Question

PLS help me with thisss:((((((​

Answer 1

Answer:

C and B

Step-by-step explanation:

(6)

$\frac{3x-7y}{8}$ × $\frac{6}{3x-7y}$ ← cancel 3x - 7y on numerator and denominator

= $\frac{1}{8}$ × $\frac{6}{1}$ = $\frac{6}{8}$ = $\frac{3}{4}$ → C

(7)

$\frac{6x}{x^2-9}$ ÷ $\frac{8}{4x-12}$

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

$\frac{6x}{(x-3)(x+3)}$ ÷ $\frac{8}{4(x-3)}$ ← cancel 8 and 4 by 4

= $\frac{6x}{(x-3)(x+3)}$ ÷ $\frac{2}{x-3}$

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

= $\frac{6x}{(x-3)(x+3)}$ × $\frac{x-3}{2}$ ← cancel x - 3 on numerator and denominator

= $\frac{6x}{x+3}$ × $\frac{1}{2}$ ← cancel 2 and 6 on numerator and denominator

= $\frac{3x}{x+3}$ → B