Questions on Word Problems: Age answered by real tutors!

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1200X+2200(X+3)=253 OR 1200X-253=-2200(X+3)
1200X+121=2200(X+3)NOW ADD THE 2 EQUATIONS
1200X-253=-2200(X+3)
---------------------
2400X-132=0
2400X=132
X=132/2400
X=.055 OR %5.5% FOR THE $1200 INVESTMENT.
.055+.03=.085 0R 8.5% FOR THE $2200 INVESTMENT.
PROOF
1200*.055+2200*.085=253
66+187=253
253=253

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this year your brother james will be 2 years from being twice as old as your sister jenny.
Let Jenny's age be "x":
Then James age is "2x-2"
------------------------------
The sum of james age and 3 times jennys age is 66. How old is Jenny?
EQUATION:
2x-2 + 3x = 66
5x = 68
x = 13 3/8 years
------------------
Cheers,
Stan H.


----------------------------------------------------

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Let g=age of gym, a=age of auditorium
g=a-21
g=a/4
a-21=a/4
4(a-21)=a
4a-84=a
3a=84
a=28
g=7
.
Ed

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WENDY'S AND DANIEL'S AGES COMBINED ARE TWICE JAMIE'S AGE. WENDY IS 8 YEARS OLDER THAN DANIEL. JAMIE'S AGE PLUS DANIEL'S AGE IS 20 YEARS. HOW OLD ARE EACH OF THEM?
-------------------------------------
Let Daniel's age be "x".
Wendy's age is "x+8".
Jamie's age is "x+8 + x" = "2x+8"
----------------------------------
EQUATION:
Jamie's age Plus Daniel's age is 20 years.
2x+8 + x = 20
3x = 12
-----------------
x = 4 (Daniel's age Now)
x+8 = 12 (Wendy's age NOW)
2x+8 = 2*4+8 = 16 (Jamie's age NOW)
========================================
Cheers,
Stan H.

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Julie is 4 times older than Aaron. In 10 years Julie will be twice as old as Aaron. How old are Julie and Aaron?
------------------------------------------
Now DATA:
Let Aaron be x yrs old.
Then Julie is 4x yrs old
-------------------------
10 years from now DATA:
Aaron will be x+10 yrs old.
Julie will be 4x+10 yrs old.
------------------------------
EQUATION:
Julie = twice Aaron's age.
4x+10 = 2(x+10)
4x + 10 = 2x + 20
2x = 10
x = 5 (Aaron's age NOW)
4x = 20 (Julie's age NOW)
=============================
Cheers,
Stan H.

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A mother is 25 years older than her daughter. Ten years ago, the daughter is 4/9 of the age of the mother was then. Find the age of each.
.
Let D = daughter's age
then
D+25 = Mother's age
(from:"A mother is 25 years older than her daughter.")
.
We derive our equation from:
"Ten years ago, the daughter is 4/9 of the age of the mother was then."
to get:
D-10 = (4/9)([D+25]-10)
D-10 = (4/9)(D+25-10)
D-10 = (4/9)(D+15)
9D-90 = (4)(D+15)
9D-90 = 4D+60
5D-90 = 60
5D = 150
D = 30 years (daughter's age)
.
D+25 = 30+25 = 55 years (Mothers age)

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: A mother is 4 times as old as her son and in twenty years will be twice as
old as her daughter, who is five years older than her brother. How old is each?
:
Let their present ages = m, b, s; for mother, brother, and sister
:
write an equation for the given statements:
"A mother is 4 times as old as her son"
m = 4b
:
"in twenty years (she) will be twice as old as her daughter,"
m + 20 = 2(s+20)
m + 20 = 2s + 40
m - 2s = 40 - 20
m - 2s = 20
:
"who is five years older than her brother."
s = (b + 5)
:
How old is each?:
:
In the equation: m - 2s = 20:
substitute 4b for m and (b+5) for s
4b - 2(b+5) = 20
4b - 2b - 10 = 20
2b = 20 + 10
2b = 30
b =
b = 15 is the brothers age
:
you can find the other ages using the equations for m and s,
Check your solutions in the statement:
"In twenty years mother will be twice as old as her daughter,
:
:
Let me know if you have any questions about this?

