algebraic word problem

Question

Sue is selling tickets for the circus. On day one she sold 4 senior tickets and 2 children's tickets and made $80. On day two she sold 5 senior tickets and 7 children's tickets making $145. how much were each senior and child's tickets? 

Answer 1

Write two equations based on the information given:

4x + 2y = 80

5x + 7y = 145

where x = senior's tickets and y = children's tickets

 

We now have two equations with two variables. We could solve this a number of ways. I will use substitution:

Solve for y in the first equation:

4x + 2y = 80

y = (80 – 4x) / 2 = 40 – 2x

 

 

Now plug y = 40 – 2x into the second equation:

5x + 7y = 145

5x + 7(40 – 2x) = 145

5x + 280 – 14x = 145

-9x = -135

x = 15

 

Now that we know x = 15, we can solve for y in either one of the original equations.

4x + 2y = 80

4(15) + 2y = 80

2y = 20

y = 10

Answer: x = 15, y = 10. (Senior’s tickets: $15, children’s tickets: $10)

You should verify that you have the right answer by substituting the values found for x and y back into the original equations.

Answer 2

Jusin I.'s answer is correct. I would just suggest using the letters a (for adult) and c (for children) as the variable names. These will be a better reminder of what the variables mean.  Here is the check that he recommends using these letters.

4s + 2c = 80        Check s = $15   c = $10

4(15) + 2(10) = 80

60  +  20 = 80

           80 = 80  We are good.

5a + 7c  =  145     Check s = $15 c = $10

5(15)  + 7(15)   = 145

   75    +  70     =  145

                 145 = 145  We check good here too. 

Answer 3

Hey Kami -- add the formulas to "even up" ... 9s + 9c = 225 ==> s+c =25 ... 2s + 2c =50 ... observe the "offset": 4s + 2c sells for $30 more ==> 2s =30 ... s=15, c=10 ... Best wishes :)