SOLUTION: Hi,
My question is to be solved using simultaneous equations.
Twelve years ago Jack was five times as old as Jill. In three years time Jill will be half of Jacks age. What are Ja
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Question 418982: Hi,
My question is to be solved using simultaneous equations.
Twelve years ago Jack was five times as old as Jill. In three years time Jill will be half of Jacks age. What are Jack and Jills present ages.
So far I have a chart representing Jack and Jill down the left with age now, age -12 and age +3 across the top.
The first equation is y-12=5(x-12) simplified to y=5x-48
The second equation is x+3=0.5(y+3) simplified to x=0.5y-1.5
That is as far as I have been able to progress. I am struggling with the fact both simplified equations still have a variable so cannot be substituted into another equation.
Regards,
Wayne
Found 3 solutions by scott8148, mananth, josmiceli:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
substitute the equivalent value from one equation into the other equation
x = .5(5x-48) - 1.5
2x = 5x - 48 - 3
51 = 3x
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Jack's age now = x
Jill's age now = y
..
12 years ago Jack = x-12
Jill = y-12
x-12=5(y-12)
x-12=5y-60
x-5y=-48......................1
y+3 = 1/2 (x+3)
2(y+3)=x+3
2y+6=x+3
2y-x=-3........................2
....
1 x -5y = -48 .............1
-1 x + 2y =-3 .............2
multiply (1)by 2
Multiply (2) by 5
2 x -10 y = -96
-5 x + 10 y = -15
Add the two equations
-3 x = -111
/-3
x=37
plug value of x in (1)
1 x -5y=-48
37+-5y =-48
-5y=-48 -37
-5y=-85
y=17
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = Jack's age now
Let = Jill's age now
given:
(1)
(2)
This is 2 equations with 2 unknowns,
so it is solvable
(1)
(1)
(1)
and
(2)
(2)
(2)
Subtract (1) from (2)
(2)
(1)
and, since
Jack is 37 and Jill is 17
check:
(2)
(2)
OK
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