The sum of three numbers is 98. The first number is 7 less than the second. The third number is 3 times the second. What are the numbers?

14, 21, and 63

Step-by-step explanation:

we can set the three unknown numbers as x, y, and z

so

x + y + z = 98

now we are given that the first (x) is seven less than the second (y)

x = y - 7

the third number (z) is three times the second (y)

z = 3y

since we have an equation for z and x in terms of y, we can plug it into the first equation and solve for y.

(y-7) + y + (3y) = 98

y - 7 + y + 3y = 98

y + y + 3y = 105

5y = 105

y=21

now that we know y, we can plug its value into either the second or the third equation to find x or z (I chose to find z first)

z = 3y

z = 3 (21)

z = 63

now we can plug in y to find x

x = y - 7

x = 21 - 7

x = 14

so the three numbers should be 14, 21, and 63