# 6.4.1 Linear Equations, PT3 Practice

6.4.1 Linear Equations, PT3 Practice 1
Question 1:
Solve the following linear equations.
(a)  4 – 3n = 5n – 4
(b) $\frac{4m-2}{10+m}=\frac{1}{2}$

Solution:
(a)
4 – 3n = 5n – 4
–3n –5n = – 4 – 4
–8n = – 8
8n = 8
n = 8/8 = 1

$\begin{array}{l}\text{(b)}\frac{4m-2}{10+m}=\frac{1}{2}\\ \text{}2\left(4m-2\right)=10+m\\ \text{}8m-4=10+m\\ \text{}8m-m=10+4\\ \text{7}m=14\\ \text{}m=\frac{14}{7}\\ \text{}m=2\end{array}$

Question 2:
Solve the following linear equations.
$\begin{array}{l}\text{(a)}\frac{x}{3}=4-x\\ \text{(b)}\frac{3\left(x-2\right)}{5}=9\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\frac{x}{3}=4-x\\ \text{}x=3\left(4-x\right)\\ \text{}x=12-3x\\ x+3x=12\\ \text{}4x=12\\ \text{}x=\frac{12}{4}\\ \text{}x=3\end{array}$

$\begin{array}{l}\text{(b)}\frac{3\left(x-2\right)}{5}=9\\ \text{}3\left(x-2\right)=9×5\\ \text{}3x-6=45\\ \text{}3x=45+6\\ \text{}3x=51\\ \text{}x=\frac{51}{3}\\ \text{}x=17\end{array}$

Question 3:
Solve the following linear equations.
$\begin{array}{l}\text{(a)}\frac{3m}{4}+15=9\\ \text{(b)}2m+8=3\left(m-2\right)\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\frac{3m}{4}+15=9\\ \text{}\frac{3m}{4}=9-15\\ \text{}\frac{3m}{4}=-6\\ \text{}3m=-6×4\\ \text{}3m=-24\\ \text{}m=\frac{-24}{3}\\ \text{}m=-8\end{array}$

$\begin{array}{l}\text{(b)}2m+8=3\left(m-2\right)\\ \text{}2m+8=3m-6\\ \text{}2m-3m=-6-8\\ \text{}-m=-14\\ \text{}m=14\end{array}$