How do you find the number of values of n that satisfy a given equation?

To find the number of values of “n” that satisfy a given equation, follow these steps:

  1. Understand the Equation: First, make sure you understand the equation clearly. Let’s take an example equation to work with:
    Example Equation: 2n – 5 = 11.
  2. Isolate the Variable: Your goal is to isolate the variable “n” on one side of the equation. In our example, add 5 to both sides of the equation to move the constant to the other side:
    2n – 5 + 5 = 11 + 5
    2n = 16.
  3. Solve for n: Now, to find the value of “n,” divide both sides of the equation by the coefficient of “n” (which is 2 in this case):
    2n / 2 = 16 / 2
    n = 8.
  4. Check Your Solution: Always check your solution by substituting it back into the original equation to ensure it satisfies the equation. In our example:
    2n – 5 = 11
    2(8) – 5 = 16 – 5 = 11.
    Since both sides of the equation are equal, n = 8 is a valid solution.

Now, let’s say you have an equation that is more complex, such as a quadratic equation:

Example Equation: n^2 – 3n + 2 = 0.

  1. Understand the Equation: This is a quadratic equation.
  2. Factor or Use the Quadratic Formula: To find the values of “n” that satisfy this equation, you can either factor it or use the quadratic formula. Factoring may be the easiest if the equation is factorable:

    n^2 – 3n + 2 = 0 factors to (n – 1)(n – 2) = 0.

  3. Solve for n: Set each factor equal to zero:

    n – 1 = 0, and n – 2 = 0.

    Solve for “n” in each equation:

    n = 1, and n = 2.

  4. Check Your Solutions: Check if these values satisfy the original equation:

    For n = 1: (1)^2 – 3(1) + 2 = 1 – 3 + 2 = 0.
    For n = 2: (2)^2 – 3(2) + 2 = 4 – 6 + 2 = 0.

Both n = 1 and n = 2 satisfy the original equation, so there are two values of “n” that satisfy the equation.

In summary, to find the number of values of “n” that satisfy a given equation, you need to isolate the variable, solve for “n,” and check your solutions to ensure they are valid. The number of solutions may vary depending on the complexity of the equation.

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