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46+X=3(10+X)
46+X=30+3X
X-3X=30-46
-2X=-16
X=-16/-2
X=8 YEARS GODFATHER WILL BE 3 TIMES THE GRANDSON'S AGE.
PROOF:
46+8=3(10+8)
54=3*18
54=54

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x+2=3y/4 or x=3y/4-2
x-2=2y/3
(3y/4-2)-2=2y/3
(3y-2*4)/4-2=2y/3
(3y-8)/4-2=2y/3
(3y-8-2*4)/4=2y/3
(3y-8-8)/4=2y/3
(3y-16)/4=2y/3 cross multiply.
3(3y-16)=4*2y
9y-48=8y
9y-8y=48
y=48 age of Edgard now.
x=3*48/4-2
x=3*12-2
x=36-2
x=34 age of Edna now.
Proof:
34-2=2*48/3
32=96/3
32=32

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Let m - Martin's age now; Let g = Gary's age now
:
Write an equation for each statement:
"If Martin was four times as old as Gary 8 years ago"
(m-8) = 4(g-8)
m - 8 = 4g - 32
m - 4g = -32 + 8
m - 4g = -24
:
" and if Martin will be twice as old as Gary 8 years hence,"
(m+8) = 2(g+8)
m + 8 = 2g + 16
m - 2g = 16 - 8
m - 2g = 8
:
Subtract the 1st equation from the above equation:
m - 2g = 8
m - 4g = -24
--------------subtraction eliminates m
0 + 2g = +32
g =
g = 16 yrs Gary's age now
:
Use the equation; m - 2g = 8 to find m
m - 2(16) = 8
m - 32 = 8
m = 8 + 32
m = 40 yrs is Martin's age now
:
:
Check solution in the statement:
"If Martin was four times as old as Gary 8 years ago and if Martin will be twice as old as Gary 8 years hence, find their ages."
40 - 8 = 4(16-8)
and
40 + 8 = 2(16+8)

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Let k = Kyle's age now.
Then the age of Kyle's aunt NOW is 3k.
The age of Kyle 5 years age is k-5.
The age of Kyle's aunt 5 years ago is 3k-5.
You need to use the following equation to set up an equation:
"Five years ago, his aunt was four times as old as Kyle"
I.e., The age of Kyle's aunt 5 years ago = 4*(The age of Kyle 5 years age)
So
3k-5=4(k-5)

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Write an equation for each statement.
:
"the sum of the ages of Ira and Sofia is 26.
I + S = 26
or
I = (26-S)
:
" the difference between 4 times Sofia's age and twice Ira's age is 8."
4S - 2I = 8
;
Substitute (26-S) for I in the above equation, find S;
4S - 2(26-S) = 8
4S - 52 + 2S = 8
4S + 2S = 8 + 52
6S = 60
S = 10
Then
I = 26 - 10
I = 16
;
:
Check solution using the statement:
"the difference between 4 times Sofia's age and twice Ira's age is 8."
4(10) - 2(16) = 8

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Mrs. Shieh was 24 years old when Ingred was born. In three years the sum of their ages will be 68 years. How old is each now?
:
We can say that S is 24 yrs older than I, therefore
S = (I+24)
:
An equation for the statement:
"In three years the sum of their ages will be 68 years."
(S+3) + (I+3) = 68
S + I + 6 = 68
S + I = 68-6
S + I = 62
:
How old is each now?
:
Substitute (I+24) for S in the above equation
(I+24) + I = 62
2I = 62 - 24
I =
I = 19 yrs
then
19+24 = 43 yrs is S
:
:
Check solution in the statement:
"In three years the sum of their ages will be 68 years."
(43+3) + (19+3) = 68

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Reggie's age = "x+5"
why? because laarni's age = "x"
Five years ago:
----------------> working eqn
substract 5 yrs from their ages and sum up to 25.
Continuing:



-------------------------> Laarnis' age
------------------> reggie's age
To check, 5 years ago: Go back working eqn



Thank you,
Jojo

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Edwin's complete solution:
Please help me solve this problems :
1. On Dave's birthday, his brother Harry is 17 yrs younger than 3 times his age. The boy's father , Tom is 12 yrs. older than twice Harry's age. If Dave is 7 years younger than his brother , how many candles are on Dave's cake?

Please help me solve this problems :

>>...his brother Harry is 17 yrs younger than 3 times his age...<<

 
H = 3D - 17

>>...The boy's father , Tom is 12 yrs. older than twice Harry's age...<<

 
T = 2H + 12

>>...Dave is 7 years younger than his brother...<<

 
D = H - 7
 

 
-3D +  H     = -17
    - 2H + T =  12
  D -  H     =  -7
 
Solve that and get D=12, H=19, T=50

----------------------

2. When Alvin divides each of 5 consecutive integers by his age, the sum of the 5 remainders he gets is 32. When Susan , several yrs. older, divides each of a different set of 5 consecutive integers by her age, the sum of the five remainders she gets is also 32. What is the sum of the ages of Alvin and Susan?

