Solve the problem. 43) The sum of two numbers is 56. The second number is three times as large as the first number. What are the numbers?
A) 11,45
B) 14,42
C) 12,36
D) 12,44

14,42

Step-by-step explanation:

x + y = 56

x = 3y

Substitute for x in the first equation:

x + y = 56

x = 3y

3y + y = 56

Combine like terms:

4y = 56

Divide both sides by 4:

y = 14

So if y = 14, and x = 3y, then x = 42.

Now substitute for both x and y in both of the original equations to prove they are the correct values.

x + y = 56

x = 42

y = 4

42 + 14 = 56

56 = 56

x = 3y

x = 42

y = 14

42 = 3*14

42 = 42

I hope this helped. I am sorry if you get this wrong.

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>