What's the question? there isn't enough info to help.
Answer:
92 attendees had activity cards
Step-by-step explanation:
Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...
7x +10(130 -x) = 1024
7x +1300 -10x = 1024 . . . . eliminate parentheses
-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300
x = 92 . . . . . . divide by 3
92 students with activity cards attended the dance.
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<em>Comment on the solution</em>
Often, you will see such a problem solved using two equations. For example, they might be ...
Let 'a' represent the number with an activity card; 'w' the number without. Then ...
- a+w = 130 . . . . the total number of students
- 7a +10w = 1024 . . . . the revenue from ticket sales
The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...
w = 130-a
and making the substitution into the second equation gives ...
7a +10(130 -a) = 1024
This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.
A. define the variables : a = abby's miles, j = james' miles
b. a = j + 4......j = a - 4
c. a + j = 28
a = j + 4
d. (j + 4) + j = 28
2j + 4 = 28
2j = 28 - 4
2j = 24
j = 24/2
j = 12
a = j + 4
a = 12 + 4
a = 16
e. check..
a + j = 28
16 + 12 = 28
28 = 28 (correct)
solution is : abby's miles (a) = 16
james' miles (j) = 12
Here is your perfect answer
