Question

Simplify quantity 18 x minus 6 over 9 x to the fifth power all over quantity 15 x plus 5 over 21 x squared

Answer 1

@jdoe0001

Answer 2

\(\bf \cfrac{\frac{18x-6}{9x^5}}{\frac{15x+5}{21x^2}}\qquad recall\implies {\color{blue}{ \cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}}}\\ \quad \\
\cfrac{18x-6}{9x^5}\cdot \cfrac{21x^2}{15x+5}\implies \cfrac{6(3x-1)}{9x^5}\cdot \cfrac{21x^2}{5(3x+1)}\\ \quad \\
\implies \cfrac{3\cdot 2(3x-1)}{3\cdot 3x^2\cdot x^3}\cdot \cfrac{3\cdot 7x^2}{5(3x+1)}\)

Answer 3

14x^2 / 15x + 1?

Answer 4

well \(\bf \cfrac{3\cdot 2(3x-1)}{3\cdot 3x^2\cdot x^3}\cdot \cfrac{3\cdot 7x^2}{5(3x+1)}\implies \cfrac{2(3x-1)\cdot 7}{x^3\cdot 5(3x+1)}\\ \quad \\
\implies \cfrac{42x-14}{15x^4+5x^3}\)

Answer 5

Would that make the answer \[14(3x - 1) \over 5x^3 (3x +1)\]
?

Answer 6

without distributing, yes

Answer 7

Does it have to be distributed?

Answer 8

no, either way will do

Answer 9

What would be the distributed form?

Answer 10

ohh ahem well, is the one I posted earlier \(\bf \cfrac{3\cdot 2(3x-1)}{3\cdot 3x^2\cdot x^3}\cdot \cfrac{3\cdot 7x^2}{5(3x+1)}\implies \cfrac{2(3x-1)\cdot 7}{x^3\cdot 5(3x+1)}\\ \quad \\
\implies \cfrac{42x-14}{15x^4+5x^3}\)