Answer:
C and B
Step-by-step explanation:
(6)
[tex]\frac{3x-7y}{8}[/tex] × [tex]\frac{6}{3x-7y}[/tex] ← cancel 3x - 7y on numerator and denominator
= [tex]\frac{1}{8}[/tex] × [tex]\frac{6}{1}[/tex] = [tex]\frac{6}{8}[/tex] = [tex]\frac{3}{4}[/tex] → C
(7)
[tex]\frac{6x}{x^2-9}[/tex] ÷ [tex]\frac{8}{4x-12}[/tex]
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
[tex]\frac{6x}{(x-3)(x+3)}[/tex] ÷ [tex]\frac{8}{4(x-3)}[/tex] ← cancel 8 and 4 by 4
= [tex]\frac{6x}{(x-3)(x+3)}[/tex] ÷ [tex]\frac{2}{x-3}[/tex]
• leave first fraction, change ÷ to × , turn second fraction ' upside down'
= [tex]\frac{6x}{(x-3)(x+3)}[/tex] × [tex]\frac{x-3}{2}[/tex] ← cancel x - 3 on numerator and denominator
= [tex]\frac{6x}{x+3}[/tex] × [tex]\frac{1}{2}[/tex] ← cancel 2 and 6 on numerator and denominator
= [tex]\frac{3x}{x+3}[/tex] → B
