SOLUTION: originally a rectangle was three times as long as it was wide. when 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an are of 90c

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: originally a rectangle was three times as long as it was wide. when 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an are of 90c      Log On


   

Question 1153710: originally a rectangle was three times as long as it was wide. when 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an are of 90cm^2. what the dimensions of the new rectangle
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
This method may seem slightly backward, but:
x, the original width
3x, the original length

%283x-2%29%28x%2B5%29=90
Solve any way you wish. Question asks for each of the two factors.
skipping steps,...
highlight_green%28x=4%29


3%2A4-2=highlight%2810%29-----length of the new rectangle
-
4%2B5=highlight%289%29------width of new rectangle

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

The original dimensions x cm wide and 3x cn long.


After changing it is (x+5) cm wide and (3x-2) cm long.


The equation


    (x+5)*(3x-2) = 90


You can solve it as a quadratic equation.


I will solve it mentally.


Multiply both sides by 3


    (3x+15)*(3x-2) = 270.


Find two integer factors of the number 270 that differ by 15-(-2) = 17.


Obviously, these factors are 27 and 10.


So, 3x-2 = 10;  3x = 10+2 = 12;  x = 12/3 = 4.


ANSWER:  The dimensions of the original rectangle are 4 cm (width) and 3*4 = 12 cm (length).