Solve these for me. i need them for a quiz.

4a)

I'm going to use the Substitution method

c = 160 + 67t 
55 + 82t = 160 + 67t replace c with 55+82t; solve for t
55 + 82t - 55 = 160 + 67t - 55
82t = 67t + 105
82t - 67t = 67t + 105 - 67t
15t = 105
15t/15 = 105/15
t = 7

Use t = 7 to find that
c = 160 + 67t
c = 160 + 67*7
c = 629
You can use the other equation as well to get the same answer for c.

It takes 7 months for the cost to be the same. That cost is $629


4b)

I'm going to use the Elimination method

Subtract the equations to get
x+y = 20
x-4y = -15

0x+5y = 35

If 5y = 35, then y = 7 after we divide both sides by 5

Plug y = 7 into the first equation and solve for x
x+y = 20
x+7 = 20
x+7-7 = 20-7
x = 13

So x = 13 and y = 7 are the solutions

Max exercises 13 hours a week while Sasha exercises 7 hours a week

4c)

I'm going to use the Substitution method

x = price for one adult ticket (in dollars)
y = price for one student ticket (in dollars)

Fact #1
4 adult tickets + 2 student tickets = 64 dollars
4x + 2y = 64

Fact #2
3 adult tickets + 3 student tickets = 60 dollars
3x + 3y = 60
3(x + y) = 60
3(x + y)/3 = 60/3
x + y = 20
x + y-x = 20-x ... subtract x from both sides
y = 20 - x

Plug that into the first equation
4x + 2y = 64
4x + 2(y) = 64
4x + 2(20-x) = 64 ... replace y with 20-x; solve for x
4x + 40-2x = 64
2x + 40 = 64
2x + 40-40 = 64-40
2x = 24
2x/2 = 24/2
x = 12

Use x = 12 to find y
y = 20 - x
y = 20 - 12
y = 8

In summary, x = 12 and y = 8, which means...
One adult ticket costs $12
One student ticket cost $8

2½ boxes

Step-by-step explanation:

From the question given above,

We understood that the machine fills 1/2 box in 1/5 min.

Thus, we can simply obtain the number boxes the machine will fill by doing the following:

The machine fills 1/2 box in 1/5 min.

Therefore, the machine will fill y box in 1 min i.e

y box = (1/2 × 1) ÷ 1/5

y box = 1/2 ÷ 1/5

y box = 1/2 × 5/1

y box = 5/2 = 2½

Therefore, the machine fills 2½ boxes in a minute.