PLease answer my question below

Question

the admission at a concert was $9.00 for adults and $4.00 for children if the total gate receipts were $775 and there were 125 people in attendance how many adult and childrens tickets were sold

Answer 1

That was a correct answer. Another way to solve it, which I think has less abstraction, so may appeal to younger students, is to use variables that relate to the problem better, and simplify the steps.

First, let the number of adults be A, and the number of children be C.

We know two things:

The total number in attendance is A + C = 125

The total price, $9 times A plus $4 times C, is 9A + 4C = 775

So C = 125 - A

Substitute this value in the money equation:

9A + 4(125 -A) = 775

Hooray! Now we have an equation with only one unknown.

9A - 4A = 775 - 500

A = 55

Plugging this back in the first equation,

55 + C = 125, or

C = 70

The student should now plug these two values into the second equation to check his work!

I think this method also shows that the solution makes common sense, and demystifies algebra.

Answer 2

$9A + $4C = $775      A + C = 125     A = 125 - C

 

$9(125 - C) + $4C = $775

 

$1125 - $9C + $4C = $775             $1125 - $5C = $775      $1125 -$1125 - $5C = $775 - $1125

 

- $5C = - $350       - $5C / - $5  = -$350 /-$5         C = 70 children     A = 125 - C     A = 125 - 70 = 55 Adults

Answer 3

Let x = number of adult tickets

      y = number of children's tickets

 

Then, x + y = 125

          9x + 4y = 775

 

Multiply the first equation by -4 to get the system:

 

-4x - 4y = -500

 9x + 4y = 775

 

Adding the equations we have 5x = 275, so x = 55.

 

Since x + y = 125 and x = 55, y = 125 - 55 = 70.

 

ANSWER:  55 adult tickets and 70 children's tickets were                     sold.