Question 1203413: Sex years ago, Jason was four times as old as Jessa. In four years, He would be twice as old jessa. How old are they now?
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
 Six years ago, Jason was four times as old as Jessa.
In four years, He would be twice as old Jessa. How old are they now?
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x = Jason's age now (in years).
y = Jessa's age now.
6 years ago
x-6 = 4(y-6), or, simplifying, x-6 = 4y - 24; or, simplifying further x - 4y = -18.
In 4 years
x+4 = 2(y+4), or, simplifying, x+4 = 2y + 8; or, simplifying further x - 2y = 4.
So, you have this system of two equations
x - 4y = -18 (1)
x - 2y = 4 (2)
To solve it, subtract equation (1) from equation (2). You will get
-2y - (-4y) = 4 - (-18)
2y = 22
y = 22/2 = 11.
Then from equation (2), x = 4 + 2y = 4 + 2*11 = 26.
ANSWER. Jason is 26 years old now; Jessa is 11 years old.
CHECK. You can check it on your own, that all conditions of the problem are satisfied,
so the answer is correct.
Solved.
Answer by MathLover1(20849)  (Show Source):
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