Math questions. please answer :)?

Question

If Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?

i get so confused with these questions..please explain

Answer 1

Let L be the age of Leah, S the age of Sue, and J the age of John.

Leah is six years older than Sue

L = S + 6

John is five years older then Leah

J = L + 5

The total of their ages is 41

J + L + S = 41

You have three equations and three unknowns. We'll use the first two equations to create substitutions for J and S in terms of L.

L = S + 6 therefore S = L - 6
J = L + 5

Now the last equation becomes

J + L + S = 41
( L + 5 ) + L + ( L - 6 ) = 41
3L -1 = 41
3L = 42
L = 42 / 3 = 14

So Leah is 14, John is 19, and Sue is 8.

Answer 2

OK, lets say Sue = S. We know the sum of their ages is 41. So...
41=S+S+6+S+6+5
(this is sue's age, plus sue's age with 6 years, plus sue's age plus 6 and 5 years)
41=3S+17
(I've added the numbers together and there are 3 S's)
41-17=3S
(I've moved the 17 to the other side of the = which makes it a negative)
24=3S
(subtracted 17 from 41)
8=S
(found out what S is by deviding both sides of the = by 3)

So Sue is 8. You can check it by added them together again. If Sue is 8, Leah must be 14, and John must be 19, total is 41! Hope this was a good explanation for you :-)

Answer 3

Leah = 6+x
Sue = x
John = 5+(6+x)
Total sum = 41

6+x+x+5+6+x=41
3x+17=41
3x=24
x=8

8+6=14
14+5=19

Hence, Sue is 8 years old, Leah is 14 years old and John is 19 years old.
Paul.

Answer 4

Let n = sue's age; n+6= leah's age; n+11 = john's age
n+6+5 = n+11

n+n+6+n+11 =41
3n + 17 =41 subtract 17 from both sides
3n=24 divide both sides by 3
n =8

Sue is 8 years old

Answer 5

Use algebra:
L=S+6
J=L+5 = (S+6)+5
L+S+J=41
(S+6)+S+[(S+6)+5] = 41
3S+17=41
S=8