Question

The length of a rectangle is twice its width. if the width is tripled and the length is decreased by 5, the area is increased by 4, find the width. D: help--

Answer 1

This one can besolved using system of equations. From the first statement we can assume that length = 2 * width. The area of rectange equals product of it's width and length. By knowing that and following to the second statement we get the second equation: (3 * width) * (length - 5) = width * length + 4. Let's mark the width as x and the length as y and write down our equations system:
$\begin{cases} y=2x \\ 3x\cdot (y-5)=xy+4 \end{cases}$

Using substituion method, lets replace every y with 2x within second equation and solve it:
$3x\cdot (2x-5)=x\cdot2x+4$
$6x^2-15x=2x^2+4$
$6x^2-15x-2x^2-4=0$
$4x^2-15x-4=0$

Now lets find the discriminant in order to solve this quadratic equation:
$D=b^2-4ac=15^2-4\cdot4\cdot(-4)=225+64=289$
$x_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{15+\sqrt{289}}{2\cdot4}=\frac{15+17}{8}=\frac{32}{8}=4$
$x_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{15-\sqrt{289}}{2\cdot4}=\frac{15-17}{8}=\frac{-2}{8}=-0.25$

The second root is negative, so we ignore it as x represents width which can't be negative.
Now, using found root, let's find y value from the first equation:
$y=2x=2\cdot4=8$

So, the width is equal 4 (and the length is equal 8).