Question

6 x + 9 / 15 x^2 all over 16 x - 12 / 10 x^4

Answer 1

2x+3/2(4x-3) correct?

Answer 2

You are missing a component on your numerator

Answer 3

x squared in front of the 2x + 3 right? x^2(2x+3)?

Answer 4

if you look at your original equation, you have x^4 components and x^2
so you can't have a solution that has the same coefficients on top and bottom

Answer 5

if you didn't have an x^4 in your equation, and instead had an x^2; then your solution would be correct :)

Answer 6

so yes?

Answer 7

You have
\[\left( \frac{ 6x+9 }{ 15x ^{2} } \right) \div \left( \frac{ 16x-12 }{ 10x ^{4} } \right)\]

which can be rewritten as
\[\frac{ (6x+9)*(10x ^{4}) }{ (16x-12)*(15x ^{2}) }\]

if you expand this out and factorise you get;
\[\frac{ 30x ^{4}(2x+3) }{ 30x ^{2}(8x-6) }\]

Cancel out, you get;
\[\frac{ x ^{2}(2x+3) }{ (8x-6) }\]

Do you understand?