Confused about how to solve simple problem...?

**EyesThatSparkle02** · September 14th 2009, 07:55 PM

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?

**Prove It** · September 14th 2009, 07:59 PM

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

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

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

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

$d = 15$ .

**I-Think** · September 14th 2009, 08:00 PM

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

**EyesThatSparkle02** · September 14th 2009, 08:17 PM

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?

**Prove It** · September 14th 2009, 11:37 PM

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.

Thread: Confused about how to solve simple problem...?

LinkBack

Thread Tools

Display

Confused about how to solve simple problem...?

Similar Math Help Forum Discussions

Confused about simple u substition problem for natural log

Simple slope problem, confused about wording

Simple math problem need help to solve

Confused, has to be something simple.

can't solve this simple problem (been trying forever)

Search Tags