Solving One-Step Equations (Negative and Positive Solutions)

Directions: Using the digits 1 to 9, at most one time each, place a digit in each box to create two equations: one where x has a positive value and one where x has a negative value.

Hint

How does the value of the number being added to x affect the value of x?
How does the value of the sum affect the value of x?

Answer

There are many answers. To get a positive solution, the number being added to x should be less than the sum. To get a negative solution, the number being added to x should be greater than the sum.

Source: Robert Kaplinsky

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7 comments

  1. Could you extend this by asking: ” Using the same digits, create an equation where x has a negative value. What do you notice?”

  2. I love the critical thinking required in these questions, but this particular question isn’t a sixth grade Common Core standard. The standard, 6.EE.7, asks students to solve multiplication and addition equation with numbers that are all “nonnegative rational numbers”. Is there a way to align this question to the standard better while still preserving the DOK level? Students do not learn operations with negative numbers until the seventh grade.

  3. X is positive:
    1 x 3 + x = 2 x 6
    3 + x = 12
    x = 9
    X is negative:
    3 x 6 + x = 2 x 5
    18 + x = 10
    x = -8

    • Interesting that you interpreted the two boxes next to each other as multiplication. I was thinking they are two digit numbers.

      Some positive solutions are:
      13 + x = 25
      24 + x = 35

      Some negative solutions are:
      25 + x = 13
      35 + x = 24

  4. Is it possible to do this and have x equal the same number in both equations? That’s originally what I thought I needed to do! 🙂

  5. Rudolf Österreicher

    There are 756 unique solutions (times 2 because you could always swap the units digits of the first number and the sum) where x is positive.
    There are equally as many where x is negative.
    So in total, there are 756*756 = 571,536 possible pairs of equations where in one x is positive and in the other x is negative.

    If you cannot repeat the digits for the two equations, then there are not 756*756 solutions, but “only” 22,680.

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