Question 85268This question is from textbook Beginning Algebra
: Please help with this.
Pat is 20 years older than his son Patrick. In two years Pat will be twice as old as Patrick. How old are they now?
What I have so far is: Patrick's current age is X, Pat's Current age is X+20. In two years Patrick's age will be X+2 and Patrick's age will be X+20+2.
So I tried X+2(Pat's future age)plus x+20+2(Patrick's future age) = 2X(Pat being twice the age of Patrick). Or X+2+X+20+2=2X which reduces to 2X+24=2X which gives 2X-2X=-24, or 0=-24. Which is not the right answer.
The back of the book gives Pat's age as 38 and Patrick's as 18.
This question is from textbook Beginning Algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Pat is 20 years older than his son Patrick. In two years Pat will be twice as old as Patrick. How old are they now?
What I have so far is: Patrick's current age is X, Pat's Current age is X+20. In two years Patrick's age will be X+2 and Patrick's age will be X+20+2.
So I tried X+2(Pat's future age)plus x+20+2(Patrick's future age) = 2X(Pat being twice the age of Patrick). Or X+2+X+20+2=2X which reduces to 2X+24=2X which gives 2X-2X=-24, or 0=-24. Which is not the right answer.
----------------
NOW DATA:
Patrick is x; Pat =x+20
--------------
2 YEARS FROM NOW DATA:
Patrick will be x+2; Pat will be x+22
EQUATION:
x+22 = 2(x+2)
x+22 = 2x+4
x=18 (Patrick's age now)
x+20 = 38 (Pat's age now)
===============
Cheers,
Stan H.
|
|
|
