Using T, C, G for Tim's, Cheryl's, and Greg's ages, respectively:

T + C = 80   (1)

T + G = 98   (2)

C + G = 94   (3)





Add (1) and (2):

T + C + T + G = 80 + 98

2T + C + G = 178



From (3), C+G = 94, substitute this in and solve for T:

2T + 94 = 178

2T = 84



T = 42  --> C = 38 (by (1))  and G = 56 (by (2))



------



Check: 

 T + C = 42 + 38 = 80  (ok)

 T + G = 42 + 56 = 98  (ok)

 C + G = 38 + 56 = 94  (ok)





 

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
 

Solution to this problem:

Tim's age plus Cheryl's age is 80.

Tim's age plus Greg's age is 98.

Cheryl's age plus Greg's age is 94. What are their ages?

Let Tim's, Cheryl's, and Greg's ages be T, C, and G, respectively

Then we get: T + C     = 80 ------ eq (i)

       Also, T     + G = 98 ------ eq (ii)

       And,      C + G = 94 ------ eq (iii)

                 C - G = - 18 ---- Subtracting eq (ii) from eq (i) ---- eq (iv)

                    2C = 76 ------ Adding eqs (iv) & (iii)

C, or 

Find Cheryl's and Greg's ages on your own! 

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!
 


You are given the sums of the three ages two at a time:

T+C = 80

T+G = 98

C+G = 94


You already have two good algebraic solutions; here is another method that might be a bit less work.


(1) Add all three equations: 2(T+C+G) = 272

(2) Divide by 2: T+C+G = 136


Now the age of each can be found by comparing this last equation to the original three equations:

G = (T+C+G)-(T+C) = 136-80 = 56

C = (T+C+G)-(T+G) = 136-98 = 38

T = (T+C+G)-(C+G) = 136-94 = 42 

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