What math is this? I am trying to figure out what payment is necessary to cover both the payment processing fees and the state's gross receipts tax.
The payment processing fee is $0.30 plus 2.9% of the purchase price plus gross receipts tax and the gross receipts tax is 1.5% of the purchase price plus the payment processing fee.
The variables are:


*

*$x$ = purchase price

*$y$ = necessary payment

*$a$ = payment processing fee 

*$b$ = gross receipts tax


The formulas are:


*

*$y=(x+a+b)$

*$a=(x+b) \cdot 0.029+0.3$

*$b=(x+a) \cdot 0.015$
The equation grows infinitely since I can't solve $a$ without solving $b$ and I can't solve $b$ without solving $a$.
I don't remember what this type of math is called so I can't research how to solve for $y$.
 A: This type of math is called "solving a system of linear equations". You should be able to find it in any sort of algebra or precalculus book. 
You should take the bottom two equations
$a=(x+b) \cdot 0.029+0.3$
$b=(x+a) \cdot 0.015$
We have $2$ equations and $2$ unknown variables ($a$ and $b$). Using the techniques of solving a system of linear equations, we should be able to reduce this to $a =$ "stuff with $x$" and $b=$ "other stuff with $x$". Then you can plug these into your first equation, and you have $y$.
Let me know if you need more details. Going to sleep now.
A: This is going to be clumsy.
$a=(x+b)\cdot0.029+0.3 ~~~~~~~~~(i)\\
b=(x+a)\cdot0.015 ~~~~~~~~~~~~~~~~~~(ii)\\
\frac{b}{0.015}=x+a\Rightarrow \boxed{a = \frac{b}{0.015}-x}$
Putting this in equation $(i)$,
$\frac{b}{0.015}-x -0.3=(x+b)\cdot0.029$
$\frac{b}{0.015}-x -0.3=b\cdot0.029+x\cdot0.029$
$x\cdot0.029+x=\frac{b}{0.015}-b\cdot0.029-0.3$
$x\cdot1.029=\frac{b}{0.015}-b\cdot0.029-0.3$
$x\cdot1.029+0.3=b(\frac{1}{0.015}-0.029)$
$x\cdot1.029+0.3=66.6376\cdot b$
$\therefore \boxed{b = x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}}$
We know, $y=x+a+b$
$y=x+\frac{b}{0.015}-x+x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}$
$y=66.66b+0.004+.0154x$
substituting, the value of $b$ will give us a linear equation in $x$ something like $ax+b = y$. I am not sure how you gonna interpret this.
A: You know $x$, right? If so, note that
\begin{align}
a&=0.029(x+b)+0.3\\
&=0.029(x+0.015(x+a))+0.3\\
&=(0.029+0.029\cdot 0.015)x+0.029\cdot0.015 a+0.3
\end{align}
Thus
$$(1-0.029\cdot0.015)a=(0.029+0.029\cdot 0.015)x+0.3$$
which means
$$a=\frac{(0.029+0.029\cdot 0.015)x+0.3}{1-0.029\cdot0.015}.$$
Now that you have $x$ and $a$, its easy to plug them in to find $b$ and then $y$.
A: The general term for this is "simultaneous equations". In this case, as Ovi notes, they are linear equations. You can write is as
$$1a+1b+-y=-x$$
$$1a-.029b=.029x+.3$$
$$.015a-1b=-.015x$$
In matrix form, that's
$\begin{bmatrix}1&1&1\\1&-.029&-1\\ .015&.015&0\end{bmatrix}\begin{bmatrix}a\\b\\y \end{bmatrix}=\begin{bmatrix}-x\\.029x+.3\\-.015x \end{bmatrix}$
If you find the inverse of that matrix on the left (a web search of "matrix calculator" should get you a page that can give you the inverse), and then do a matrix multiplication of the inverse and $\begin{bmatrix}-x\\.029x+.3\\-.015x \end{bmatrix}$ will get you $a$, $b$, and $y$.
