?

Answer 3

no, sorry, that was confusing the way i posted it. it is a^2(x)+b^2(y)=1

Answer 4

i fixed it sorry, i didnt see what i had typed. my bad.

Answer 5

np. Re-writing to find solution then:

(a^2)x + (ab)y = 0 adding the negative of equation #1 to equation #2
(a^2)x + (b^2)y = 1

(a^2)x + (ab)y = 0
[(b^2) - (ab)]y = 1 y = 1/[(b^2) - (ab)]

Using equation #1, re-written as: x = -by/a

x = (-b/a)/[(b^2) - (ab)] = 1/(a^2 - ab)

Answer 6

All good now?

Answer 7

I put these expressions for x and y back into the original equations and they work.

Answer 8

So that's it.

x = 1/(a^2 - ab)

and

y = 1/(b^2 - ab)

Any questions on that?

Answer 9

(a^2)x + (ab)y = 0 adding the negative of equation #1 to equation #2
(a^2)x + (b^2)y = 1
can you explain this i dont understand what happened

Answer 10

I made the coefficients on "x" the same so when you subtract equation #1 from equation #2, the "x" term disappears. The right side is just 1 - 0 = 1

Answer 11

so you multiply the first equation by a?

Answer 12

yes, because I am using Gaussian elimination and I have to isolate the variables. I just chose "x" so that I came down to one equation in one variable, "y".

Answer 13

ok. so I was trying to solve this earlier, and i got to here:
-by^2+aby+1=0
and then i tried to plug it into the quadratic equation, with
-b="a"
ab="b"
1="c"
would this be a different way to solve it or is this completely off?

Answer 14

Where did -by^2 + aby + 1 = 0 come from?

Answer 15

[(b^2) - (ab)]y = 1 you had it like this...
i had not factored out a y and my signs were opposite on both sides (i had multiplied by -1 for some reason) so mine was like this
-by^2+aby=-1
so then i added the 1 to both sides and it looked like a quadratic

Answer 16

Before you get too far afield with your explanation, the key to solving this is to recognize that you have 2 equations in 2 variables, so you would be using either elimination or substitution.

Answer 17

$$ax+by=0\\by=-ax\\a^2x+b^2y=1\\b^2y=1-a^2x\\-abx=1-a^2x\\(a^2-ab)x=1\\x=\frac1{a(a-b)}\\by=-\frac{a}{a(a-b)}=-\frac1{a-b}$$

Answer 18

Just saw your explanation on where your equation came from. That would be an error because there is no "y^2" term. It all "x" and "y" to the first power.

Answer 19

... therefore \(x=\frac1{a(a-b)}\), \(y=-\frac1{b(a-b)}\).

Answer 20

When you are dealing with undefined constants like "a" and "b", it is usually easier to use elimination than substitution. But no quadratic here. Make sense now?

Answer 21

@oldrin.bataku it would be good to take the negative out of the numerator for y to keep a consistent "1" in the numerator for both "x" and "y".

Answer 22

Just switch the order of subtraction in the denominator for "y".

Answer 23

@tcarroll010 it makes no difference, but you could do \(y=\frac1{b(b-a)}\)

Answer 24

okay i understand this better now. thank you

Answer 25

True, it can be written either way, but if you keep "1" in the numerator, then you will be able to see a consistent "- ab" term in the denominator for both "x" and "y" and there is a good comparison to be made on how "x" and "y" differ. Anyway, the problem has already been solved.

Answer 26

Take a look at my 6th post.

Answer 27

There is a pleasing visual difference of just a^2 for one and b^2 for the other in the denominator.

Answer 28

Good luck in all of your studies and thx for the recognition! @Studentc14

Answer 29

you're welcome, thank you actually, you were very helpful :)

Answer 30

Remember, Gaussian elimination is usually the way to go over substitution. And the key was 2 equations in 2 variables. If you get that far in the future, you will generally have clear sailing.

Answer 31

thanks for the tips!

Answer 32

The most aesthetic solution IMHO would probably be something like \(x=\frac1ak\), \(y=\frac1bk\) where \(k=\frac1{a-b}\).

Answer 33

It was a pleasure working with you and I hope I see you again in the problems!