Problema Solution
There are three numbers whose sum is 45. The ratio of the sum of the first two numbers to the third number is 8/7. THe difference between the first number and the second number is 8. Find the three numbers.
Answer provided by our tutors
Set up the equation.
Equation 1: A + B + C = 45
Equation 2: (A + B) / C = 8/7
Equation 3: A = B + 8
Plug (B + 8) into equation 1 and solve for C
Equation 1: A + B + C = 45
(B + 8) + B + C = 45 (Combine like terms)
2B + 8 + C = 45 (Subtract 2B + 8 from both sides)
C = 45 - 2B - 8 (Combine like terms
C = 37 - 2B
Now plug (B + 8) into equation 2 for A and use (37 - 2B) for C
Equation 2: (A + B) / C = 8/7
((B + 8) + B)/(37 - 2B) = 8/7
Multiply both sides by 7
7*((B + 8) + B)/(37 - 2B) = 8
Multiply both sides by (37-2B)
7*((B + 8) + B) = 8*(37 - 2B)
Simplify
7B + 56 + 7B = 296 - 16B
Combine like terms
14B + 56 = 296 - 16B
Add 16B to both sides
30B + 56 = 296
Subtract 56 from both sides
30B = 240
Divide both sides by 30
B = 8
Now plug 8 into equation 3 for B
Equation 3: A = B + 8
A = 8 + 8
A = 16
Now plug the values into equation 1 and solve for C
Equation 1: A + B + C = 45
16 + 8 + C = 45
C = 21
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