How would I solve this?
"The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle."
Let l denote the length of the rectangle and w the width. The you know:
$\displaystyle 2l + 2w = 40$
$\displaystyle 2\cdot (2l)+2\cdot \left(\frac12w\right) = 40+16$
That means you have to solve for l and w:
$\displaystyle l+w=20$
$\displaystyle 4l+w=56$
I've got l = 12 and w = 8
$\displaystyle P=2l+2w, $where l = length and w = width
$\displaystyle 40cm=2l+2w$
and
$\displaystyle 56cm=2*2l+\frac{2w}{2}$
$\displaystyle 56cm=4l+w$
$\displaystyle 40cm-2w=2l$
$\displaystyle 20cm-w=l$
$\displaystyle 56cm=4(20cm-w)+w$
$\displaystyle 56cm=80cm-4w+w$
$\displaystyle -24cm=-3w$
$\displaystyle w=8cm$
now plug $\displaystyle w=8cm$ into the original equation
$\displaystyle 40cm=2(8cm)+2w$
$\displaystyle 24cm=2w$
$\displaystyle w=12cm$