How many possible solutions does these system of equations have ?
Algebra
Level
3
If X+ Y <= 49 and X >= 0; Y >= 0 X,Y are whole numbers. Above is a system of equations, how many values of (X,Y) satisfy all of the above equations ?.
The answer is 1275.
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1 solution
Since X and Y are >= 0 , the minimum value of (X+Y) is 0 and as (X+Y) <= 49 the maximum value of (X+Y) is equal to 49.
So when X+Y = 0 there is only one point (X,Y) that satisfies all the equations namely (X,Y) =(0,0).
Similarly when (X+Y) = 1 there are two points which satisfy these equations namely (1,0) and (0,1)
When (X+Y) = 2 there are three solutions which satisfy these equations namely (0,2), (1,1) and (2,0)
So on for each value of (X+Y) = n where n ranges between 0 and 49 has 1,2,3,4,5...50 possible feasible points that satisfies these equations.
So the total number of points which satisfy the inequality (X+Y) <= 49 given that X and Y are not fractions and are greater than or equal to 0 are
1 + 2+ 3 + 4....50 = 51*50/2 = 1275.
Feasible