# Confused about how to solve simple problem...?

• Sep 14th 2009, 08:55 PM
EyesThatSparkle02
Confused about how to solve simple problem...?
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?
• Sep 14th 2009, 08:59 PM
Prove It
Quote:

Originally Posted by EyesThatSparkle02
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?

If $d$ is the distance between the two cities then

$d = 10 + \frac{1}{3}d$

$\frac{2}{3}d = 10$

$d = \frac{3}{2}\cdot 10$

$d = 15$.
• Sep 14th 2009, 09:00 PM
I-Think
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
$X=10+\frac{1}{3}X$
$\frac{2}{3}X=10$
$X=15$

Edit: Beaten by Prove It, I'm slow tonight
• Sep 14th 2009, 09:17 PM
EyesThatSparkle02
Thanks guys,

New Question: (Similar)
If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?
• Sep 15th 2009, 12:37 AM
Prove It
Quote:

Originally Posted by EyesThatSparkle02
Thanks guys,

New Question: (Similar)
If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?

Let $P_1 = x$ be the perimeter of one of the squares (length of one of the pieces of wire).

Its side length is therefore $L_1 = \frac{1}{4}x$.

Thus its area is $A_1 = L_1^2$

$= \left(\frac{1}{4}x\right)^2$

$= \frac{1}{16}x^2$.

Let $P_2 = 5 - x$ be the perimeter of the other piece.

Clearly $P_1 + P_2 = 5$.

Its side length is therefore $L_2 = \frac{1}{4}(5 - x)$.

Thus its area is $A_2 = \left[\frac{1}{4}(5 - x)\right]^2$

$= \frac{1}{16}(5 - x)^2$

$= \frac{1}{16}(25 - 10x - x^2)$

$= -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16}$.

We also know that $A_2 = 4A_1$

Therefore

$-\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16} = \frac{1}{4}x^2$

$0 = \frac{5}{16}x^2 + \frac{5}{8}x - \frac{25}{16}$

$0 = x^2 + 2x - 5$

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2(1)}$

$= \frac{-2 \pm \sqrt{24}}{2}$

$= \frac{-2 \pm 2\sqrt{6}}{2}$

$= -1 \pm \sqrt{6}$.

Since you can not have a negative length

$x = -1 + \sqrt{6}$.

Remembering that the other length $= 5 - x$

$= 5 - (-1 + \sqrt{6})$

$= 5 + 1 - \sqrt{6}$

$= 6 - \sqrt{6}$.

Put this into the calculator and you will know how many cm this length is.