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:
[tex]\begin{cases} y=2x \\ 3x\cdot (y-5)=xy+4 \end{cases}[/tex]
Using substituion method, lets replace every y with 2x within second equation and solve it:
[tex]3x\cdot (2x-5)=x\cdot2x+4[/tex]
[tex]6x^2-15x=2x^2+4[/tex]
[tex]6x^2-15x-2x^2-4=0[/tex]
[tex]4x^2-15x-4=0[/tex]
Now lets find the discriminant in order to solve this quadratic equation:
[tex]D=b^2-4ac=15^2-4\cdot4\cdot(-4)=225+64=289[/tex]
[tex]x_{1}=\frac{-b+\sqrt{D}}{2a}=\frac{15+\sqrt{289}}{2\cdot4}=\frac{15+17}{8}=\frac{32}{8}=4[/tex]
[tex]x_{2}=\frac{-b-\sqrt{D}}{2a}=\frac{15-\sqrt{289}}{2\cdot4}=\frac{15-17}{8}=\frac{-2}{8}=-0.25[/tex]
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:
[tex]y=2x=2\cdot4=8[/tex]
So, the width is equal 4 (and the length is equal 8).
