SOLUTION: Mrs. Kwok is 6 years more than 3 times as old as her daughter. Eight years ago she was 2 years less than 9 times as old as her daughter was then. Find each of their ages. And

Question 24020: Mrs. Kwok is 6 years more than 3 times as old as her daughter. Eight years ago she was 2 years less than 9 times as old as her daughter was then. Find each of their ages.
And also:
The sum of 7 times Paul's age and 4 times Kent's age is 111. Kent is 1 year less than four times as old as Paul is. Find each of their ages.
I would like to see it worked out step by step, please.
Answer by Paul(988) (Show Source):
You can put this solution on YOUR website!
1. Mrs Kwok = 3x+6
Daughter's age =x

Eight years ago = 3x+6-8 (mrs Kwok)
Eight years ago = x-8 (daughter's age)

3x+6-8=9(x-8)-2
3x-2=9x-72-2-2
-6x-2=-74
-6x=-72 Divide both sides by -6
x = 12

3(12)+6=42

Hence Mrs. Kwok is 42 years old and her daughter is 12 years old.

2. Paul's age = x
Kent's age = y

7x+4y=111
4y=111-7x
y=27.75-1.75x (subsitution)

"Kent is 1 year less than four times as old as Paul is"
Kent = y
Paul = 4x-1

y=4x-1 (substitute)
(27.75-1.75x)=4x-1 simplify
28.75=5.75x
x=5

y=27.75-1.75(5)
y=19

Hence Paul is 19 and Kent is 5.

Check: 4(19) + 7(5) = 111
Paul.