8 + 8 = 16
16 - 1 = 15
or
7 + 7 = 14
14+1 = 15
Answer:
James bought 11 good tickets and 5 bad tickets.
Step-by-step explanation:
Given that:
Cost of each good ticket = $8
Cost of each bad ticket = $5
Total amount spent = $113
Total tickets bought = 16
Let,
x be the number of good tickets bought
y be the number of bad tickets bought
x+y=16 Eqn 1
8x+5y=113 Eqn 2
Multiplying Eqn 1 by 5
5(x+y=16)
5x+5y=80 Eqn 3
Subtracting Eqn 3 from Eqn 2
(8x+5y)-(5x+5y)=113-80
8x+5y-5x-5y=33
3x=33
Dividing both sides by 3

Putting x=11 in Eqn 1
11+y=16
y=16-11
y=5
Hence,
James bought 11 good tickets and 5 bad tickets.
Answer:
No
There are infinitely many solutions to the system.
{36x+21y=24312x+7y=81
Multiply the second equation by 3.
3(12x+7y36x+21y=81)=243
Subtract this new equation from the first equation.
36x+21y−(36x+21y0+0=243=243)=0 true
Answer: 20 apples, because if you say 8 apples=$4, then that means 2 apples =$1, so $10=20 apples
Step-by-step explanation: