What is (9p^2-1)/(pq-2q)*(3p-6)/(1-3p) ?

The solution to (9p^2-1)/(pq-2q)*(3p-6)/(1-3p) is -(3(3p+1))/q

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simplify (9p^2-1)/(pq-2q)*(3p-6)/(1-3p)

simplify

\frac{9 p ^{2} - 1}{pq - 2 q} \cdot \frac{3 p - 6}{1 - 3 p}

- \frac{3 ( 3 p + 1 )}{q}

Solution steps

\frac{9 p ^{2} - 1}{pq - 2 q} \cdot \frac{3 p - 6}{1 - 3 p}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( 9 p ^{2} - 1 ) ( 3 p - 6 )}{( pq - 2 q ) ( 1 - 3 p )}

Factor

9 p^{2} - 1 : (3 p + 1) (3 p - 1)

= \frac{( 3 p + 1 ) ( 3 p - 1 ) ( 3 p - 6 )}{( pq - 2 q ) ( 1 - 3 p )}

(pq - 2 q) (1 - 3 p) = - (pq - 2 q) (3 p - 1)

= \frac{( 3 p + 1 ) ( 3 p - 1 ) ( 3 p - 6 )}{- ( pq - 2 q ) ( 3 p - 1 )}

Cancel the common factor:

3 p - 1

= \frac{( 3 p + 1 ) ( 3 p - 6 )}{- ( pq - 2 q )}

Factor

3 p - 6 : 3 (p - 2)

= \frac{( 3 p + 1 ) \cdot 3 ( p - 2 )}{- ( pq - 2 q )}

Factor out common term

q

= \frac{( 3 p + 1 ) \cdot 3 ( p - 2 )}{- q ( p - 2 )}

Cancel the common factor:

p - 2

= \frac{( 3 p + 1 ) \cdot 3}{- q}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{( 3 p + 1 ) \cdot 3}{q}

= - \frac{3 ( 3 p + 1 )}{q}

3 x^{2} + 12 x - 135

- 5 \cdot 20

\frac{c}{4} + 10 = 22

1. 3^{- 5}

x^{2} - 10 x - 9 = 0

What is (9p^2-1)/(pq-2q)*(3p-6)/(1-3p) ?
The solution to (9p^2-1)/(pq-2q)*(3p-6)/(1-3p) is -(3(3p+1))/q