MATH SOLVE

3 months ago

Q:
# Find all pairs of natural numbers which can be the solution to the equation: x+y=11

Accepted Solution

A:

Answer:How many solution does ++=11x+y+z=11 have where ,,x,y,z are non-negative integers. In light of the restrictions, its clear that ,,β{0,1,2,..11}x,y,zβ{0,1,2,..11}. So, at face value I would assign a value for xand determine the different combinations that y and z can hold. For example,For =0x=0, we have +=11y+z=11. With writing them out I found that there are 1212 different assigned combinations for y and z that satisfy the equation. For =1x=1, I got 1111. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is 1+2+3+4+5+6+7..+12=781+2+3+4+5+6+7..+12=78. I was wondering if there is an easier method perhaps with combinations equation (,)C(a,b)..?Step-by-step explanation: