Use digits 1-9 to create a true statement. Be sure to verify (prove) that the left side of the equation is the same as the right side of the equation.

Question

Use each digit only once. Each under score is a blank box. _x*2 + _x*3 + _x*2 - _x*3 = _x*2 + _x*3 - _x*2

Answer 1

If I understand this correctly you are filling in the underscores with digits and the * you use is to denote an exponent such as x² = x*2 (as your write it).

So we have __x² + __x³ + __x² - __x³ = __x² + __x³ - __x²

To make this always true we need to make the coefficients of __x² and __x³ equal on both sides of the equation.

We have (__ + __)x² + (__ - __)x^{3 =} (__ - __)x² + __ x³

We now have two equations (replacing __ with letters we get)

a + b = c - d and...

e - f = g where no two a, b, c, d, e, f, g are the same digit

now we can plug in

a = 4, b = 3 , c = 8, d = 1 and...

e = 7, f = 5, g = 2

So we have ax² + ex³ + bx² - fx³ = cx² + gx³ - dx²

where 4x² + 7x³ + 3x² - 5x³ = 8x² + 2x³ - 1x²

^==> So we have 7x² + 2x³ = 7x² + 2x³

which is always equal

Answer 2

ax^2 + bx^3 + cx^2 - dx^3 = ex^2 + fx^3 - gx^2

group like terms

(a+c)x^2 + (b-d)x^3 = (e-g)x^2 + fx^3

a+c = e-g

b-d = f

one solutiion is a=1, c=2, e=7, g=4, b=9 d=3 f= 6

another solutiion is a=2, c =3, e=9, g=4, b=6, d=1, f= 5

there may be many other solutions

Answer 3

Given: __x² + __x³ + __x² - __x³ = __x² + __x³ - __x²

ax² + bx³ + cx² - dx³ = ex² + fx³ - gx²

(b-d)x³ + (a+c)x² = fx³ + (e-g)x²

For this equation to be true:

b-d = f

a+c = e-g

Constraint: each number 1 to 9 is used at most once as a coefficient.

We may find all solutions using exhaustive enumeration:

numbers = (1,2,3,4,5,6,7,8,9)

for a in numbers:

for b in numbers:

for c in numbers:

for d in numbers:

for e in numbers:

for f in numbers:

for g in numbers:

if ((b-d)==f) & ( (a+c)==(e-g) ):

answers = [a,b,c,d,e,f,g]

res = [i for n, i in enumerate(answers) if i not in answers[:n]]

if len(res) == 7:

print (a,", ",b,", ",c,", ",d,", ",e,", ",f,", ",g)

There are 168 solutions !!

Here are the first few: