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 $y$.

  • $\begingroup$ Thank you to those who have gone beyond the question of what type of math this involves and are helping me understand my specific "word problem." $\endgroup$ May 28, 2019 at 7:38

4 Answers 4


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.

  • $\begingroup$ I don't see any way to reduce the equations, they depend on each other. $\endgroup$ May 28, 2019 at 5:35
  • 2
    $\begingroup$ @user5507535 If you look up "system of linear equations", you will see there is a way. I can show you how tomorrow if you wish. $\endgroup$
    – Ovi
    May 28, 2019 at 5:39
  • $\begingroup$ Ok. This is how I think one would simplify the equation. I still don't know how to "get rid of" $a$ and $b$. Maybe one can't. $((a-0.3)/0.029))-b=x=(b/0.015)-a$ $\endgroup$ May 28, 2019 at 6:40

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 $(i)$,

$\frac{b}{0.015}-x -0.3=(x+b)\cdot0.029$

$\frac{b}{0.015}-x -0.3=b\cdot0.029+x\cdot0.029$




$x\cdot1.029+0.3=66.6376\cdot b$

$\therefore \boxed{b = x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}}$

We know, $y=x+a+b$



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.


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 $y$.

  • $\begingroup$ I used this approach in a spreadsheet and the result for my example $x=49.99$ was $y=52.54$ but upon performing the reverse a $y$ value of 52.54 equaled an $x$ value of 49.93 rounding all values to pennies at the final result. I don't see why that would happen. $\endgroup$ May 28, 2019 at 7:27

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 $y$.


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