Let either one of their ages be A.
Let the 5 consecutive integers be I1,I2,I3,I4,I5
Let the remainders when these integers are divided
by A be R1,R2,R3,R4,R5

If none of the consecutive integers is A or
or a multiple of A, then the 5 remainders will 
themselves be 5 consecutive integers.

Let's see if there exist 5 consecutive integers 
which the remainders could be, so as to have the sum 32.






No, that is not possible, for n doesn't come out
to an integer.

That means one of 5 consectutive integers must
be a multiple of A, say kA

Whichever one of the 5 is kA, The remainder left when
dividing by A will of course be 0.  

Any of the 5 remainders of members of the 5 that are 
less than kA, will be consecutive integers, the largest 
of which must be A-1, then the one below that will be
A-2, and so on.

Any of the 5 remainders that are greater than the multiple
of A will be consecutive integers 1,2,3, and so on

Then we have 5 cases to consider, that is, when kA comes
1st, 2nd, 3rd, 4th, and 5th.

         I1    I2    I3    I4    I5     R1   R2   R3   R4  R5   Sum  
case 1.  kA   kA+1  kA+2  kA+3  kA+4     0    1    2    3   4    10
case 2. kA-1   kA   kA+1  kA+2  kA+3    A-1   0    1    2   3   A+5
case 3. kA-2  kA-1   kA   kA+1  kA+2    A-2  A-1   0    1   2    2A
case 4. kA-3  kA-2  kA-1   kA   kA+1    A-3  A-2  A-1   0   1  3A-5    
case 5. kA-4  kA-3  kA-2  kA-1   kA     A-4  A-3  A-2  A-1  0  4A-10
 
We set each of those sums on the far right equal to 32 and see which,
if any, have an integer solution:

Case 1. This is out because 10 does not equal 32.

Case 2.  A+5 = 32
           A = 27

That's a whole number, so one of their ages can be 27.

Case 3.  2A = 32 
          A = 16

That's a whole number, so one of their ages can be 16.

Case 4. 3A-5 = 32
        2A   = 37
         A   = 27/2

That is not a whole number, so Case 4 is out.

Case 5. 4A-10= 32
        4A   = 42  
         A   = 21/2

That is not a whole number, so Case 5 is out.

So Alvin is 16 and Susan is 27, making the sum of their ages 43.

Now there is no way to tell whether Susan's 5 consecutive
integers were 

26,27,28,29,30, or 53,54,55,56,57, or 80,81,82,83,84 or etc.

There is also no way to tell whether Alvin's 5 consecutive
integers were 

14,15,16,17,18, or 30,31,32,33,34, or 46,47,48,49,50 or etc.
 
However he must be 16 and she must be 27.

Edwin

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= Dave's age now
= Harry's age now
= Tom's age now
(1)
(2)
(3)
--------------------
Replace in (1) with in (3)
(1)


Subtract from both sides


-------------------
(3)


There are 12 candles on Dave's cake
Note that I didn't need to know the father's age to solve this
-------------------
The 2nd one may be too tough for me

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I have been trying to solve this question but I cant figure it out: The sum of the present ages of Eric and his father is 58 years. In 10 years, his father will be twice as old as Eric will be at that time. Find their present ages.
------------------------
EQUATIONS:
Present: F + E = 58
Future: (F+10) = 2(E+10)
------------------------
Rearrange:
F + E = 58
F -2E = 10
-------------
Subtract the 2nd equation from the 1st to solve for E:
3E = 48
E = 16 yrs old (Eric's age now)
Since F+E= 58, F=42 (Father's age now)
========================
Cheers,
Stan H.

