Answer:
12 and 14
Step-by-step explanation:
Let the even integers be x and x+2
three times the larger number is expressed as 3(x+2)
30 more than the smaller one is x + 30
Equating both expressions
3(x+2) = x +30
Find x
3x+6 = x+30
3x-x = 30-6
2x = 24
x = 12
The second integer is 12+2 = 14
Hence the required integers are 12 and 14
I rounded 9.27 repeating to 9.28 to make it easier to solve. It would be 232/25 as an improper fraction and 9 and 7/25 as a mixed number.
Answer : C (24 - 6) ÷ 3
(almost 100% sure this is correct lol)
The intersection is at (2,-1)
