Only integer solutions are accepted in an diophantine equation
If x = y, then the equation is true for any integer pair. Otherwise, the equation is not equivalent and is impossible.
A division equation in which the numerator is an integer multiple of the denominator.
insert the answer in the equation, replacing the variable, and see if it still makes sense.
The person or program that solves the equation does.
Y = 3.
Yes, it is an integer sequence.
You're allowed to do that, yes. Whether or not it "solves" any particular equation depends on the equation itself.
There is no such thing as "solving integers". You can solve an equation, which means finding all the unknowns in that equation, but you can't solve an integer.
The integer number of dollars that you owe your friend is 19. A strange question!
The exclamation point in a math equation symbolizes the factorial function. The factorial of an integer > 0 is the product of that integer and all of the integers between 1 and that integer. For instance 7! is 7 * 6 * 5 * 4 * 3 * 2 * 1, or 5040. The special case of 0! is defined as 1.
ok this is what it looks like to me...I definitely could be wrong. one integer is three more than twice another integer if one integer is three more than another number (x+3=y) Then you just multiply "the other number" by 2. (x+3=2*y) Sum of two integers is 36....so add them (x+y=36) Solve that equation for one number so...(y=36-x) Then you would plug that equation into the first equation where there is a y: (x + 3 = 2*(36-x))
If we write the problem as 6x-5=x2, then we can write in the form of ax2+bx+c: -x2+6x-5. Then we can use the quadratic equation, x = (-b ± √(b2 - 4ac))/2a, and put in our own values to get the equation x = 3 ± 2. Therefore, x1= 1 and x2=5.
no it depends on if the first number in the equation is bigger or smaller than the second ex. 2-3= -1
17 is not an equation and so there can be no "solution of 17". There is, therefore, no possible answer to the question.
y=-(2/3)x + (n) where n can be any integer.
None. -42 is a single integer, not an equation nor an inequality. So there are no solutions.
In normalized scientific notation all numbers are written in the form a x 10^b (a times ten raised to the power of b) where a is a nonzero single-digit integer and b is an integer.
Fermat's last theorem states that the equation xn + yn = zn has no integer solutions for x, y and z when the integer n is greater than 2. When n=2, we obtain the Pythagoras theorem.
There are 120 solutions.
have changed your equation from 19x=29 mod16 to 19x=16 mod29 as it is impossible to have 29 mod 16 in modular arithmetic/it makes no sense. Anyway, any modular equation a=b mod m can be rewritten as a-b=km where k is an integer. or in other words, m divides a-b without remainder. Applying this to the equation above we get (19x-16)/29=k. Now just stick it into a calculator and replace x with integers until the equation gives out an integer, in this case x turns out to be 10.
Let the unknown integer by n. 2n = 6 + 2n2 : 2n2 - 2n + 6 = 0 For a quadratic equation an2 + bn + c = 0 then the equation has no real roots if b2 - 4ac < 0 : In the example b2 - 4ac = 4 -48 = -44 as this < 0 then the original expression cannot have a real integer solution.