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John is 4 years younger than her brother. Five years from now, the sum of their ages will be 32. Find John's present age.
-------------------
John's a girl?
J = John's age
B = brother's age
----------------
John is 4 years younger than her brother.
So, B = J + 4 eqn 1
--------------
Five years from now, the sum of their ages will be 32.
(J+5) + (B+5) = 32
J + B = 22 eqn 2
Substitute for B from eqn 1
J + (J + 4) = 22
2J + 4 = 22
2J = 18
J = 9
B = 9+4
B = 13

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Bob is 20 years older than Ben. In 10 years, Bob will be twice as old as Ben. How old is each now?
.
Let x = age of Ben
then because "Bob is 20 years older than Ben"
x+20 = age of Bob
.
Our equation is derived from: "In 10 years, Bob will be twice as old as Ben."
(x+20) + 10 = 2(x+10)
x + 30 = 2x + 20
30 = x + 20
10 years = x (age of Ben)
.
x+20 = 10 + 20 = 30 years (age of Bob)

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anns age is seven-tenths of bills age.
a = b
or
a = .7b
:
in 4 years anns age will be eight-elevenths of bills age.
a + 4 = (b+4)
:
Replace a with .7b
.7b + 4 = (b+4)
;
Multiply both sides by 11 to get rid of the annoying denominator
7.7b + 44 = 8(b+4)
;
7.7b + 44 = 8b + 32
:
44 - 32 = 8b - 7.7b
:
12 = .3b
b =
b = 40 yr is Bill
then
.7(40) = 28 yrs is Ann
:
:
Check solution in the statement:
"in 4 years anns age will be eight-elevenths of bills age"
32 = *44

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Remember the following:

Number of tickets sold:
, five times as Ashley
, # of tickets Ashley sold
, maria sold 4 more than Ashley
Therefore:
-----------> eqn 1

--------->
----------------> # of tickets sold by Ashley
----------> # of tickets sold by Jill
---------> # of tickets sold by Maria
In doubt? Go back eqn 1:


Thank you,
Jojo

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Let's make for all reference.
1st condition:
, Sue's age
, leah being 6 years older than Sue
2nd condition:
, John being 5 years older than Leah
Equating to totasl age:
--------> working eqn

------------->
years old (Sue)
Leah= , 6 years older than Sue
John= , 5 years older than Leah
In doubt? go back to our condition & working eqn.


Thank you,
Jojo

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let n="number of years ago"

(24-n)/(12-n)=4 __ multiplying by 12-n __ 24-n=48-4n __adding 4n-24 __ 3n=24 __ dividing by 3 __ n=8

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The word problem translates into this equation

where x is the number of years ago that scott was 4 times as old as shannon



Start with the given equation.


Distribute.


Subtract from both sides.


Add to both sides.


Combine like terms on the left side.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is which means that 8 years ago scott was 4 times as old as shannon

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.5H+D/3=11
(H+6)+(D+6)=40
H+D+12=40
H+D=40-12
H+D=28
H=28-D NOW REPLACE H BY (28-D) IN THE FIRST EQUATION
.5(28-D)+D/3=11
14-.5D+D/3=11
-.5D+D/3=11-14
(3*-.5D+D)/3=-3
(-1.5D+D)/3=-3
-.5D=-9
D=-9/-.5
D=18 ANSWER FOR DAISY'S AGE NOW.
H+18+12=40
H+30=40
H=40-30
H=10 AGE OF HENRY NOW.
PROOF:
.5*10+18/3=11
5+6=11
11=11

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Let Chan be C years old.
His cousin, Zeke (we'll call him that), is Z years old.
"Chan is 4 times older than Zeke."
1.
In 8 years, Chan will be (C+8) years old and Zeke will be (Z+8) years old.
"In 8 years Chan will be two times older than Zeke."
2.
2.
Substitute eq. 1 into 2 and solve for Z.
2.



Then from eq. 1,
1.


Check the answer.
In eight years, Chan will be 24 and Zeke will be 12.

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The present age of a father exceeds 3 times the present age of his son by 2 years. Six years ago, the age of the father exceeded 4 times the age of the son by 4 years. Find the present ages of the son and the father.
.
Let F = present age of father
and S = present age of son
.
Since we have two unknowns we'll need to find two equations.
.
From:"The present age of a father exceeds 3 times the present age of his son by 2 years. " we get equation 1:
F = 3S + 2
.
From:"Six years ago, the age of the father exceeded 4 times the age of the son by 4 years." we get equation 2:
F-6 = 4(S-6) + 4
.
Using the value of F from equation 1, substitute it into equation 2 and solve for S:
F-6 = 4(S-6) + 4
(3S + 2)-6 = 4S - 24 + 4
3S + 2 -6 = 4S - 24 + 4
3S - 4 = 4S - 20
-4 = S - 20
16 years = S (age of son)
.
Use the above, and substitute it into equation 1 and solve for F:
F = 3S + 2 = 3(16) + 2 = 50 years (age of father)

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.
Let x = James age now
.
Then, from the statement of the problem, we formulate our equation:
x+30 = 8(x-12)
x+30 = 8x - 96
30 = 7x - 96
126 = 7x
18 = x
.
Therefore, James is 18 years old.

