proportions - heavy algebra needed

**zachd77** · December 28th 2012, 02:54 PM

Found this problem looking in my old textbook:

If

proportions - heavy algebra needed-codecogseqn-10.gif

, find the numerical value of the ratio

proportions - heavy algebra needed-codecogseqn-11.gif

.

I started cross multiplying & I got stuck quite easily. How do I finish this?

**abender** · December 28th 2012, 03:05 PM

Use a common denominator.

**zachd77** · December 28th 2012, 03:24 PM

I agree with your advice, but would I then solve for a/b?

**abender** · December 28th 2012, 03:59 PM

$\frac{4a-9b}{4a} = \frac{a-2b}{b}$

$\frac{4a-9b}{4a}\cdot \frac{b}{b} = \frac{a-2b}{b} \cdot \frac{4a}{4a}$

$\frac{4ab-9b^2}{4ab} = \frac{4a^2-8ab}{4ab}$

$\frac{4a^2-8ab}{4ab} - \frac{4ab-9b^2}{4ab} = 0$

$\frac{4a^2-8ab-4ab+9b^2}{4ab}=0$

$\frac{4a^2-12ab+9b^2}{4ab}=0$

$4a^2-12ab+9b^2=0$

$(2a-3b)(2a-3b)=0$

$2a=3b\neq0$

$a=1.5b$

$\frac{a}{b} = \frac{1.5b}{b} = 1.5$

**zachd77** · December 28th 2012, 04:45 PM

thanks I remember attempting this problem a few years ago, didn't turn out well.

**ibdutt** · December 28th 2012, 09:26 PM

Alternatively:
(4a-9b)/4a= (a-2b)/b Just cross multiply and we get
4ab – 9b^2 = 4a^2 – 8 ab
Or 4a^2 – 12 ab + 9 b^2 = 0
( 2a )^2 – 2 ( 2a) ( 3b) + ( 3b)^2 = 0
( 2a – 3b)^2 = 0
That gives a = 1.5 b

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