Splitting $~x~$ and $~y~$ and solving for $~y~$

Question

Solve for $~y~$:

$$3xy+5y=2x+7$$

I have to do this for as an assignment going into Calculus, the problem is the teacher wants us to research how to do the problems on our own, and I don't know what I should be looking for.

Should I start by attempting to factor, and if so what?

Any and all tips would be greatly appreciated.

Answer 1

I'll do a similar problem

$$8xy+11y=22x+7$$

Notice that on the LHS of this problem we have

$$8xy+11y$$

which contains the variable $y$ in both terms. Let's break this up by writing

$$8xy+11y=y(8x+11)$$

Then, we can write

$$8xy+11y=y(8x+11)=22x+7$$

from which, if $8x+11\neq 0$, we can divide both the middle term and RHS term by $8x+11$ to form

$$y=\frac{22x+7}{8x+11}$$

and have therefore solved for $y$ (observe that the above representation is valid provided that $8x+11\neq 0$).

You should follow the exact same procedure for your problem.

1. Factor out the $y$ on the LHS of the equation.
2. Divide by what $y$ is factored with (observing that this will produce a similar condition in which a specific value of $x$ is invalid).
3. Then arrive at a solution which contains $y$ on one side of the equation and $x$ plus constants on the other side of the equation.

Answer 2

Solve for $y$ in your case is very easy: in fact you have only to pick up $y$ from $3xy+5y=y(3x+5)$. So the equation now is: $y(3x+5)=2x+7$ and: $$y=\frac{2x+7}{3x+5}$$