How can i prove that there is a pair of natural numbers a and b such that the next relationship is true,

To prove this, I solved for the answer, to see if there was a pair of natural numbers that fit.

ab=22+ \frac{4-a-b}{6}

6ab = 132 + 4 - a - b

6ab = 136 - a - b

We know that an even # times an even # is even, odd times odd is odd, and odd times even is even, but even minus an odd number minus an even number is odd. So if a and b are even, then the result is even, if they're odd, then multiply by 6, they become even, and an even minus an odd minus an odd becomes even again. Basically, a and b must both either be even or both be odd.

Another thing is that ab ≤ 22, because if we try to keep

(4-a-b)/6

positive, the only posible numbers are 0, 1, 2, 3, and 4, and plugging them in doesn't give us the answer, so

(4-a-b)/6

must come out negative, which means ab must be smaller than 22.

Knowing this, I guessed through a list of numbers, and the numbers 7 & 3 go into the equation and come out right.

21 = 22 + (-6/6)

21 = 22 - 1

21 = 21

7 and 3 are natural numbers.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x