How do you know if an equation has integer solutions?

If a divides b, then the equation ax=b has exactly one solution that is an integer. If a does not divide b, then the equation ax=b has no solution that is an integer.

What is integer solution for inequality?

When solving inequalities there will be a range of answers because any numbers represented by the range are acceptable, and there are an infinite amount of solutions to inequalities. For example, if , then any number that is bigger than 3 is a possible answer, from any decimal slightly bigger than 3 to infinity.

How do you find the solution and not a solution?

How to determine whether a number is a solution to an equation

1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

What are the integers between 3 and 4?

There are no integers, i.e. whole numbers, between 3 and 4, or between 4 and 3.

How many integers are there?

Zero. Decimal numbers and fractions are not included in integers. There are infinite integers.

How many solutions does x + y + z = 15?

Hence, the number of solutions of the equation x + y + z = 15 in the nonnegative integers subject to the restrictions that 1 ≤ x ≤ 5 is (2) We wish to solve the equation x + y + z = 15 in the nonnegative integers subject to the restrictions that x ≥ 2 and y ≤ 3.

Is (x + y) + x – y = 2 x?

That would mean ( x + y) + ( x − y) = 2 x is also odd, which contradicts the existence of integer solutions. Answering this requires no advanced math knowledge at all. The difference between every square is an odd number, and since 2002 is even, then y must be an even number less than x.

What is the number of solutions of equation 1 in nonnegative integers?

The number of solutions of equation 1 in the nonnegative integers is the number of ways to select which two of the 16 symbols (the 14 ones and the two addition signs) will be addition signs, which is However, the restriction that 1 ≤ x ≤ 6 ⇒ 0 ≤ w ≤ 5. Thus, we must remove those solutions in which w ≥ 6. Assume w ≥ 6.

Is there any other solution for x<>1?

For each x>0, there is at least one solution y = x. and it’s the only solution. To see if there’s any other solution for x<>1, we can rewrite the equation, taking logs, as This function clearly has two “branches”, one for 0<1 and one for x>1. For each x>0, there is at least one solution y = x. and it’s the only solution.