Question

How do you solve `2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2-7f)`?

Answer 1

`46/27`

Explanation:

expand all brackets:

`2f(5f-2) = 10f^2-4f`

`-10(f^2-3f+6) = -10f^2+30f-60`

`-8f(f+4) = -8f^2-32`

`4(2f^2-7f) = 8f^2-28f`

input expanded brackets into equation:

`10f^2-4f-10f^2+30f-60 = -8f^2-32+8f^2-28f`

collect like terms on either side:

`10f^2 - 10f^2 - 4f + 30f - 60 = -8f^2+8f^2-28f+32`

cancel out terms that add to `0`:

`cancel(10f^2) cancel(- 10f^2) - 4f + 30f - 60 = cancel(-8f^2)cancel(+8f^2)-28f+32`

add like terms:

`26f - 60 = -28f + 32`

add `28f:`

`54f - 60 = 32`

add `60:`

`54f = 92`

divide by `54:`

`f = 92/54 = 46/27`