Question 229739
<hr />FIRST, we need variables.  Once we define variables, it's much easier to turn this word problem into a math problem.


Let a = first side
Let b = second side
Let c = third side


NOW, we can turn the words into equations:


"a triangle has a perimeter of 165 cm."
a + b + c = 165


"the first side is 65cm less than twice the second side."
a = 2b - 65


"the third side is 10cm less than the second side."
c = b - 10


Before we finish, I have to ask:  Who writes problems like this???  Pointless problems like these are why kids don't like math!  Ugh.  Drives me crazy.  It's a shame, because solving math problems really does have a certain satisfaction, once you learn how.  [Okay.  I'm done.  Back to the problem.]


If we could get this to have only one variable, we could solve it.  Substitution to the rescue!
a + b + c = 165 (rewrote equation from above)
(2b - 65) + b + (b - 10) = 165 (substituted "a" and "c" from equations above)


See how that works?  Let's solve it.
2b - 65 + b + b - 10 = 165 (dropped the parentheses, because there was nothing to distribute, not even a minus sign)
4b - 75 = 165 (combined like terms)
4b = 240 (added 75 to both sides)
b = 60 (divided both sides by 4)


We found the second side!  We can find the first and third sides using those equations from above:
a = 2b - 65
a = 2*60 - 65
a = 120 - 65
a = 55


c = b - 10
c = 60 - 10
c = 50


All done.


<hr />For online tutoring, or worked problems by e-mail, please visit <a href="http://www.mathlocker.com"><br />Math Locker.com<img src=" http://www.mathlocker.com/images/math_locker_logo_w_bg.png" /></a><hr /><br />