Question

PLS help me with thisss:((((((​

Answer 1

C and B

Explanation:

(6)

`(3x-7y)/(8)` ×
`(6)/(3x-7y)` ← cancel 3x - 7y on numerator and denominator

=
`(1)/(8)` ×
`(6)/(1)` =
`(6)/(8)` =
`(3)/(4)` → C

(7)

`(6x)/(x^2-9)` ÷
`(8)/(4x-12)`

Factorise the denominators of both fractions

x² - 9 = x² - 3² = (x - 3)(x + 3) ← difference of squares

4x - 12 ← factor out 4 from each term

= 4(x - 3)

Then rewrite as

`(6x)/((x-3)(x+3))` ÷
`(8)/(4(x-3))` ← cancel 8 and 4 by 4

=
`(6x)/((x-3)(x+3))` ÷
`(2)/(x-3)`

• leave first fraction, change ÷ to × , turn second fraction ' upside down'

=
`(6x)/((x-3)(x+3))` ×
`(x-3)/(2)` ← cancel x - 3 on numerator and denominator

=
`(6x)/(x+3)` ×
`(1)/(2)` ← cancel 2 and 6 on numerator and denominator

=
`(3x)/(x+3)` → B