algebra problem, i need help?

Question

A group of 54 people are traveling by bus. Each child (c) has 2 piece of luggage and each adult (a) has 3 pieces of luggage. The total number of luggage is 111.

the question is...

How many children are on the bus? How many adults?

if you can, explain to me how you did it.

Answer 1

let

c = child

a = adult

54 = total child and adult

2c = child luggage

3a = adult luggage

111 = total child and adult luggage

- - - - - - - - - - - -

c + a = 54- - - - - - - - Equation 1
2c + 3c = 111- - - - - Equation 2
- - - - - - - - - - -

Substitute method equation 1

c + a = 54

c + a - c = - c + 54

a = - c + 54

Substitute the a value into equation 2
- - - - - - - - - - - - - - -

2c + 3a = 111

2c + 3(- c + 54) = 111

2c + ( - 3c + 162) = 111

2c - 3c + 162 = 111

- c + 162 = 111

- c + 162 - 162 = 111 - 162

- c = - 51

Multiply - c by - 1 to make c positive

- (- 1)(c) = - (-1)(51)

- ( - c) = - (- 51)

c = 51. . .number of childres

Insert the c calue into equation 1

- - - - - - - - - - - - -

c + a = 54

51 + a = 54

51 + a - 51 = 54 - 51

a = 3. . . . .number of adults

Insert the a value into equation 1

- - - - - - - - - - - - -- - - -- - - - - - - -

Check for equation 1

c + a = 54

51 + 3 = 54

54 = 54

- - - - - - - - - -

Check for equation 2

2c + 3a = 111

2(51) + 3(3) = 111

102 + 9 = 111

111 = 111

- - - - - - - - - - -

Both equations balance

Theer are 51 children and 3 adults

- - - - - - - - - -s-

Answer 2

total of people, childrena and adults = 54

c + a = 54

total number of luggage = 111

2c + 3a = 111 -->children had 2, adults had 3

now isolate a variable in the first equation
c + a = 54
c = 54 - a

substitute in that equation into the second equation for c.

2c + 3a = 111
2(54 - a) + 3a = 111 --->distribute the 2
108 - 2a + 3a = 111 --->add -2a and 3a together
108 + a = 111 ---> subtract 108 from both sides
a = 3


now that you have the number of adults, u can solve for the children, c, by substituting a back in the first equation.

a = 3
c + a = 54
c + 3 = 54
c = 51

so you have 51 children and 3 adults



check your answers by subbing them into the equations
c + a = 54
51 + 3 = 54
54 = 54

2c + 3a = 111
2(51) + 3(3) = 111
102 + 9 = 111
111 = 111

Answer 3

Let c = the number of children
a = the number of adults

54 = total number of passenger

2c = the total pieces of luggage of the children
3a = the total pieces of luggage of adults
111 = the total number of luggages

Equation:
c + a = 54 (eq. 1)
2c + 3a = 111 (eq. 2)

From eq. 1

c + a = 54 ( children plus adults = 54 passengers)
c = 54 - a => (eq. 3)

Substitute the value of c to eq. 2
2c + 3a = 111
2(54 - a) + 3a = 111
108 - 2a + 3a = 111
Add - 108 to both sides
a = 111 - 108
a = 3 ===> number of adults

c + a = 54
c + 3 = 54
c = 54 - 3
c = 51 ===> number of children

To check:

2c+ 3a = 111
2(51) + 3(3) = 111
102+ 9 = 111
111 = 111

Answer 4

Let the number of children is, c
and Let the number of adults is, a
Since total number of passengers is 54,
therefore a+c=54 (Equation 1)

Now every child is carrying 2 piece of luggage therefore 'c' children will carry 2c piece of luggage
Similarly since every adult is carrying 3 piece of luggae therefore 'a' adults will carry 3a pieces of luggage
Since total number of luggage pieces is 111, therefore luggage carried by adults and children will become
2c+3a =111 (Equation 2)

Now if you solve both equation 1 and 2
you will get a=3 and c=51 which means there are 3 adults and 51 children

:-)

Answer 5

Let
x be the number of children and
y be number of adults

The number of children and adults total is 54:
x + y = 54
x = 54 - y

The pieces of luggage can be formalized as follows:
2x + 3y = 111

since x can be written as 54 - y
2(54-y) + 3y = 111
108 -2y + 3y = 111
108 + y = 111
y = 3

So 3 adults and 51 children

Answer 6

(1) c + a = 54
(2) 2c + 3a = 111

multiply both sides of equation (1) by 2:
(3) 2c + 2a = 108

subtract (2) - (3):
a=3

solve for c in (1):
c + 3 = 54
c = 51

verify in (2):
(2 * 51) + (3 * 3) = 102 + 9 = 111

so,
51 children and 3 adults

Answer 7

c + a = 54

2c + 3a = 111

Solve the first equation for c

c = 54 - a

Substitute

2(54 -a) + 3a = 111

Distribute

108 - 2a + 3a = 111

Combine like terms

108 + a = 111

Subtract 108

a = 3

Substitute 3 into either equation

c + 3 = 54

c = 51

There are 3 adults and 51 children

Answer 8

51 children = 102 pieces of luggage

3 adults = 9 pieces of luggage

102 + 9 = 111

Answer 9

no of peaople = 54
let no of childeren be x let no of adults be y
by given condition x+y = 54 so x+54 - y ........................1
by the condition of luggage
2x+3y = 111.........................2
substituting 1 in 2
2(54 - y) + 3y = 111
on simplyfying
y = 3
and x = 51
so no of adults = 3
no of childeren = 51

Answer 10

C + A = 54
2C +3A = 111
C=54-A
2(54-A) +3A =111
108 -2A +3A =111
A = 3
C=51