Originally Posted by

**JeffM** Let's start with two really basic ideas in algebra.

a + b = c + d means that two unknown numbers have a sum equal to the sum of another two unknown numbers. In other words, the two sides of the equation represent **the same number.**

Now if I multiply (a + b) by e and e is any number except 1, the number represented by e(a + b) is different from a + b, which is the same as c + d so it is absolutely **FALSE** that e(a + b) = c + d. To maintain equality, I must multiply both sides of the equation by e.

In short $a + b = c + d \implies e(a + b) = e(c + d).$ First week of algebra. As I said, truly basic stuff.

Now fractions are a nuisance, and it is frequently useful to get rid of them as quickly as possible. To do that we can use this fundamental propert of numbers

$b \ne 0 \implies b * \dfrac{a}{b} = a.$

We actually learn that in arithmetic: $3 * \dfrac{2}{3} = \dfrac{\cancel 3}{1} * \dfrac{2}{\cancel 3} = 2.$

So the way to get rid of a fraction in an equation is to multiply both sides of the equation by the **denominator** of that annoying fraction.

Let's go.

$\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4}.$

To get rid of the fraction $\dfrac{17}{4},$

I multiply **BOTH SIDES** of the equation by the relevant fraction's denominator, which is **4**.

$4 * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 4 * \left ( 7y - \dfrac{17}{4} \right ) \implies$

$\dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - \dfrac{\cancel 4 * 17}{\cancel 4} \implies \dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - 17.$

All the fractions are gone from the right hand side of the equation. Now let's get rid of (36 / 8). The **denominator** is 8 so we multiply **both sides** of the equation by **8**.

$8 * \left (\dfrac{36y}{10q - 8} - \dfrac{36}{8} \right ) = 8 * (28y - 17) \implies$

$\dfrac{288y}{10q - 8} - \dfrac{\cancel 8 * 36}{\cancel 8} = 224y - 136 \implies \dfrac{288y}{10q - 8} - 36 = 224y - 136.$

We have one fraction left, namely $\dfrac{288y}{10q - 8}$, which has a denominator of $10q - 8.$

So we multiply both sides of the equation by that denominator just as we did when the denominators were expressed as numerals.

$(10q - 8) * \left ( \dfrac{288y}{10q - 8} - 36 \right ) = (10q - 8) * (224y - 136) \implies$

$\dfrac{\cancel {(10q - 8)} * 288y}{\cancel {10q - 8}} - 36(10q - 8) = 224y(10q - 8) - 136(10q - 8) \implies$

$288y - 360q + 288 = 2240qy - 1792y - 1360q + 1088.$

Now you have no fractions to deal with. But you complain that equation is not what your explanation gave. It turns out, however, that we can factor both sides of the equation by 8 and divide both sides by 8 as follows.

$8 * (36y - 45q + 36) = 8(280qy - 224y - 170q + 136) \implies$

$\dfrac{\cancel 8 * (36y - 45q + 36)}{\cancel 8} = \dfrac{\cancel 8(280qy - 224y - 170q + 136) }{\cancel 8} \implies$

$36y - 45q + 36 = 280qy - 224y - 170q + 136.$

That is very tedious. A somewhat less tedious way is to find the least common multiple of all the fractions in the equation and multiply by that least common multiple. That gets rid of all the fractions in one go.

Now obviously the least common multiple of 4 and 8 is 8. And the least common multiple of 8 and (10q - 8) is their product of 8(10q - 8).

$\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4} \implies$

$8(10q - 8) * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 8(10q * 8) * \left ( 7y - \dfrac{17}{4} \right ) \implies$

$\dfrac{8 \cancel {(10q - 8)} * 9y}{\cancel {10q - 8}} - \dfrac{\cancel 8(10q - 8)9}{\cancel 8} = 8(10q - 8)(7y) - \dfrac{8(10q - 8)(17)}{4} \implies$

$72y - 9(10q - 8) = 56y(10q - 8) - \dfrac{\cancel 4 * 2 * (10q - 8) * 17}{\cancel 4} \implies$

$72y - 90q + 72 = 560qy - 448y - 34(10q - 8) \implies$

$72y - 90q + 72 = 560qy - 448y - 340q + 272.$

It is obvious that both sides of the equation can be divided by 2 because every numeral is even giving

$36y - 45q + 36 = 280qy - 224y - 170q + 186.$

And we reach the same result.

Sorry to take baby steps all the way, but skipping steps sometimes leads to confusion.