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# Leah is 6 years older than Sue. John is 5 years older than Leah. Total combined age is 41. How old is Sue?

What is the algebraic equation

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Leah is 6 years older than Sue. John is 5 years older than Leah. Total combined age is 41. How old is Sue?

Sue is X

Leah is X+6

John is X+11 (5 years older than Leah who is 6 years older than Sue).

You can use

X+X+6+X+11 = 41

-11 on both sides gets you

X+X+6+X = 30

-6 on both sides gets you

3x = 24

Divide each side by 3 to get  the value of X (Sue's age)

x = 8

Cynthia N. | I take the stress out of learning math, and make it seem easy!I take the stress out of learning math, ...
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Usually the last question in the word problem determines the unknown quantity X.  In this case, we want to find out Sue's age??

Let x = Sue's age

Let x+6 = Leah's age

Let (x+6) + 5 = John's age

The ages of Sue, Leah and John add up to 41 years, therefore:

x + (x+6) + [(x+6) + 5] = 41

Combine Like Terms:

3x + 17 = 41

Subtract 17 from both sides to maintain the equality:

3x = 24

Divide both sides by 3:

x = 8 years old = Sue's age

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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Mental math approach (without solving equations):

Since Leah is 6 years older than Sue, and John is 6 years older than Leah, John is 11 years older than Sue. Taking 6 years away from Leah age, and 11 years away from John's age, then the three would have the same age as Sue. Therefore, Sue's age = (41-6-11)/3 = 8 years old

Mental math approach is quite often used in math contest such as in Mathcounts Countdown round.

I was in Mathcounts! I also trained a competitor a couple of years ago. What a great program.

Lynne R. | Tutoring in Math (General Math thru Algebra 11)Tutoring in Math (General Math thru Alge...
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Leah is 6 years older than Sue, John is 5 years older than Leah. Total combined age is 41. How old is Sue?

Leah’s Age + Sue’s Age + John’s Age = 41

We must then determine what to fill in for our unknowns. First, We must determine which is our ultimate unknown. There is usually one in which, if we knew this one unknown, we could most easily figure out the others using this number (or unknown). Start by really looking at your problem. Which age would enable you to figure out the other ages most easily. This will be X. (It’s also a good hint that in the given problem, both Sue and Johns Ages were referenced in terms of Leah’s Age, when looking for what is your X (ultimate unknown). It all become’s pretty simple once we have Leah’s Age, so Leah’s Age is X.

Leah’s Age = X

In this problem, if we knew Leah’s age...we know that Leah is six years older than Sue…we would subtract 6 years from Leah’s age to give us Sue’s age.

Sue’s Age =    Leah’s Age – 6    or    X - 6

We know John is 5 years older than Leah…we would add 5 years to Leah’s age to get John’s age.

John’s Age =   Leah’s Age + 5     or     X + 5

Now, back to the wordy problem….with substitutions…This will give us the equation you are looking for.

Wordy Problem

Leah’s Age + Sue’s Age + John’s Age = 41

New Substitutions

X      +      X – 6     +      X + 5     =     41

Combine Like Terms

3X – 1 = 41

Solve for X = 14

Remember solving for X is not necessarily solving the question to your problem. X is Leah’s Age. You’ve been asked to find Sue’s Age. Leah is 6 years older than Sue…Sue is 6 years younger than Leah (X – 6). Leah’s Age = X = 14…….Sue is X – 6 years old

or

14 – 6 = 8 years old

Uma L. | Math and Entrance Exam TutorMath and Entrance Exam Tutor
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Leah is 6 years older than Sue. John is 5 years older than Leah. Total combined age is 41. How old is Sue?

Leah's age = 6 years + Sue's age. John's age = 5 years + Leah's age. Combined age of Leah + Sue + John = 41 years

Now let us their ages as; Leah's age = L years, Sue's age = S years and John's age = J  years for simplicity (you could also assume, x,y, z or a,b,c or any three variables for that matter! but L, S and J would make most sense and easy to identify with their respective names)

L = 6 + S ,  J= 5 + L   and L+ S + J = 41

The question asked for Sue's age. Let us therefore solve for S

We already have L in terms of S (L= 6+S)

We need J in terms of S. We know that J = 5 + L = 5 + 6 +S (from the above expression)

J = 11 + S. Now we have even J in terms of S

so we can plug these values in the third equation

L+ S + J = 41

6 + S + S + 11 +S = 41

6 + 11 + S +S +S = 41

17 + 3S = 41

Subtract 17 from both sides

17 + 3S - 17 = 41-17

3S = 24

Divide both sides by 3

3S/3 = 24/3

S = 8  OR Sue's age = 8 years

Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)
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Let 'L' represent Leah's age, 'S' represent Sue's age, and 'J' represent John's age.

