How would I put this into an equation?

Question

Adult tickets for a play cost $4 and child tickets cost $1. If there were 23 people at the performance and the theater collected $80.00 form ticket sales how many children attended the play?

Answer 1

let

c = number of children tickets

a = number of adults tickets

23 = number od children and adults

1c = value of the children tickets

4c = value of the adults tickets

80 = total collected

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c + a = 23- - - - - - - -Equation 1
1c + 4a = 80- - - - - -Equation 2
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Substitute Method equation 1

c + a = 23

c + a - c = - c + 23

a = - c + 23

Substitute the a value into equation 2

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1c + 4a = 80

1c + 4(- c + 23) = 80

1c + ( - 4c + 92) = 80

1c - 4c + 92 = 80

- 3c + 92 = 80

- 3c + 92 - 92 = 80 - 92

- 3c = - 12

- 3c / - 3 = - 12 / - 3

c = - 12 / - 3

c = 4. . .Number of children

Insert the c value into equation 1

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c + a = 23

4 + a = 23

4 + a - 4 = 23 - 4

a = 19. . . .Number of adults

insert the a value into equation 1

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Check for equation 1

c + a = 23

4 + 19 = 23

23 = 23

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Check for equation 2

1c + 4c = 80

1(4) + 4(19) = 80

4 + 76 = 80

80 = 80

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Both equations balance

There are 4 children and 19 adults

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Answer 2

The equation of a without delay Line is given via, y = mx + c the place m is the gradient c is the y-intercept looking m, m = (4 - 2)/(seventy 9 - 40 9) = 2 / 30 = a million/15 The equation is subsequently, y = (a million/15) x + c replace x = 40 9 and y = 2 to locate c, 2 = (a million/15) (40 9) + c c = 2 - 40 9/15 c = (30 - 40 9)/15 c = - 19/15 subsequently, y = (a million/15) x - 19/15 15y = x - 19

Answer 3

let number of adults be x
let number of children be y

first given total number of ppl is =23

x + y = 23; --eqn 1

price of tickets for an adult is $4
price of tickets for a child is $1
theater collected $80

4x + y = 80; --eqn 2
eqn 2 - eqn 1
3x = 57
x = 19
substitute x = 29 to eqn 1

19 + y = 23
y = 4;

x = 19 and y = 4

Answer 4

If # of adults= X
then, # of children=(23-X)

4*X + 1*(23-X) = 80
4X + 23 - X = 80
3X=80-23
X=57/3=19

therefore # of adults= 19
# of children= 23-19 = 4

Answer 5

a = number of adults
23 - a = number of children

4a + (1)(23-a) = 80
4a + 23 - a = 80
3a =57
a = 19

So there's 19 adults and 4 kids

19(4) + 4(1) = 80 It checks.

Answer 6

Let C = no. of children
A = no of adult

A + C = 23 then A = 23 - C -------eq'n 1

4A + 1C = 80 -------------------------eq'n 2

substituting A in eq'n 2 hence,

4(23-C) + C = 80
92 - 4C + C = 80
-3C = 80 - 92
-3C = -12
C = 4 the number of children who attended the play

Answer 7

let no of adults = x
and no of kids = y
total number of pple = x+y = 23
total sales = US80
adult sales = 4 X x
kids sales = 1 X y
total sales = adult sales + kids sales = 80
i.e. 4x + y = 80
this leaves a simultaneous equation
x + y=23
4x + y=80
when sloved,
x=19, y=4
no of kids is 4

Answer 8

x = number of adults
y = number of children

4x + y = 80
x + y = 23 (simultaneous equations)

3x = 57
x = 19

Sub in 1st eqn:
4x + y = 80
76 + y = 80
y = 4

Check in 2nd eqn.
x + y = 23
19 + 4 = 23 (correct)

4 CHILDREN ATTENDED THE PLAY

Answer 9

A = # of adults
C = # of childen

A + C = 23
4A + 1C = 80

Now use what you learned about solving simultaneous equations.

3A = 80 - 23 = 57
A = 19
A + C = 23
19 + C = 23
C = 4

Answer 10

let there be x children and y adults.therefore
the required equation is x*1+y*4=80
but y=23-x
so your final equation is=
x*1+(23-x)*4=80
hope i'm clear.............