Solving Implicit Equations for y in Terms of x

[Fuz]

How do you go about solving implicit equations for y in terms of x that look like these?

y² + yx = 1

and

y³ + yx = 1

or even more complicated implicit equations.

I'm taking AP Calculus AB this year and am just curious how this is done.

Here are the solutions from Wolframalpha:
http://www.wolframalpha.com/input/?i=y^2+%2B+yx+%3D+1
http://www.wolframalpha.com/input/?i=y^3+%2B+xy+%3D+1

 

[Number Nine]

You just need to play with the function until you find an answer. Use the usual rules of arithmetic. You aren't going to encounter any situations in AP Calculus where you would need to solve a hugely complex implicit function for y.

 

[Fuz]

Can you show me the steps for one of them? I've already tried solving them myself, but got nowhere.

 

[Mentallic]

Can you show me the steps for one of them? I've already tried solving them myself, but got nowhere.

By solve I assume you mean to express the implicit function in the form y=f(x). Well, for the first you can use the quadratic formula or complete the square (it's basically equivalent) and for the second you'll need to use the cubic formula.

 

[CellCoree]

Solving implicit equations for y in terms of x can be a challenging task, but there are a few strategies that can be used to approach these types of equations. The first step is to rearrange the equation so that all terms containing y are on one side of the equation and all terms containing x are on the other side. In the given examples, this would result in the equations being rearranged to y2 + yx - 1 = 0 and y3 + yx - 1 = 0.

From here, there are a few different methods that can be used to solve for y in terms of x. One approach is to use the quadratic formula for the first equation, which can be applied to equations of the form ax2 + bx + c = 0. For the second equation, a cubic formula can be used, which is a bit more complex but can still be applied.

Another method is to use numerical methods, such as Newton's method or the bisection method, to approximate the solutions for y in terms of x. These methods involve plugging in different values for x and using them to find the corresponding values for y, gradually getting closer to the exact solutions.

In more complicated implicit equations, it may be necessary to use a combination of these methods or even more advanced techniques, such as the method of undetermined coefficients or the method of elimination. It's important to carefully analyze the equation and choose the most appropriate method for solving it.

In calculus, you will also learn about implicit differentiation, which can be used to find the derivative of y with respect to x in implicit equations. This can also be a useful tool in solving for y in terms of x, as it can help determine the general shape of the graph and any critical points.

Overall, solving implicit equations for y in terms of x requires careful analysis, a good understanding of algebraic techniques, and the use of appropriate methods for the specific equation at hand. Practice and familiarity with these methods will help you become more comfortable with solving these types of equations.