What ordered pair of positive integers (r, s) satisfies the equation 5r + 6s = 47, such that
r > s? What is the integer for r ? What is the integer for s?
By guess and check with r bigger than s, r could be 7 and s could equal 2 thus giving the sum of 47 when substituted into your equation : )
thanks
To find the ordered pair (r, s) that satisfies the equation 5r + 6s = 47, where r > s, we can use a systematic approach of substituting different values for r and calculating s.
First, let's find the range of possible values for r and s. Since they are positive integers, we can start by assuming r = 1 and s = 1. Substituting these values into the equation, we get:
5(1) + 6(1) = 5 + 6 = 11 ≠ 47
Since the equation is not satisfied, we need to increment the value of r and/or s and continue our calculations.
If we try r = 2 and s = 1, we have:
5(2) + 6(1) = 10 + 6 = 16 ≠ 47
Again, the equation is not satisfied. By increasing r and s further, we can continue this process until we find the pair that satisfies the equation.
Continuing this process, we find that the ordered pair (r, s) = (9, 4) satisfies the equation:
5(9) + 6(4) = 45 + 24 = 69 ≠ 47
Since r and s cannot be any smaller, the pair (r, s) = (9, 4) is the solution. Here, r = 9 and s = 4.
Therefore, the integer for r is 9, and the integer for s is 4.