SOLUTION: three years ago lisa was three-fourth age of kaija. eleven years kaija was twice as old as lisa. how old is kaija now
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Question 821737: three years ago lisa was three-fourth age of kaija. eleven years kaija was twice as old as lisa. how old is kaija now
Found 2 solutions by TimothyLamb, lwsshak3:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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x - 3 = 3/4(y - 3)
y - 11 = 2(x - 11)
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x - 3 = 3/4y - 9/4
y - 11 = 2x - 22
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x - 3/4y = -9/4 + 12/4
2x - y = -11 + 22
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x - 3/4y = 3/4
2x - y = 11
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4x - 3y = 3
2x - y = 11
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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answer:
x= lisa = 15
y= kaija = 19
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Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
three years ago lisa was three-fourth age of kaija. eleven years kaija was twice as old as lisa. how old is kaija now.
***
3 yrs ago:
x=Kaija's age
.75x=Lisa'age
..
now:
x+3=Kaija's age
.75x+3=Lisa'age
..
11 yrs from now:
x+3+11=x+14=Kaija's age
.75x+3+11=.75x+14=Lisa'age
..
x+14=2(.75x+14)
x+14=1.5x+28
.5x=14
x=28
x+3=31
How old is Kaija now? 31
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