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Musya8 [376]
2 years ago
6

Billy Jo's school is selling tickets to a Fall Festival. On the first day of ticket sales the school sold 3 adult ticket and 8 s

tudent tickets for a total of $72. The school took in $152 on the second day by selling 7 adult tickets and 16 student tickets. How much is a student ticket?
Mathematics
1 answer:
denis23 [38]2 years ago
6 0

Answer:

Y = 6

Step-by-step explanation:

On the first day of ticket sales the school <em><u>sold 3 </u></em>adult ticket and <em><u>8 student </u></em>tickets for a <em><u>total of $72</u></em>. The school took in <em><u>$152</u></em> on the second day by selling<em><u> 7 adult tickets and 16 student tickets.</u></em> How much is a student ticket?  

Day 1: 3x + 7y = 72

Day 2: 7x + 16y = 152

This becomes a system of equations.

3x+8y=72

3x+8y+−8y=72+−8y (Add -8y to both sides)

3x=−8y+72

3x/3 =  −(8y+72)/3

X = (-8/3) y + 24

Substitute  (-8/3) y + 24 for x in7x+16y=152:

7(-8/3 y + 24) + 16y = 152

-8/3 y + 168 = 152 (Simplify both sides of the equation)

-8/3 y + 168 − 168 = 152 + (−168) (Add -168 to both sides)

-8/3 y = -16

(-8/3 y)/(-8/3 y) = -16/ -8/3 y (Divide both sides by (-8)/3)

<u><em>y = 6</em></u>

<em>If you wanted to keep going...</em>

Substitute 6f or y in x = (-8/3) y + 24

(-8/3) y + 24  =  (-8/3) (6) + 24

SImplify: x = 8

<em>X = 8 and </em><u><em>Y = 6</em></u>

<u><em></em></u>

Hope this helps! <em>♥</em>

<u><em>~A.W.E.</em></u><em>S.W.A.N~</em>

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Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac
melomori [17]
Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ b+c=20
Equation [2] ⇒ 1.50b+5c=54.50

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; b and c, that we need to solve, then this becomes a simultaneous equation problem. 

First, rearrange Equation [1] to make either b or c the subject:
b+c=20
b=20-c

Then we substitute b=20-c into Equation [2]
1.50b+5c=54.50
1.50(20-c)+5c=54.50
30-1.50c+5c=54.50
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c=7

Now we know the value of c which is c=7, substitute this value into b=20-c we have b=20-7=13

Answer:
Numbers of books = 13
Numbers of CDs = 7
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