Respuesta :
The cost of one adult ticket is $7.5 and one child ticket is $3.5
Julia will pay $48 for 5 adult tickets and 3 child tickets.
Step-by-step explanation:
Let,
Cost of one adult ticket = x
Cost of one child ticket = y
According to given statement;
3x+5y=40 Eqn 1
x+5y=25 Eqn 2
Subtracting Eqn 2 from Eqn 1
[tex](3x+5y)-(x+5y)=40-25\\3x+5y-x-5y=15\\2x=15[/tex]
Dividing both sides by 2
[tex]\frac{2x}{2}=\frac{15}{2}\\x=7.5[/tex]
Putting x=2.5 in Eqn 1
[tex]3(7.5)+5y=40\\22.5+5y=40\\5y=40-22.5\\5y=17.5[/tex]
Dividing both sides by 5
[tex]\frac{5y}{5}=\frac{17.5}{5}\\y=3.5[/tex]
The cost of one adult ticket is $7.5 and one child ticket is $3.5
5 adult tickets = 5*7.5 = $37.5
3 child tickets = 3*3.5 = $10.5
Total = 37.5+10.5 =$48
Julia will pay $48 for 5 adult tickets and 3 child tickets.
Keywords: linear equation, subtraction
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a. The cost of an adult ticket is equal to $7.5.
b. The cost of an child ticket is equal to $3.5.
c. Ju-lia will pay $48 for 5 adult tickets and 3 child tickets.
- Let the cost of a child ticket be b.
- Let the cost of an adult ticket be a.
Given the following data:
- Rico's expenses = $40.
- Sasha's expenses = $25.
Translating the word problem into an algebraic expression, we have;
For Rico:
[tex]3a + 5c =40[/tex] ....equation 1.
For Sasha:
[tex]a + 5c=25[/tex] ....equation 2.
Making a the subject of formula in eqn. 2, we have:
[tex]a=25-5c[/tex] ....equation 3.
Substituting eqn. 3 into eqn. 1, we have:
[tex]3(25-5c)+5c=40\\\\75-15c+5c=40\\\\35=10c\\\\c=\frac{35}{10}[/tex]
c = $3.5
For the value of a:
[tex]a=25-5c\\\\a=25-5(3.5)\\\\a=25-17.5[/tex]
a = $7.5
c. Cost of 5 adult tickets and 3 child tickets:
[tex]5a + 3c\\\\5(7.5)+ 3(3.5) = 37.5 + 10.5\\\\5a + 3c = 48[/tex]
Therefore, Ju-lia will pay $48 for 5 adult tickets and 3 child tickets.
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