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3 years ago

A small art museum charges $5 for an adult ticket and $3 for a student

Mathematics

1 answer:

Shkiper50 [21]3 years ago

5 0

There were 37 student tickets and 52 adult tickets

Step-by-step explanation:

Let x be the number of student tickets and y be the adult tickets

Then

According to given statement that total 89 tickets were sold

x+y=89 Eqn 1

And

$5 for an adult ticket and $3 for student ticket ,and total cost $371

3x+5y=371 Eqn 2

From eqn 1

$x=89-y$

Putting in eqn 2

$3(89-y)+5y=371\\267-3y+5y=371\\267+2y=371$

Subtracting 267 from both sides

$267+2y-267=371-267\\2y=104$

Dividing both sides by 2

$\frac{2y}{2}=\frac{104}{2}\\y=52$

Putting y=52 in eqn 1

$x+52=89\\x=89-52\\x=37$

Hence,

There were 37 student tickets and 52 adult tickets

Keywords: Linear Equation, Variable

Learn more about linear equation at:

#LearnwithBrainly

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3 years ago

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joe has a piggy bank with $8.90 split upon nickels, dimes, and quarters. The piggy bank contains 76 coins in all. The number of

kozerog [31]

Answer:

There are 38 dimes, 22 nickels and 16 quarters.

Step-by-step explanation:

Let n, d and q represent the # of nickels, dimes and quarters respectively.

Then n + d + q = 76

The value of a nickel is $0.05; that of a dime is $0.10, and that of a quarter is $0.25.

Thus, the value of n nickels is $0.05n (and so on).

The total value of the coins is $0.05n + $0.10d + $0.25q = $8.90.

d = n + q allows us to eliminate d.

First, n + d + q = 76 becomes n + (n + q) + q = 76, and second:

$0.05n + $0.10(n + q) + $0.25q = $8.90. Here we have succeeded in eliminating d from two different equations, and now we have these two different equations in two unknowns (n and q), which is solvable.

Simplifying both equations, we get:

2n + 2q = 76 and

5n + 10n + 10q + 25q = 890, or 15n + 35q = 890

Let's use the substitution method of solving linear equations:

Rewrite 2n + 2q = 76 as n + q = 38, or n = 38 - q. Substituting this result into the second equation, we get:

15(38 - q) + 35 q = 890, or

570 - 15q + 35q = 890, or

570 + 20 q = 890. Then 20q = 890 - 570 = 320, and q = 320/20 = 16.

There are 16 quarters. Thus, the number of nickels is n = 38 - 16 = 22.

Finally, since n + d + q = 76, 22 + d + 16 = 76, or:

22 + d = 60, or d = 60 - 22 = 38.

There are 38 dimes, 22 nickels and 16 quarters.

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