# Solve Proportions using cross products

• Jan 6th 2013, 08:50 PM
BrokenRitual
Solve Proportions using cross products
My sister thinks these problems are unsolvable for x. I can't find any examples in my book. I know how to solve basic proportion problems, but these two are hard.

1. x : x-3 = x+4 : x

2. x+1 : 6 = x-1 : x

Thanks!
• Jan 6th 2013, 11:52 PM
chiro
Re: Solve Proportions using cross products
Hey BrokenRitual.

Try getting rid of the denominators and collect the terms.

For the first one we have x^2 + 4x =x^2 - 3x or 7x = 0 => x = 0 but this is a contradiction since x is on the denominator and can't be zero. So no solution exists.

You try the same kind of technique for the second one.
• Jan 7th 2013, 07:10 AM
Soroban
Re: Solve Proportions using cross products
Hello, BrokenRitual!

Why do you say these are "hard"?

Quote:

$\displaystyle [1]\;\;x : x-3 \:=\: x+4 : x$

We have: .$\displaystyle \frac{x}{x-3} \:=\:\frac{x+4}{x}$

Cross product: .$\displaystyle (x)(x) \:=\:(x-3)(x+4)$

n . . . . . . . . . . . . .$\displaystyle x^2 \:=\:x^2 + x - 12$

. . . . . . . . . . . . . . . $\displaystyle 0 \:=\:x - 12$

. . . . . . . . . . . . . . . $\displaystyle x \:=\:12$

Quote:

$\displaystyle [2]\;\; x+1 : 6 \:=\: x-1 : x$

We have: .$\displaystyle \frac{x+1}{6} \:=\:\frac{x-1}{x}$

Cross product: .$\displaystyle (x+1)(x) \:=\:6(x-1)$

n . . . . . . . . . . . . . $\displaystyle x^2 + x \:=\:6x-6$

. . . . . . . . . . .$\displaystyle x^2 - 5x + 6 \:=\:0$

. . . . . . . . .$\displaystyle (x-2)(x-3) \:=\:0$

. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:2,\:3$