math is good bro

Question

Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay

$86.  

Two adults and three children must pay

$61.  

Find the price of the​ adult's ticket and the price of a​ child's ticket.

Answer 1

3A + 4C = 86

2A + 3C = 61 where A is the price of the adult ticket and C is the child's price.

Suggest elimination method:

Multiply 1st eqn by -2 and 2nd eqn by 3

-6A - 8C = -172

6A + 9C = 183

C = $11

Then substitute: 3A + 4(11) = 86

3A = 42

A = $14

Answer 2

First, let A refer to the price of an adult ticket, and C refer to the price of a child's ticket.

Our first sentence gives us that 3 adult tickets and 4 children tickets costs $86. This gives us

the following equation:

3A + 4C = 86.

Our first sentence gives us that 2 adult tickets and 3 children tickets costs $61. This gives us

the following equation:

2A + 3C = 61.

Therefore, we have the following system of equations,

3A + 4C = 86

2A + 3C = 61.

For this case, elimination will be a fine approach.

Multiplying the second equation by -1 gives

3A + 4C = 86

-2A + -3C = -61.

Then, adding the two equations together, we get

3A + (-2A) + 4C + (-3C) = 86 + (-61).

A + C = 25.

Then, we can isolate for A,

A = 25 - C.

Finally, we can substitute this expression for A in one of our equations.

I'll plug in A in the first equation:

3A + 4C = 86

3(25 - C) + 4C = 86

75 - 3C + 4C = 86

75 + C = 86

C = 11.

Finally, we know A = 25 - C, and C = 11. Thus,

A = 25 - C

A = 25 - 11

A = 14.

Thus, our prices are: $14 for adults, $11 for children.

One more thing: To double check that your answer is actually correct,

plug in A = 14 and C = 11 into both equations:

3A + 4C = 3 * 14 + 4 * 11 = 42 + 44 = 86

2A + 3C = 2 * 14 + 3 * 11 = 28 + 33 = 61.

This last part may not be strictly necessary, but is an excellent exercise to prove to yourself that your answer makes sense!

Answer 3

Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $86.

Two adults and three children must pay $61.

Find the price of the​ adult's ticket and the price of a​ child's ticket.

Let P_A = price of an adult ticket

N_A = number of adults

P_C = price of a child ticket

N_C = number of children

and C_T = total cost

The total cost is just a "weighted" sum of the adult cost plus the child cost:

N_A * P_A + N_C * P_{C =} C_T

For the 1st case, we substitute the numbers given

3 * P_A + 4 *P_C = 86 (1)

Similarly, for the second case

2 * P_A + 3 * P_C = 61 (2)

Here we simultaneously have 2 equations in 2 unknowns, P_A and P_C (a classic situation). It is solvable by substitution.

First we rearrange equation #2 to solve for P_A:

2*P_A = 61 - 3*P_C

P_A = (61 - 3*P_C) / 2

Now "substitute" this expression into equation #1

3 * (61 - 3*P_C) / 2 + 4*P_C = 86

3*61/2 - 3*3*P_C / 2 + 4*P_C = 86

3*61/2 + (-9/2 + 4)*P_C = 86

-0.5*P_C = 86 - 3*61/2 = 86 - 91.5 = -5.5

P_C = -5.5 / -0.5 = 11

Now we can put the value of PC again into equation #1 and solve for P_A

3 * P_A + 4 * P_C = 3 * P_A + 4 * 11 = 86

3 * P_A = 86 - 44 = 42

P_A = 42 / 3 = 14

To check if you are right, substitute the values for P_A & P_C into equation #2

2 * P_A + 3 * P_C = 61

2 * 14 + 3 * 11 = 61

28 + 33 = 61

61 = 61

First we solved the equ #2 for P_A and substituted into equ #1 to get P_C. Try doing it in reverse.

Try solving equ #1 for P_C and substituted into equ #1 to get P_A. It won't matter, You get the same answers.

That's the beauty of solving "simultaneous equations." ........... :-)