# 3.2 Algebraic Expressions II, PT3 Focus Practice

Question 6:

Solution:
$\begin{array}{l}-4\left(5x-3\right)+7x-1\\ =-20x+12+7x-1\\ =-13x+11\end{array}$

Question 7:
$\left(3x-2y\right)-\left(x+4y\right)$

Solution:
$\begin{array}{l}\left(3x-2y\right)-\left(x+4y\right)\\ =3x-2y-x-4y\\ =2x-6y\end{array}$

Question 8:

Solution:
$\begin{array}{l}-3\left(2a-4b\right)-\frac{1}{3}\left(6a-15b\right)\\ =-6a+12b-2a+5b\\ =-8a+17b\end{array}$

Question 9:
$\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)$

Solution:
$\begin{array}{l}\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)\\ =\overline{)-\frac{2}{3}x}+2y-3z\overline{)+\frac{2}{3}x}+3y-4z\\ =5y-7z\end{array}$

Question 10:
$\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)$

Solution:
$\begin{array}{l}\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)\\ =\frac{1}{2}a+3bc-\frac{3}{5}-\frac{2}{5}bc+\frac{2}{5}a\\ =\frac{5a+4a}{10}+\frac{15bc-2bc}{5}-\frac{3}{5}\\ =\frac{9a}{10}+\frac{13bc}{5}-\frac{3}{5}\end{array}$