**4.2.1 Linear Equations I, PT3 Practice 1**

**Question 1:**

Solve the following linear equations.

*n*= 5

*n*– 4 (a) 4 – 3

$\frac{4m-2}{10+m}=\frac{1}{2}$
(b)

Solution:Solution:

**(a)**

4 – 3

*n*= 5*n*– 4–3

*n*–5*n*= – 4 – 4 –8

*n*= – 8 8

*n*= 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.

$\begin{array}{l}\text{(a)}\frac{x}{3}=4-x\\ \text{(b)}\frac{3\left(x-2\right)}{5}=9\end{array}$

$\begin{array}{l}\text{(a)}\frac{x}{3}=4-x\\ \text{(b)}\frac{3\left(x-2\right)}{5}=9\end{array}$

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

**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:Solution:

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

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

**Question 4:**

Solve the following linear equations.

$\begin{array}{l}\text{(a)}11+\frac{2x}{3}=9\\ \text{(b)}\frac{x-5}{3}=\frac{x}{6}\end{array}$

Solution:Solution:

$\begin{array}{l}\text{(a)}11+\frac{2x}{3}=9\\ \text{}\frac{2x}{3}=9-11\\ \text{}\frac{2x}{3}=-2\\ \text{}2x=-6\\ \text{}x=\frac{-6}{2}\\ \text{}x=-3\end{array}$

$\begin{array}{l}\text{(b)}\frac{x-5}{3}=\frac{x}{6}\\ \text{}6\left(x-5\right)=3x\\ \text{}6x-30=3x\\ \text{}6x-3x=30\\ \text{}3x=30\\ \text{}x=10\end{array}$

**Question 5:**

Solve the following linear equations.

$\begin{array}{l}\text{(a) 4}\left(2x-3\right)=24\\ \text{(b)}\frac{y}{2}-\frac{y+4}{3}=5\end{array}$

Solution:Solution:

$\begin{array}{l}\text{(a) 4}\left(2x-3\right)=24\\ \text{}8x-12=24\\ \text{}8x=24+12\\ \text{}8x=36\\ \text{}x=\frac{36}{8}\\ \text{}x=4\frac{1}{2}\end{array}$

$\begin{array}{l}\text{(b)}\frac{y}{2}-\frac{y+4}{3}=5\\ \text{}\frac{y\times 3}{2\times 3}-\frac{2\left(y+4\right)}{3\times 2}=5\\ \text{}\frac{3y-2\left(y+4\right)}{6}=5\\ \text{}3y-2y-8=30\\ \text{}y=30+8\\ \text{}y=38\end{array}$