sarah is twice as old as john. six years ao, sarah was 4 times as old as john was then. how old is john now?
First step in algebraic word problems is to represent values :) If we let x be the age of John now, what is the age of Sarah?
2x?
Good :) Okay, so what would be John's age SIX YEARS AGO ?
2x-6? :)
uh-uh, that's ROSE's age six years ago. What about John?
Remember, John's age now is x.
omg fail
um well idk
forget I said anything... that's Sarah
Sarah's age 6 years ago was 2x - 6
lol XD i was so confused
Sorry, rosie, I was putting you in the problem XD
lol
gah
But anyway, John's age now is x, Sarah's age now is 2x. Six years ago, their ages should be...?
so sarahs age is six yrs ago was 2x-6
yup. What about John?
and john is 4x-6?
No... John's age now is x. so six years ago, it should be...? (if you're 20 now, six years ago it would be 14...)
x-6?
Yes, precisely :D
John's age six years ago = x - 6 Sarah's age six years ago = 2x - 6 Now, to quote your problem: Six years ago, Sarah was 4 times as old as John was then. How do you translate that in math terms?
4(2x-6)(x-6)?
or is it 4(x-6)
Okay, I'll translate, you solve... :) Trust me, the only difficult thing in these word problems is finding what problem to solve... when you get it, it's usually easy :) Now... Six years ago... (this only means we'll be using their ages six years ago) Sarah 2x - 6 was 2x - 6 = 4 times 2x - 6 = 4() as old as John was then 2x - 6 = 4(x - 6) Now just solve for x... which we know to be John's age now :D
oh my god thank you so much. this is from the book for the sats. i was studying so yea :)
Thank me by telling me your answer :3
ok so i got 9. is that correct?
Bingo :)
yay finally got it. thanks so much :)
No problem :)
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