# 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 $$y$$. • 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." May 28, 2019 at 7:38 ## 4 Answers 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. • I don't see any way to reduce the equations, they depend on each other. May 28, 2019 at 5:35 • @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. – Ovi May 28, 2019 at 5:39 • 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)-aMay 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 $$(i)$$, $$\frac{b}{0.015}-x -0.3=(x+b)\cdot0.029$$ $$\frac{b}{0.015}-x -0.3=b\cdot0.029+x\cdot0.029$$ $$x\cdot0.029+x=\frac{b}{0.015}-b\cdot0.029-0.3$$ $$x\cdot1.029=\frac{b}{0.015}-b\cdot0.029-0.3$$ $$x\cdot1.029+0.3=b(\frac{1}{0.015}-0.029)$$ $$x\cdot1.029+0.3=66.6376\cdot b$$ $$\therefore \boxed{b = x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}}$$ 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. 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 $$y$$. • I used this approach in a spreadsheet and the result for my examplex=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. 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 $$y$$.