Since Leah is 6 years older than Sue, then we arrive at the following:   L = S + 6

And since John is 5 years older than Leah, we arrive at the following:   J = L + 5

With this, we are given that   S + L + J = 41

Since we are trying to solve for S (Sue's age), we want all parts of the equation to be in terms of S...

...to do that, we substitute S + 6 for L, and for J = L + 5 we also substitute S + 6 for the L

(i.e., J = L + 5 = ( S + 6 ) + 5 = S + ( 6 + 5 ) = S + 11

So,

S + L + J = 41          substitute S + 6 for L and S + 11 for J

S + ( S + 6 ) + ( S + 11 ) = 41

S + S + S + 6 + 11 = 41

3*S + 17 = 41               subtract 17 from both sides of the equation

3S + 17 - 17 = 41 - 17

3S = 24                        divide both sides of the equation by 3 to solve for S

3S/3 = 24/3      ===>   S = 8

Thus, Sue is 8 years old.

Ali M. | Friendly and High Quality TutoringFriendly and High Quality Tutoring
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Let x= age of sue then

x+6=age of Leah

And 5+(x+6) =x+11= age of John so the total age is x+(x+6)+(x+11)=41

So 3x+17=41so 3x=24 so x=age of sue= 8

Alyssa F. | Highly Experienced Elementary Tutor/TeacherHighly Experienced Elementary Tutor/Teac...
4.7 4.7 (192 lesson ratings) (192)
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Leah is 6 years older than Sue, John is 5 years older than Leah. Total combined age is 41. How old is Sue?

Sue's age (S) is Leah's age minus 6 (L - 6), so S = L - 6.

John's Age (J) is 5 + Lea's age (L), so J = L + 5

Leah (L) is Sue's age + 6, or L = S + 6

Since their total combined age is 41, then Sue's age + John's age + Leah's age = 41, or

L - 6 + L + 5 + S + 6 = 41. Combine the L's and the numbers to get 2L + 5 + S = 41. Since S = L - 6, you can substitute "L - 6" for "S" to eliminate a variable, which would give you 2L + 5 + L - 6 = 41. Combine the L's and numbers again and you get 3L - 1 = 41. Add 1 to both sides to clear the one on the left: 3L - 1 + 1 = 41 + 1, or 3L = 42. Divide both sides by 3 to find L: 3L/3 = 42/3. L = 14. CHECK - does 14 x 3 = 42? If it does, you're doing great. If not you made a mistake somewhere and should start by redoing the division, then working backwards from there, if necessary.

Since L = 14, then Leah is 14 years old. To find S and J, you simply substitute: S (Sue's age) = L - 6 = 14 - 6 = 8. J (John's age) = L + 5 = 14 + 5 = 19, so Leah is 14, Sue is 8, and John is 19. CHECK - add 14 + 8 + 19. It has to equal 41, their combined ages. 14 + 8 = 22. 22 + 19 = 41. You did it right! If those numbers do NOT add up to 41, double check your addition, then work backwards from there, if necessary.

Mike M. | An Engineer by Profession...A Tutor by PreferenceAn Engineer by Profession...A Tutor by P...
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First, assign variables to the age of each person. For example:

Leah = L

Sue = S

John = J

Next, write the given information in equation form:

Leah is 6 years older than Sue --> L=S+6

John is 5 years older than Leah --> J=L+5

Total combined age is 41 --> 41=L+S+J

Then, replace the variables L and J in the last equation with their corresponding equations:

41=(S+6)+S+(L+5)

To solve for S, we need to go one more step and replace the L in the equation above with its corresponding equation:

41=(S+6)+S+((S+6)+5)

Finally, we simplify the equation to get S on one side:

41=S+6+S+S+6+5 --> 41=3S+17 --> 24=3S --> 8=S  Sue is 8 years old.