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.
Let x = number of years until father will be 3 times as old as son
.
From the statement of the problem, we get our equation:
41+x = 3(9+x)
41 + x = 27 + 3x
41 = 27 + 2x
14 = 2x
7 = x
.
Therefore, in 7 years the man will be three times as old as the son.

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Your initial equations look good!
3x+2y = 44
y = 2x-8 Substitute this into the first equation.
3x+2(2x-8) = 44 Simplify.
3x+4x-16 = 44
7x-16 = 44
7x = 60
x = 60/7
y = 2x-8
y = 2(60/7)-8
y = 120/7 - 56/7
y = 64/7
Well, mathematically, these are the solutions.
Was there anything in the instructions that led you to believe that the answers should be integers, besides the fact that you would normally expect integer answers in problems like this?

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Let m = Maya's age now
Let d = David's age now
;
Write an equation for each statement;
"In 15 years, Maya will be twice as old as David is now."
m + 15 = 2d
:
"In 15 years, David will be as old as Maya will be in 10 years from now."
d + 15 = m + 10
d = m + 10 - 15
d = m - 5
:
how old are they now?
;
Substitute (m-5) for d in the 1st equation
m + 15 = 2(m-5)
m + 15 = 2m - 10
15 + 10 = 2m - m
m = 25 yrs is m's present age
:
d = m - 5
d = 25 - 5
d = 20 is d's present age
;
:
Check solutions in the statement:
"In 15 years, David will be as old as Maya will be in 10 years from now."
d + 15 = m + 10
20 + 15 = 25 + 10; confirms our solutions

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A is eleven times as old as B.
A = 11B
-------------
in a certain number of years, A will be five times as old as B
A+x = 5(B+x)
-------------
and in five years after that she will be three times as old as B.
A+x+5 = 3(B+x+5)
-----------------
how old are they now?
-------
Rearrange the three equations with variables A, B, and x
A - 11B + 0x = 0
A - 5B -4x = 0
A -3B - 2x = 10
---------------
Solve the systems of three equations by any method to get:
A = 22
B = 2
x = 3
-----------
A is 22 yrs. old
B is 2 yrs. old
======================
Chers,
Stan H.

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Six years ago bob was four times as old as Mary.
In four years he will be twice as old as Mary.
How old are they?
-----------
Let Mary's age now be "m".
Let Bob's age now be "b".
---------------------------
6 yrs. ago equation:
(b-6) = 4(m-6)
------------------------------
4 yrs. from now equation:
b = 2m
--------------------------
Substitute to solve for "m".
(2m-6) = 4(m-6)
2m-6 = 4m-24
2m = 18
m = 9 yrs old (Mary's age now)
-------------------
Since b = 2m, b = 2*9 = 18 yrs old (Bob's age now)
===========================
Cheers,
Stan H.

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Six years ago bob was four times as old as Mary. In four years he will be twice as old as Mary. How old are they?
B= Bob's age now
M= Mary's age now
You did it wrong you have to start like this:
B-6=4(M-6)
B+4=2(M+4)
solving we have
B=26 and M=11

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2.54 cm = 1 in __ times 12 __ 30.48 cm = 1 ft __ .3048 m = 1 ft __ squaring __ .0929 m^2 = 1 ft^2 (approx)

to convert between the two; multiply ft^2 by .0929 to get m^2, or divide m^2 by .0929 to get ft^2

20 m^2/.0929 = 215.29 ft^2 (approx)

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Let j be joe's age and s be sally's age.
Given

Also given

Substituting from above yields




Joe is 30
sally is 30-15 = 15

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B - Bob's age
F- Bob's father's age
1.
Four years ago, Bob was B-4 years old and his father was F-4.
2.
Now you have two equations with two unknowns, you can solve it.
Substitute eq. 1 into eq. 2 and solve for B.
2.




Now use equation 1 to solve for F.
1.


Bob is 12 and his father is 36.
Four years ago, Bob was 8 and his father was 32.

