Respuesta :
Part A: Let x be the cost of each student and y be the cost of each teacher.
Part B: 25x + 2y = 97.5 (1)
32x + 3y = 127 (2)
Part C: Solve simultaneously.
3(1) - 2(2): 75x - 64x = 292.5 - 254
11x = 38.5
x = 38.5/11 = $3.50
Since x = $3.50, Â 32(3.5) + 3y = 127
112 + 3y = 127
3y = 15
y = $5
Part D: It makes sense, but let's check with the first equation to make sure it satisfies the first equation.
LHS = 25(3.5) + 2(5) = 87.5 + 10 = $97.50, which is true.
Hence, each student costs $3.50 and each teacher costs $5
Part B: 25x + 2y = 97.5 (1)
32x + 3y = 127 (2)
Part C: Solve simultaneously.
3(1) - 2(2): 75x - 64x = 292.5 - 254
11x = 38.5
x = 38.5/11 = $3.50
Since x = $3.50, Â 32(3.5) + 3y = 127
112 + 3y = 127
3y = 15
y = $5
Part D: It makes sense, but let's check with the first equation to make sure it satisfies the first equation.
LHS = 25(3.5) + 2(5) = 87.5 + 10 = $97.50, which is true.
Hence, each student costs $3.50 and each teacher costs $5
A: Let Each Student's share be s
   Let Each Teacher's share be t
B: 1st group : 25s + 2t = 97.50 --eqn 1
   2nd group : 32s + 3t = 127.0 --eqn 2
C :Â (25s + 2t = 97.5)*3 = 75s + 6t = 292.5 --eqn3
    (32s + 3t = 127.0)*2 = 64s + 6t = 254.0 --eqn4
    Subtracting eqn4 from eqn3-->
    11s = 38.5
    so , s = 3.5
    25*3.5 + 2t = 97.5 (from eqn1)
    so, t = 5
D: Each student has 3.5$ and Each teacher has 5$ in their share.
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   Let Each Teacher's share be t
B: 1st group : 25s + 2t = 97.50 --eqn 1
   2nd group : 32s + 3t = 127.0 --eqn 2
C :Â (25s + 2t = 97.5)*3 = 75s + 6t = 292.5 --eqn3
    (32s + 3t = 127.0)*2 = 64s + 6t = 254.0 --eqn4
    Subtracting eqn4 from eqn3-->
    11s = 38.5
    so , s = 3.5
    25*3.5 + 2t = 97.5 (from eqn1)
    so, t = 5
D: Each student has 3.5$ and Each teacher has 5$ in their share.
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