Directions: Use the digits 1 to 9, at most one time each, to create an equation where x has the smallest (or greatest) possible value.

### Hint

### Hint

Where do you want to put the smaller (or larger) numbers?

How does choosing the coefficient or constant affect the variable’s value?

How does choosing the coefficient or constant affect the variable’s value?

### Answer

### Answer

The smallest solution I’ve found so far is x = -70

(9/1)(3x + 8) + 4x = 6x + 2

(9/1)(3x + 8) + 4x = 6x + 2

The largest solution I’ve found so far is x = 8:

(4/8)(6x + 2) + 1x = 3x + 9

Source: Daniel Luevanos

Here is an equation where x=20.

(9/3)(1x+8)+2x=6x+4

Here is an equation where x=61

(1/7)(8x+2)+5x=6x+9

We found another way to get x = 61.

(2/7)(4x + 1) + 5x = 6x + 9

x=4

(2/4)(3/1)=1x-3

I think the solution above where the solution is x = -70 is actually a case where 25x = -70, which would make x = -2.8. (9/1)(3x + 8) + 4x = 6x + 2 gives us 27x + 72 + 4x = 6x + 2, which then simplifies to 31x + 72 = 6x + 2, which further simplifies to 25x = -70. Did I miss something in my calculations?

We tried this problem today and found that x did not = -70 either. I think it was a calculation error.

One of my students found several ways to make the equation true for any number, so that you could use 100 or 1,456,918 and it would still be true:

9/3(1x + 2) + 4x = 7X + 6

6/3(2x + 4) + 1x = 5x + 8

6/2(1x + 3) + 4x = 7x + 9

Another student tried to find the lowest possible and got x = -44.

7/8(1x + 4) + 5x = 3x + 9

If you define smallest to be closest to zero, then 8/2(9x + 1) + 6x = 3x + 4 is zero.