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Three years ago, Rica was thrice as old as Dandy. Three years hence, she will be twice as old. Find their present ages.
-----------------------
Let Rica's present age be "r". Let Dandy's present ate by "d".
-----------------------
EQUATIONS:
Three years ago: r-3 = 3(d-3)
Three years hence: r+3 = 2(d+3)
-----------------
Rearrange:
r - 3d = -6
r - 2d = 3
--------------
Subtract 1st from 2nd to get:
d = 9 (Dandy's age NOW)
-----------
Substitute into r-2d = 3
r -(2*9) = 3
r = 21 (Rita's age NOW)
================
Cheers,
Stan H.

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Let x = yrs when richie is 2/3's aimee's age
.
14+x = (2/3)(26+x)
42+3x = 2(26+x)
42+3x = 52+2x
42+x = 52
x = 10 years

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1. Erwin is 5 times older than john and John is twice as old as his sister, Alice.

Erwin=5j
John=j; 2a
Alice=a; 1/2j

Use j, since all of the people can be expressed in terms of j.
In 2 years time, the sum of their ages will be 58.
5j+2+j+2+1/2j+2=58

Combine like terms
6.5j+6=58

Subtract 6
6.5j=52

Divide by 6.5
j=8

Erwin=5(8)
John=(8)
Alice=1/2(8)

5(8)+(8)+1/2(8)+6=58
40+8+4+6=58

Erwin=40
*John=8*
Alice=4
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2. John is twice as old as his friend peter. Peter is 5 years older than Alice.

John=2p
Peter=p; a+5
Alice=p-5

Use p as the variable.
In 5 years, John will be three times as old as Alice.
2p+5=3(p-5+5)

Simplify
2p+5=3p

Subtract 2p from both sides
5=p

John=2(5)
Peter=(5)
Alice=(5)-5

John=10
*Peter=5*
Alice=0
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3. Reggie is 5 years older than Lani.

Reggie=L+5
Lani=L

Five years ago, the sum of their ages is 25.
(L+5)-5+(L)-5=25

L+L-5=25
2L-5=25
2L=30
L=15

Reggie=(15)+5
Lani=(15)

Reggie=20
Lani=15

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Rita was 25 when her first son was born. Today the sum of their ages is 105. How old is Rita and her son?

Let the son's age today be S

>>...Rita was 25 when her first son was born...<<

This tells us that Rita is 25 years older than her son,
therefore,

Rita's age today is (S+25)

>>...Today the sum of their ages is 105...<<

Rita's age today +  Son's age today = 105

(S+25) + S = 105

Solve that and get S = 40.

So the son is 40 and Rita is 25 years older or 65.

Checking:

40 years ago the son was just born and Rita was 25.

Edwin

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Let L = Leah's age, S = Sue's age, and J = John's age, then, from the problem description, you can write:
1) L = S+6 "Leah (L) is 6 years older (+) than Sue (S)".
2) J = L+5 "John (J) is 5 years older (+) than Leah (L)".
3) L+S+J = 41 "The total (sum) of their ages is 41 years.
So you now have a system of three equations with three unknowns (L, S, and J).
You can solve this system of equations by first eliminating one of the variables to leave you with two equations with two unknowns, then eliminate another variable to leave you with one equation with one unknown.
Here's how it works:
Start with:
L+S+J = 41 Substitute L = S+6 (Equation 1) to get:
(S+6)+S+J = 41 Now substitute J = L+5 (equation 2) to get:
(S+5)+S+(L+5) = 41 But, since you want to find Sue's (S) age, replace the L here with L = S+6 (equation 1) to leave you with:
(S+6)+S+((S+6)+5) = 41 Now you have one equation with one unknown (S) which you can easily solve by combining like-terms:
3S+17 = 41 Subtract 17 from both sides of the equation,
3S = 24 Finally, divide both sides by 3 to leave you with:
S = 8
So Sue is 8 years old.

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Let S be Sue's age, L be Leah's age and J be John's age
Given


Find Sue's age (S)

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they would be 80

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----"Luisa is 5 years younger than her sister Sigrid"----

Luisa=L
Sigrid=L+5

----"8 years from now she will be four-fifths as old as her sister will be then"----

L+8=4/5(L+5+8)

Combine like terms on the right side
L+8=4/5(L+13)

Distribute on the right side
L+8=4/5L+10.4

Subtract 8 from both sides
L=4/5L+2.4

Subtract 4/5L from both sides
1/5L=2.4

Multiply both sides by 5
L=12

Luisa=L
Sigrid=L+5

----"How old is each now?"----

Louisa=12
Sigrid=17