i really need to finish a lot of word problems by monday and i really need help pleaseeeee

Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $40

 and same-day tickets cost $35

. For one performance, there were 70

 tickets sold in all, and the total amount paid for them was $2600

. How many tickets of each type were sold?



Answer 1

Suppose that there are two types of tickets to a show: advance and same-day.

Lets say number of advance tickets ,x

number of same-day tickets,y

Advance tickets cost $40

total cost of advance tickets $40x

same-day tickets cost $35

total cost of same-day tickets $35y

For one performance, there were 70 tickets sold in all

x+y=70 (equation#1)

and the total amount paid for them was $2600

40x+35y=2600 (equations#2)

How many tickets of each type were sold?



Solve equation#1 and equation#2 simultaneously by substitution or elimination method

Substitution method is easiest for these equations.

So , if x+y=70

y=70-x plug this in equation#2

40x+35y=2600

40x+35(70-x)=2600

Solve for x

40x+2450-35x=2600

40x-35x=2600-2450

5x=150

x=30

plug this value in equation#1 or equation#2 and solve for y(equation#1 is the easiest to solve )

x+y=70

30+y=70

y=70-30

y=40

Therefore,40 same-day tickets were sold and 30 advanced tickets were sold.

Answer 2

Let x= advance tickets

let 70-x = same day tickets

set up an equation

40x + 35(70–x) = 2600

40x + 2450 - 35x = 2600

combine like terms

5x + 2450 = 2600

subtract 2450 from both sides of the equation

5x + 2450 - 2450= 2600-2450

5x = 150

Divide both sides of the equation by 5

x = 30

so, there were 30 advance tickets and 70 - 30 = 40 same day tickets.