Question

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Consider the functions given below. VIEW FILE ATTACHED

Answer 1

Answer:  see below

Step-by-step explanation:

$P(x)=\dfrac{2}{3x-1}\qquad \qquad Q(x)=\dfrac{6}{-3x+2}$

P(x) ÷ Q(x)

$\dfrac{2}{3x-1}\div \dfrac{6}{-3x+2}\\\\\\=\dfrac{2}{3x-1}\times \dfrac{-3x+2}{6}\\\\\\=\large\boxed{\dfrac{-3x+2}{3(3x-1)}}$

P(x) + Q(x)

$\dfrac{2}{3x-1}+ \dfrac{6}{-3x+2}\\\\\\=\dfrac{2}{3x-1}\bigg(\dfrac{-3x+2}{-3x+2}\bigg)+ \dfrac{6}{-3x+2}\bigg(\dfrac{3x-1}{3x-1}\bigg)\\\\\\=\dfrac{2(-3x+2)+6(3x-1)}{(3x-1)(-3x+2)}\\\\\\=\dfrac{-6x+4+18x-6}{(3x-1)(-3x+2)}\\\\\\=\dfrac{12x-2}{(3x-1)(-3x+2)}\\\\\\=\large\boxed{\dfrac{2(6x-1)}{(3x-1)(-3x+2)}}$

P(x) - Q(x)

$\dfrac{2}{3x-1}- \dfrac{6}{-3x+2}\\\\\\=\dfrac{2}{3x-1}\bigg(\dfrac{-3x+2}{-3x+2}\bigg)- \dfrac{6}{-3x+2}\bigg(\dfrac{3x-1}{3x-1}\bigg)\\\\\\=\dfrac{2(-3x+2)-6(3x-1)}{(3x-1)(-3x+2)}\\\\\\=\dfrac{-6x+4-18x+6}{(3x-1)(-3x+2)}\\\\\\=\dfrac{-24x+10}{(3x-1)(-3x+2)}\\\\\\=\large\boxed{\dfrac{-2(12x-5)}{(3x-1)(-3x+2)}}$

P(x) · Q(x)

$\dfrac{2}{3x-1}\times \dfrac{6}{-3x+2}\\\\\\=\large\boxed{\dfrac{12}{(3x-1)(-3x+2)}}$