The greatest mathematical achievement of the ancient world was the development of Greek mathematics. Around 300 B.C., Greek mathematicians developed many of the concepts and functions we still use today.
They developed the fields of arithmetic, algebra, and geometry, along with terms and concepts such as equation, variable, coordinate plane, and proof.
Their contributions are so powerful that this period is called the golden age of Greek mathematics. Many modern scientists credit early Greek mathematicians with their inspiration to explore science.
In this article, we will look at one particular contribution from ancient Greeks: geometrical proofs. We will examine what they are, how to do them, and how they can be applied in daily life.
X cannot be 0
In this equation, there are no variables that equal zero. The question asks what the value of x is, so assuming that x is not a variable and is instead a number, x would have to be some sort of number that does not equal zero.
Since the equation asks for what the value of X is, X cannot be one of the variables in the equation. The only thing that can be left is a number.
The question also asks what x could be, so assuming that x is not a variable and is instead a number, it would have to be some sort of number that does not equal zero or any other number in the problem.
This makes sense because if x was 0 or any other number in the problem then the whole problem would be 0 which would make the answer undefined.
X cannot be 2
If the number 2 were subtracted twice, the result would be zero. Two minus two equals zero. Since the answer given was three hundred and thirteen, this number could not be two.
Subtracting two twice would give a negative value. Since the answer given was positive, this could not be the answer. We can rule out two as an option because of this.
X must be a number other than two, then. We can also assume that X is not one thousand or higher, as that would result in a very large difference between 2x – 2x – 2 and 3(213).
If X were one thousand or higher, then 2x – 2x – 2 would be one thousand plus two times one thousand plus two, which would make it much larger than three hundred and thirteen. This does not fit with the answer given.
X must be either negative or positive
The answer to this math problem can only be a negative or positive value, since the question asks for the value of X , not -X .
The equation 2x-2x-2=3(213) is asking you to solve for x , the variable that stands in for the unknown number. In this case, x could be any number, including -1, -2, 1, or 2.
So why does this equation only give you positive and negative answers? Let’s break it down.
First off, we know that x has to be a number between -2 and 2, because all of the other numbers in the equation are positive. We also know that x cannot be 3 because then the entire equation would change.
Since we already know that 2x-2x-2=3(213), then we can eliminate all of the negatives. This leaves us with only positive values for x .
By process of elimination, the value of X is 3
So, if 2x – 2x – 2 = 3(213), then x must be equal to 3, because there is no value of x that makes the equation true when substituting 3 for x .
By checking the remaining values of x , you can see that none of them make the equation true. For example, if x was 5, then 2(5) – 5 – 5 would be 0, which is not equal to 3(213).
So, since no other values of x make the equation true when substituting 3 for x , we can determine that the value of X is 3. This was done through process of elimination!
Note: To do this problem in a different way, you could first solve for X by solving 2x – 2x – 2 = 0. Then, you would check to see if 0 equals X by substituting 0 into the original equation. If not, then X=Nth variable.
Confirm using algebraic methods
After checking your work using the calculator, you can also use algebra to confirm your answer.
To do this, you must solve for the variable x in the equation 2x – 2x – 2 = 3(213). This is done by subtracting twice the number in front of the x from both sides of the equation.
Then, you must solve for x in an equation that uses only integers (no fractions or decimals). You can do this by multiplying both sides of the equation by 3 and then adding 3 to get an integer.
For example, if you had solved for x in the above example, you would have gotten x = 6. This is because you added 3 to both sides of the equation when solving for x.
The solution is x = 3
The solution for this equation is x = 3. This equation is read as “if 2 times 2 minus 2 minus 2 equals 3 times 213, then x equals 3.”
We know that the left side of the equation equals 3 times 213, so we can substitute that into the right side of the equation. Then we can solve for x to get our solution of x = 3.
So, how did they get this solution? First, they factored the left side of the equation to get (2x – 2) * (2x – 2) = 6*213. Then, they solved each of these equations for x and substituted them into the original equation. This gave them the solution of x = 3.
Check your work
Even if you think you have the correct answer, it is important to check your work. Make sure that you have all of the steps correct and that you have answered the question asked.
It is easy to get frustrated and just say you have the right answer, but then later find a mistake. It is better to take some extra time now than to spend even more time correcting mistakes later.
It is especially important to check your work when solving complex algebra problems. A common mistake people make is assuming that two variables are equal when they are actually not.
For example, in a problem where one variable equals half of another variable, if you solve for one of them, they will be equal, but on the test or in real life application, only one will be correct.
Checking your work helps prevent this confusion and errors.
