Hello, alaricepent!

Make a sketch, then baby-talk your way through it . . .

Suppose the length of a rectangle is twice the width.

If we increase both the width and length of the rectangle by 2 units,

the area is increased by 34 square units.

Find the width of the rectangle.

. . $\displaystyle (a)\;3 \qquad (b)\;8 \qquad (c)\;10 \qquad (d)\;5 \qquad(e)\text{ none of these}$ Code:

2x
*-------------*
| |
x | | x
| |
*-------------*
2x

Let $\displaystyle x$ = width

Then $\displaystyle 2x$ = length.

. . The area is: .$\displaystyle A_1 \:=\:(2x)(x) \:=\:2x^2$

Code:

2x+2
*-------------*---*
| : |
| : |
x+2 | : | x+2
* - - - - - - * |
| |
*-----------------*
2x+2

Now $\displaystyle x+2$ = width

and $\displaystyle 2x+2$ = length

. . The area is: .$\displaystyle A_2 \;=\;(x+2)(2x+2) \:=\:2x^2+6x+4$

The new area is 34 more than the original area:

. . $\displaystyle 2x^2 + 6x + 4 \;=\;2x^2 + 34$

So we have: .$\displaystyle 6x \:=\:30 \quad\Rightarrow\quad x \:=\:5$

The width is 5 . . . answer (d)