2x + 10 /3 = 20/6..
solve this.
Answers
Answer:
2x + 10 /3 = 20/6..
2x=20/6 - 10/3
=20/6 -20/6
=0
2x=0
x=0
Answer:
If 2=6, 3=12, 4=20, 5=30 and 6=42, what does 9 equal to, 56, 81, 72 or 90?
My original answer was collapsed, and the answer police said that is because my answer needs editing. There were several things mentioned that said I should explain why my answer is correct. I thought I did so, but let me explain in more detail. A comment on my answer informed me that this problem comes from an area (a "growing" area - horror of horrors!) known as "logical reasoning". Here I will show, using actual logical reasoning why my answer is correct and the answer police are all wet.
Here's your question:
"If 2=6, 3=12, 4=20, 5=30 and 6=42, what does 9 equal to, 56, 81, 72 or 90?"
Detail about why my answer is correct:
1. Your first equation implies that 4=0.
Proof:
If 2=6, then 2-2=6-2, because of the compatibility of equality in arithmetic with the subtraction operator. Since 2-2=0, it follows from your first equation that 0=6-2, because of transitivity of equality and the substitution rule for equality. Since 6-2=4, it then follows that 0=4, by the substitution rule for equality.
2. Your second equation implies that 9=0.
Proof:
If we assume that 3=12. Then 12=3, by the symmetry property of equality. Thus, because equality is compatible with subtraction, 12-3=3-3=0. By transitivity of equality and since 12-3=9, and by the substitution rule, it follows from the given assumption that 9=0.
3. Your first two equations imply that 4=9.
Proof:
Assume that 2=6 and 3=12. By 1. above, we get 4=0, and by 2. above, we get that 9=0, so by symmetry and transitivity of equality, we have 4=9.
4. From your equations, anything can be proven, because you are beginning with false assumptions.
Here's my original answer that was collapsed:
The question makes no sense. Let me explain:
1. In the integers, 2 is not 6, so whatever in the world you meant, you do not mean for a rational human being to solve a problem involving integer arithmetic.
2. In modular arithmetic, if 2=6 (modn)2=6 (modn) , then 4=0 (modn)4=0 (modn) , so that nn divides 44 in integer arithmetic. That is, n∈{1,2,4}n∈{1,2,4} . If also 3=6(mod n)3=6(mod n) , then nn must also divide 33 , i.e. n∈{1,3}n∈{1,3} . But then nn is in the intersection of the sets {1,2,4}{1,2,4} and {1,3}{1,3} , whence n=1n=1 . But in modular arithmetic, say arithmetic modulo nn , if n=1n=1 , then for every integer xx and every integer yy , we have x=yx=y . But this would be rather uninteresting arithmetic, so since you are an interesting person, no doubt, you would not ask such an uninteresting question.
3. In the real numbers, in modulo 1 arithmetic, the interesting problems are diophantine problems, and your question does not ask about a non-integer real number, so we would be led back to 2, making the problem given just a silly one.
4. In mathematics, you may treat numerals as merely symbols in a language, but to do so without saying what you mean indicates a disingenuous disposition. Why would you do this?
5. Other answers given in this thread are treating your problem like a kind of substitution cypher, but if that's what you mean, then you have used equal signs completely inappropriately. Please stop doing such things. If you meant "If 2 maps to 6, 3 maps to 12, 4 maps to 20, 5 maps to 30 and 6 maps to 42, what does 9 map to, 56, 81, 72 or 90?" Then why not ask your question in that way? That is, why not say what you really meant? In this case, there's no need for me to answer your question because then the answer is that 9 can map to anything. You have not given enough conditions on the kind of mapping you intend so that the problem has a unique solution, but you have asked the question with a rhetorical slant that suggests that there is exactly one solution. This is disingenuous, to say the least.