# If a, b, x, and y are real numbers where ax + by = 3 ax^2 + by^2 = 7 ax^3 + by^3 = 16

#### NovErdy

##### New member
Hey, it's me again
Can you help me solve this?

If a, b, x, and y are real numbers where
ax + by = 3
ax^2 + by^2 = 7
ax^3 + by^3 = 16
ax^4 + by^4 = 42
What is ax^5 + by^5 = ?

I don't know where to start.

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#### ksdhart2

##### Senior Member
Seems to me like it's a standard system of equations, just with a lot more to keep track of and a lot more ways to get bogged down in all the details. That said, there's four equations in four variables, so you can solve it in whatever usual way you use to solve systems of equations. For a problem like this, I prefer to use substitution and elimination. Let's start with just the first equation and see what we can come up with:

$$\displaystyle ax + by = 3 \implies ax = 3 - by \implies x = \dfrac{3 - by}{a}$$

That seems promising. What if we plugged that into the second equation?

$$\displaystyle a\left(\dfrac{3 - by}{a}\right)^2 + by^2 = 7 \implies \dfrac{(3 - by)^2}{a} + by^2 = 7 \implies \cdots$$

You try finishing this part here and see what you get. At the end of this step, I'd managed to find an expression for a in terms of the only b and y, which I could then plug into the expression for x and simplify it some. Where does all this lead you? What do you think you'd do next?

#### Harry_the_cat

##### Senior Member
This is how I approached the problem:

$$\displaystyle (ax+by)(x+y) = (ax^2 +by^2) +(a+b)xy = 7 + (a+b)xy$$ … as it turns out, you don't actually need this!

$$\displaystyle (ax^2+by^2)(x+y) = (ax^3 + by^3) +ax^2y + by^2x = (ax^3 + by^3) + xy(ax+by) = 16 +3xy$$ .... (*)

Rewriting LHS, we have

$$\displaystyle 7(x+y) = 16 +3xy$$ ….....…(1)

$$\displaystyle (ax^3+by^3)(x+y) = (ax^4+by^4) + xy(ax^2 +by^2)$$ .... (**)

giving

$$\displaystyle 16(x+y) = 42 + 7xy$$ ..........(2)

I then let A=x+y and B=xy, and simultaneously solved for A and B using equations (1) and (2).

Then used a similar expansion to (*) and (**) to bring in the fifth powers, and substitute values of A and B for x+y and xy respectively.

... I'll let you do all that ...

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#### NovErdy

##### New member
For a problem like this, I prefer to use substitution and elimination.
Thanks for the suggestion, but substitution method seems hard for me, and I think there is more efficient way to do this kind of question, and that's what I was searching for.

This is how I approached the problem:

$$\displaystyle (ax^2+by^2)(x+y) = (ax^3 + by^3) +ax^2y + by^2x = (ax^3 + by^3) + xy(ax+by) = 16 +3xy$$ .... (*)

Rewriting LHS, we have

$$\displaystyle 7(x+y) = 16 +3xy$$ ….....…(1)

...

I then let A=x+y and B=xy, and simultaneously solved for A and B using equations (1) and (2).

Then used a similar expansion to (*) and (**) to bring in the fifth powers, and substitute values of A and B for x+y and xy respectively.
Thanks for your suggestion, actually I think this is the clever way to do the question, and I finally figured out the solution of the problem.

...and actually I've tried this before xD, but I want to know how to solve it. And I guess I can't use WolframAlpha in an incoming math competition

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#### NovErdy

##### New member
Thanks for all the help, I really appreciate it . Btw, wish me luck in my upcoming local math competition

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#### Harry_the_cat

##### Senior Member
NovErdy, what did you get for the answer? Was it 20?

#### NovErdy

##### New member
NovErdy, what did you get for the answer? Was it 20?
Yup, with x + y = -14 and xy = -38.

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#### Harry_the_cat

##### Senior Member
Yup, with x + y = -14 and xy = -38.

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Great! Thats what I got too.