4.2.2 Linear Equations I, PT3 Practice

Question 6:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}5a-16=a\\ 5a-a=16\\ 4a=16\\ a=4\end{array}$

(b)
$\begin{array}{l}6+\frac{2}{3}\left(-9p+12\right)=p\\ 6-6p+8=p\\ -6p-p=-8-6\\ -7p=-14\\ p=2\end{array}$

Question 7::
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}a-2=\frac{a}{5}\\ 5a-2=a\\ 4a=2\\ a=2\end{array}$

(b)
$\begin{array}{l}\frac{b-3}{4}=\frac{2+b}{5}\\ 5b-15=8+4b\\ 5b-4b=8+15\\ b=23\end{array}$

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Question 8:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}x=-24-x\\ 2x=-24\\ x=-12\end{array}$

(b)

Question 9:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}5p=8p-9\\ 5p-8p=-9\\ -3p=-9\\ p=3\end{array}$

(b)
$\begin{array}{l}3q=\frac{20-13q}{4}\\ 12q=20-13q\\ 12q+13q=20\\ 25q=20\\ q=\frac{4}{5}\end{array}$

4.2.1 Linear Equations I, PT3 Practice 1

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

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

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

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

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:

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

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

4.1 Linear Equations I

4.1 Linear Equations I

4.1.1 Equality
1. An equation is a mathematical statement that joins two equal quantities together by an equality sign ‘=’.
Example: km = 1000 m

2.
If two quantities are unequal, the symbol ‘≠’ (is not equal) is used.
Example: 9 ÷ 4 ≠ 3

4.1.2 Linear Equations in One Unknown
1. A linear algebraic term is a term with one unknown and the power of unknown is one.
Example: 8x, -7y, 0.5y, 3a, …..

2.
A linear algebraic expression contains two or more linear algebraic terms which are joined by a plus or minus sign.
Example:
3x – 4y, 4+ 9, 6x – 2y + 5, ……

3.
A linear equation is an equation involving numbers and linear algebraic terms.
Example:
5x – 4 = 11, 4x + 7 = 15, 3y – 2 = 7

4.1.3 Solutions of Linear Equations in One Unknown
1. Solving an equation is a process of finding the values of the unknown in the equation.
2. The number that satisfies the equation is called the solution or root of the equation.
Example 1:
+ 4 = 12
x = 12 – 4 ← (When +4 is moved to the right of the equation, it becomes –4)
= 8

Example 2:
– 7 = 11
x = 11 + 7 ← (When –7 is moved to the right of the equation, it becomes +7)
= 18

Example
3:

$\begin{array}{l}8x=16\\ x=\frac{16}{8}←\overline{)\begin{array}{l}\text{when the multiplier 8 is moved}\\ \text{to the right of the equation, it}\\ \text{becomes the divisor 8}\text{.}\end{array}}\\ x=2\end{array}$

Example
4:
$\begin{array}{l}\frac{x}{5}=3\\ x=3×5←\overline{)\begin{array}{l}\text{the divisor 5 becomes the}\\ \text{multiplier 5 when moved}\\ \text{to the right of the equation}\text{.}\end{array}}\\ x=15\end{array}$