Rearranging Equations: How to Express y as a Function of x?

[beamthegreat]

I'm slightly embarrassed to ask such an easy question but how do you rearrange the following equation such that y=f(x)?

x²+2xy-3y²=0

I always have trouble with these types of problems when there are multiple "y"'s in an equation especially when they have different powers. Can someone please help?

 

[Mark44]

I'm slightly embarrassed to ask such an easy question but how do you rearrange the following question such that y=f(x)?

x²+2xy-3y^2=0

I always have trouble with these types of problems when there are multiple "y"'s in an equation especially when they have different powers. Can someone please help?

The left side factors very easily. You will get two relationships with x as a function of y.

 

[beamthegreat]

Can you clarify what you mean? If I factor the equation I get (x-y)(x+3y)=0 but I want it in a y=f(x) form.

 

[nuuskur]

It does not satisfy the definition of a function, why would you need to express it as one?
At any rate, I would go about expressing y by either completing the square or factorising and then expressing y.

 

[beamthegreat]

Can you show how its done? I want to express it in a y=f(x) format since there is another function that intersects with this one.
Do you know any other way to find the coordinate that it intersects without having to express it in a y=f(x) format?

 

[nuuskur]

Yes, you have 2 canonic equations. You can express y or x from either one, plug it into the other equation and solve the resulting equation. Assuming those 2 sets intersect, the result will not be an empty set.
Again, the first is not a function, it's some sort of a relation between the two variables. It is a set, there can be intersection between sets.

 

[PeroK]

I'm slightly embarrassed to ask such an easy question but how do you rearrange the following equation such that y=f(x)?

x²+2xy-3y²=0

I always have trouble with these types of problems when there are multiple "y"'s in an equation especially when they have different powers. Can someone please help?

First, consider x as fixed, then you have a quadratic equation in y. Simply solve that quadratic and you have y expressed in terms of x. If there is more than one value of y for a given x, then this will not be a function, but you could make it a function by restricting the values of y to be unique for each x.

 

[beamthegreat]

First, consider x as fixed, then you have a quadratic equation in y.

I'm not sure if I understand you. The problem still persist even if you try to rearrange the equation such that x=f(y).

Also, if you want to rearrange the equation Vi(t)+1/2g(t)^2=D for t such that t equals a function of Vi and g, how do you do it?

Can someone actually demonstrate it?

 

[Mark44]

I'm not sure if I understand you. The problem still persist even if you try to rearrange the equation such that x=f(y).

Also, if you want to rearrange the equation Vi(t)+1/2g(t)^2=D for t such that t equals a function of Vi and g, how do you do it?

Can someone actually demonstrate it?

Your notation is confusing, especially Vi(t), which could be interpreted as Vi being a function of t, or V times i, which is a function of t.
I'm going to assume you mean this: v_it + (1/2)gt² = D. This equation is quadratic in t, so you can use the Quadratic Formula to solve for t. To make it a bit easier, write the equation in this form: gt² + 2v_it - D = 0.

In this form, a = g, b = 2v_i, and c = -D. Just plug these into the Quadratic Formula to get t.

 

[WWGD]

It may be possible to rotate the plane to get rid of the mixed term.

 

[beamthegreat]

Your notation is confusing, especially Vi(t), which could be interpreted as Vi being a function of t, or V times i, which is a function of t.

I'm going to assume you mean this: v_it + (1/2)gt² = D. This equation is quadratic in t, so you can use the Quadratic Formula to solve for t. To make it a bit easier, write the equation in this form: gt² + 2v_it - D = 0.

In this form, a = g, b = 2v_i, and c = -D. Just plug these into the Quadratic Formula to get t.

Alright, thanks.

What do do you do if you can't use the quadratic equation?

For example, how do you rearrange the following equation:

y=x^5+x^3+x^2+x+4 so that x=f(y)

 

[PeroK]

I'm not sure if I understand you. The problem still persist even if you try to rearrange the equation such that x=f(y).

Also, if you want to rearrange the equation Vi(t)+1/2g(t)^2=D for t such that t equals a function of Vi and g, how do you do it?

Can someone actually demonstrate it?

It's really not very difficult. $x^2 +2xy - 3y^2 = 0 \ \ iff \ \ (x-y)(x+3y)=0 \ \ iff \ \ y = x \ \ or -\frac{x}{3}$

And that's y expressed in terms of x. As explained above, this is not a function of x.

If you want y = f(x), where (x, f(x)) is a solution to the equation, then you have two obvious options:

$y = f(x) = x$ and $y = f(x) = -\frac{x}{3}$

 

[Mark44]

Alright, thanks.

What do do you do if you can't use the quadratic equation?

For example, how do you rearrange the following equation:

y=x^5+x^3+x^2+x+4 so that x=f(y)

I don't believe it's possible for this one. Did you make it up?

 

[beamthegreat]

I don't believe it's possible for this one. Did you make it up?

Yes I made it up. Is it really impossible to find the inverse of this equation?

 

[WWGD]

The inverse/implicit function theorems give conditions under which a local inverse (may)exist(s).

 

[beamthegreat]

The inverse/implicit function theorems give conditions under which a local inverse (may)exist(s).

I see. What about other equations that has an inverse function?

For example, y=x^3+x^2 has an inverse function according to wolfram but how the heck do you solve it? Is there a rule/method you can use to solve these types of problems? What are these problems called anyway?

 

[HallsofIvy]

There are formulas for solving quadratic, cubic, and quartic equations. It has been proven that there cannot be a formula, involving only roots, addition, multiplication, addition, and subtraction, for solving polynomials of degree 5 or higher.

 

[WWGD]

There are formulas for solving quadratic, cubic, and quartic equations. It has been proven that there cannot be a formula, involving only roots, addition, multiplication, addition, and subtraction, for solving polynomials of degree 5 or higher.

Do you mean for _all_ quintics (sorry if you do, I am no expert in this area)? I think if the Galois group of the polynomial is solvable, then you can use those operations only.

 

[HallsofIvy]

I meant a general formula.. Yes, of course, there exist solutions to some polynomials of any degree.

 

[WWGD]

Ah, sorry, I am kind of new to that area.