Question

The lenght of a rectangle is 3 times the width. The lenght is decreased by 2 cm, and the width is increased by 3cm. The area of the new rectangle is 70cm^2 find the dimensions of the original rectangle

Answer 1

The two equations you need to solve are:

\[l=3w\]
\[(l-2)(w-3)=70\]

Answer 2

oops, i meant (l-2)(w+3)=70

Answer 3

the choices are 4*12,6*12,5*11,4*16

Answer 4

L = 3w
(L-2)(w+3)=70
(3w-2)(w+3)=70
3w^2+7w-6=70
3w^2+7w-6-70=0
3w^2+7w-76=0
(w-4) (3w+19)=0
w-4=0 or 3w+19=0
w=4 or w=-19/3

Throw out the negative solution to get w = 4

So the width is 4 and the length is L = 3w = 3*4 = 12

So the dimensions are 4 by 12

Answer 5

Solve the following simultaneous equations for L and W\[{L = 3 W, (L - 2) (3 W + 3) = 70}\]There are two solutions for L and W. The solution of interest is:\[\left\{L\to \frac{1}{2} \left(-1+\sqrt{305}\right),W\to \frac{1}{6} \left(-1+\sqrt{305}\right)\right\}=\{L\to 8.232,W\to 2.744\} \]\[(L-2)(3W+3)=\left(-2+\frac{1}{2} \left(-1+\sqrt{305}\right)\right) \left(3+\frac{1}{2} \left(-1+\sqrt{305}\right)\right)=70 \]\[\frac{L}{W}\text{=}\frac{\frac{1}{2} \left(-1+\sqrt{305}\right)}{\frac{1}{6} \left(-1+\sqrt{305}\right)}=\frac{1}{2}*\frac{6}{1}=3 \]

Answer 6

It says that the width is increased by 3 cm, not that 3 times the width is increased by 3cm.

Answer 7

Sorry.
Blew two out of two word problems tonight.

Pre dimension changes;
W=W
L =3W
Post dimension changes:
W=W+3
L =3W-2
product of width and length is the area
(W+3)(3W-2)=70
3 W^2+7 W-76=0
And so forth . . . .