The distance between two cities is 10km more than one third of the distance between them?
I understand the answer is 15 km - but how do i put the answer in an algebraic equation?
Let the distance between the two cities be X
X is 10km more than a third the distance between them
Hint: When they say 1/3 distance between them, they're referring to X
Algebraically
$\displaystyle X=10+\frac{1}{3}X$
$\displaystyle \frac{2}{3}X=10$
$\displaystyle X=15$
Edit: Beaten by Prove It, I'm slow tonight
Let $\displaystyle P_1 = x$ be the perimeter of one of the squares (length of one of the pieces of wire).
Its side length is therefore $\displaystyle L_1 = \frac{1}{4}x$.
Thus its area is $\displaystyle A_1 = L_1^2$
$\displaystyle = \left(\frac{1}{4}x\right)^2$
$\displaystyle = \frac{1}{16}x^2$.
Let $\displaystyle P_2 = 5 - x$ be the perimeter of the other piece.
Clearly $\displaystyle P_1 + P_2 = 5$.
Its side length is therefore $\displaystyle L_2 = \frac{1}{4}(5 - x)$.
Thus its area is $\displaystyle A_2 = \left[\frac{1}{4}(5 - x)\right]^2$
$\displaystyle = \frac{1}{16}(5 - x)^2$
$\displaystyle = \frac{1}{16}(25 - 10x - x^2)$
$\displaystyle = -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16}$.
We also know that $\displaystyle A_2 = 4A_1$
Therefore
$\displaystyle -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16} = \frac{1}{4}x^2$
$\displaystyle 0 = \frac{5}{16}x^2 + \frac{5}{8}x - \frac{25}{16}$
$\displaystyle 0 = x^2 + 2x - 5$
$\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2(1)}$
$\displaystyle = \frac{-2 \pm \sqrt{24}}{2}$
$\displaystyle = \frac{-2 \pm 2\sqrt{6}}{2}$
$\displaystyle = -1 \pm \sqrt{6}$.
Since you can not have a negative length
$\displaystyle x = -1 + \sqrt{6}$.
Remembering that the other length $\displaystyle = 5 - x$
$\displaystyle = 5 - (-1 + \sqrt{6})$
$\displaystyle = 5 + 1 - \sqrt{6}$
$\displaystyle = 6 - \sqrt{6}$.
Put this into the calculator and you will know how many cm this length is.