.2119=x2
.2056=x1
I solved this using a calculator, and it does look like those two equations are both true if you round to the 2nd decimal spot at least.
(0.040x - 0.0025)/ 0.0025 = 2.29
X1 came out to be .2056. This X, if rounded up, should be equal to the other X in the second equation. The first equation is verified by the second equation usually, using a number of manipulations or methods.
The second equation, 3.39 is normally a truth statement. When you see an equation and that equal sign, it reads as a logic statement. Thus they want you to assume the equation is true, but of course they do not provide you what X=?, so generally the student attempts to solve for X by using the two equations together.
The first equation is rather complicated to demonstrate this concept in action, so I will reduce the complexity.
2x+5=9
3x=6
Assuming X is a real number (it's not imaginary or i), and presupposing/assuming that both equations are true at the same time without conflicts, then X=2 for both.
3*2=6
2*2+5=9
How to arrive at the X=? number is the difficulty usually. If 3x=6 and x is a real number variable, then merely reverse manipulate the equation. Force x to be alone on one side, the left side, and divide six by 3, moving the 3 from the left to the right. I do not know your current status, so I have to cover some of the basics, even if it is already mastered, since I do not know the student's current strengths/weaknesses.
x=6/3
To put it in words, what number multiplied against 3 equals six. To test if this works for the entire equation set, put x=2 and substitute it into the first (or more complicated) equation.
When the equation writes =9 and then =6, assume it is a truth statement that is true. Unless the question specifically asks to verify whether the equation is correct or not.
The other option is that there is a typo in the question, since it uses several decimal places.
For the first complicated equation, I would write it down on paper to check my logic and for any mistakes. Even with a calculator, the order of those operations can be manipulated, but it is easy to make mistakes. Which is why the second equation is much simpler and is normally the one you want to solve for X with. Once you have a tentnative "what does X equal", then you can plug it into/substitute it into the first complicated equation, to test whether the equation and statement is still true.
Due to the way the first equation is formatted, it is extremely complicated, perhaps overly so, if it is trying to teach a simple algebra concept of how to deal with X variables.
If there is additional context about this question, feel free to provide it, as I cannot guarantee I know what they are trying to test you on, with the current limited data.
P.S. Thinking about it some more, those decimals will also convert to whole fractions, perhaps it is testing you on fraction and order of operations too. To convert .0025 to a fraction, do this operation 1/.0025, and that computes to be =400.
1/.04=25
That means the conversion of .0025 can be directly lead to or be replaced by 1/400. And .04=1/25
To test it, just divide to see if it equals the decimal.
[(1/25*x- (1/400)]/ (1/400)
That may or may not help simplify it. If the context is that the student cannot use a calculator, then you would have to use basic fractions to get the answer by hand.
Kenneth S.
11/14/17