SOLUTION: Richard is 5 years older than Sheila. Three more than four times sheila’s age is 2 less than 3 times Richard’s age. How old is each person?
Algebra ->
Customizable Word Problem Solvers
-> Age
-> SOLUTION: Richard is 5 years older than Sheila. Three more than four times sheila’s age is 2 less than 3 times Richard’s age. How old is each person?
Log On
Question 1107200: Richard is 5 years older than Sheila. Three more than four times sheila’s age is 2 less than 3 times Richard’s age. How old is each person? Found 3 solutions by ikleyn, stanbon, josmiceli:Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
Richard is 5 years older than Sheila. Three more than four times sheila’s age is 2 less than 3 times Richard’s age. How old is each person?
~~~~~~~~~~~~~~~~~~~~
R = S + 5, (1)
4S + 3 = 3R - 2. (2)
Sub (1) into (2) to get
4S + 3 = 3*(S+5) - 2 ====>
4S + 3 = 3S + 15 - 2 ====> S = 15 - 2 - 3 = 10.
Answer. Sheila is 10, Richard is S + 5 = 10+5 = 15 years old.
``````````````````````````````````````````````````
I edited my error/typo in the previous version.
You can put this solution on YOUR website! Richard is 5 years older than Sheila. Three more than four times sheila’s age is 2 less than 3 times Richard’s age. How old is each person?
------
Equations:
R = S + 5
4S + 3 = 3R-2
-------
Substitute for "R" and solve for "S":
4S+3 = 3(S+5)-2
4S+3=3S + 13
S = 10 yrs
----
R = S + 5 = 15 yrs.
----------
Cheers,
Stan H.
----------
You can put this solution on YOUR website! Let = Richard's age now
Let = Sheila's age now
-------------------------------
(1)
(2)
-------------------------------
(2)
and
(1)
Multiply both sides of (1) by
and subtract (2) from (1)
(1)
(2)
--------------------------
and
(1)
(1)
(1)
----------------------
15 = Richard's age now
10 = Sheila's age now
----------------------------
check:
(1)
(1)
and
(2)
(2)
(2)
OK
(