6.4.1 Linear Equations, PT3 Practice - Mathematics

6.4.1 Linear Equations, 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}

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

Solution:

(a)

4 – 3n = 5n – 4

–3n –5n = – 4 – 4

–8n = – 8

8n = 8

n = 8/8 = 1

\begin{matrix} (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{matrix}

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

Question 2:

Solve the following linear equations.

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

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

Solution:

\begin{matrix} (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{matrix}

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

\begin{matrix} (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{matrix}

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

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:

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

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

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