4.2.2 Linear Equations I, PT3 Practice

Posted on March 4, 2020 by user

Question 6:
Solve each of the following equations:

$\begin{array}{l} (a) 5 a - 16 = a \\ (b) 6 + \frac{2}{3} (- 9 p + 12) = p \end{array}$

Solution:
(a)

$\begin{array}{l} 5 a - 16 = a \\ 5 a - a = 16 \\ 4 a = 16 \\ a = 4 \end{array}$

(b)

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

Question 7::
Solve each of the following equations:

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

Solution:
(a)

$\begin{array}{l} a - 2 = \frac{a}{5} \\ 5 a - 2 = a \\ 4 a = 2 \\ a = 2 \end{array}$

(b)

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

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Question 8:
Solve each of the following equations:

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

Solution:
(a)

$\begin{array}{l} x = - 24 - x \\ 2 x = - 24 \\ x = - 12 \end{array}$

(b)

$\begin{array}{l} y + \frac{5}{4} (4 - 2 y) = - 4 \\ 4 y + 20 - 10 y = - 16 (\times 4) \\ - 6 y = - 16 - 20 \\ - 6 y = - 36 \\ y = 6 \end{array}$

Question 9:
Solve each of the following equations:

$\begin{array}{l} (a) 5 p = 8 p - 9 \\ (b) 3 q = \frac{20 - 13 q}{4} \end{array}$

Solution:
(a)

$\begin{array}{l} 5 p = 8 p - 9 \\ 5 p - 8 p = - 9 \\ - 3 p = - 9 \\ p = 3 \end{array}$

(b)

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

4.2.1 Linear Equations I, PT3 Practice 1

Posted on July 5, 2017 by user

4.2.1 Linear Equations I, PT3 Practice 1

Question 1:

Solve the following linear equations.

(a) 4 – 3n = 5n – 4

(b)

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

Solution:

(a)

4 – 3n = 5n – 4

–3n –5n = – 4 – 4

–8n = – 8

8n = 8

n = 8/8 = 1

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

Question 2:

Solve the following linear equations.

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

Solution:

$\begin{array}{l} (a) \frac{x}{3} = 4 - x \\ x = 3 (4 - x) \\ x = 12 - 3 x \\ x + 3 x = 12 \\ 4 x = 12 \\ x = \frac{12}{4} \\ x = 3 \end{array}$

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

Question 3:

Solve the following linear equations.

$\begin{array}{l} (a) \frac{3 m}{4} + 15 = 9 \\ (b) 2 m + 8 = 3 (m - 2) \end{array}$

Solution:

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

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

Question 4:

Solve the following linear equations.

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

Solution:

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

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

Question 5:

Solve the following linear equations.

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

Solution:

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

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

4.1 Linear Equations I

Posted on July 1, 2017 by user

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: 1 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, 4x + 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:

x + 4 = 12

x = 12 – 4 ← (When +4 is moved to the right of the equation, it becomes –4)

x = 8

Example 2:

x – 7 = 11

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

x = 18

Example 3:

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

Example 4:

$\begin{array}{l} \frac{x}{5} = 3 \\ x = 3 \times 5 \leftarrow \begin{array}{l} the divisor 5 becomes the \\ multiplier 5 when moved \\ to the right of the equation . \end{array} \\ x = 15 \end{array}$