Question

How do you divide `\frac { 5a } { 12b c ^ { 2} } \div \frac { 15a ^ { 2} } { 18b ^ { 2} c ^ { 2} }`?

Answer 1

`b/(2a)`

Explanation:

`(5a)/(12bc^2)÷(15a^2)/(18b^2c^2)`

`=(5a)/(12bc^2 )× (18b^2c^2)/(15a^2)`

`=(cancel(5a))/(12cancel(b) cancel (c^2))× (18cancel (b^2) color(red)(b) cancel(c^2))/(cancel(15a^2) color(red)(3a))`

`=(18b)/(12×3a)`

`=b/(2a)`

Answer 2

The result is `b/(2a)`.

Explanation:

Use this rule when dividing fractions:

`color(red)a/color(blue)bdivcolor(green)c/color(purple)dqquad=>qquadcolor(red)a/color(blue)bxxcolor(purple)d/color(green)c`

And use this rule to combine them:

`color(red)a/color(blue)bxxcolor(purple)d/color(green)cqquad=>qquad(color(red)axxcolor(purple)d)/(color(blue)bxxcolor(green)c`

Now here's our problem:

`color(white)=color(red)(5a)/color(blue)(12bc^2)divcolor(green)(15a^2)/color(purple)(18b^2c^2)`

`=color(red)(5a)/color(blue)(12bc^2)xxcolor(purple)(18b^2c^2)/color(green)(15a^2)`

`=(color(red)(5a)xxcolor(purple)(18b^2c^2))/(color(blue)(12bc^2)xxcolor(green)(15a^2))`

`=(color(purple)(color(magenta)90color(red)ab^2c^2))/(color(blue)(12bc^2)xxcolor(green)(15a^2))`

`=(color(purple)(color(magenta)90color(red)ab^2c^2))/color(blue)(color(turquoise)180color(green)(a^2)bc^2)`

`=(color(purple)(color(red)cancelcolor(magenta)90^color(black)1color(red)ab^2c^2))/color(blue)(color(red)cancelcolor(turquoise)180^color(black)2color(green)(a^2)bc^2)`

`=(color(purple)(color(red)ab^2c^2))/color(blue)(color(black)2color(green)(a^2)bc^2)`

`=(color(purple)(color(red)ab^2color(red)cancelcolor(blue)(c^2)))/color(blue)(color(black)2color(green)(a^2)bcolor(red)cancelcolor(blue)(c^2))`

`=(color(purple)(color(red)ab^2))/color(blue)(color(black)2color(green)(a^2)b)`

`=(color(purple)(color(red)ab^color(red)cancelcolor(purple)2))/color(blue)(color(black)2color(green)(a^2)color(red)cancelcolor(blue)b)`

`=(color(purple)(color(red)ab))/color(blue)(color(black)2color(green)(a^2))`

`=(color(purple)(color(red)color(black)cancelcolor(red)ab))/color(blue)(color(black)2color(green)(a^color(red)cancelcolor(green)2))`

`=color(purple)b/(color(black)2color(green)a)`

This is the simplified result. Hope this helped!