Simultaneous Linear Equations

Algebraic Solution:

This involves:

1. Substitution Method
2. Elimination Method

Substitution Method:

Example 2.1.1:

Use the substitution method to solve the following pairs of simultaneous equations

(a) q + 2r = 8;          r – \( \frac{q}{2} \) + 1 = 0

(b) y – \( \frac{x}{4} \) = 3;         3y + x = 23

Solution:

(a) q + 2r = 8;          r – \( \frac{q}{2} \) + 1 = 0

⇒ Simplify \( \scriptsize r \: – \: \normalsize \frac {q}{2} \scriptsize \: + \: 1 = 0 \)

⇒ Re-write as   \( \scriptsize r\: -\: \normalsize \frac {q}{2} \scriptsize = \: -1 \)

Multiply both sides by 2

⇒    2r – q = -2         …………(1)

From the question we write down the other equation

 ⇒   q + 2r = 8          …………(2)

Make q the subject

⇒     q = 8 – 2r         ………….(3)

Substitute for q in equation (1)

⇒    2r – q = -2         …………(1)

⇒    2r – (8 – 2r) = -2

Expand brackets

⇒     2r – 8 + 4r = -2

⇒     6r – 8 = -2

Collect like terms

⇒    6r = -2 + 8

⇒    6r = 6

Divide both sides by 6

⇒     r = 1

Substitute for r in equation (3)

⇒     q = 8 – 2r         ………….(3)

⇒    q = 8 – 2(1)

⇒      q = 8 – 2

⇒      q = 6

Check:

2r + q = 8

2(1) + 6 = 8

2 + 6 = 8

8 = 8

(b) y – \( \frac{x}{4} \) = 3;         3y + x = 23

⇒ Simplify \( \scriptsize y\: – \: \normalsize \frac {x}{4} \scriptsize = 3 \)

Multiply both sides by 4

⇒ 4y – x = 12         ……….(i)

From the question, write down the other equation

⇒ 3y + x = 23         ………………..(ii)

Make x the subject of the equation

⇒         x = 23 – 3y              …………….(iii)

Substitute for x in equation (i)

⇒ 4y – x = 12         ……….(i)

⇒ 4y – (23 – 3y) = 12

open brackets

⇒      4y – 23 + 3y = 12

⇒      4y + 3y = 12 + 23

⇒      7y = 35

divide both sides by 7

⇒      \(\frac{\not{7}y}{\not{7}} = \frac{35}{7}\)

⇒        y = 5

Substitute for y in equation (iii)

⇒         x = 23 – 3y              …………….(iii)

⇒ x = 23 – 3(5)

⇒ x = 23 – 15

⇒ x = 8

Check:

3y + x = 23

3(5) + 8 = 23

15 + 8 = 23

23 = 23

Elimination Method:

Example 2.1.2:

Use elimination method to solve the following pairs of simultaneous equations:

(a) p + r = \( \frac{1}{4} \);              5p + 2r = 2

(b) 3p – 2r = 21;       4p + 5 r = 5

Solution:

(a) p + r = \( \frac{1}{4} \);              5p + 2r = 2

First, let’s eliminate r. We achieve this by getting a get a common coefficient of r in both equations.

⇒         5p + 2r = 2          ………….(i)

Divide both sides by 2

⇒     \( \frac{5}{2} \scriptsize p + r = 1 \)     ………….(ii)

 ⇒     p + r = \( \frac{1}{4} \)          …………..(iii)

To eliminate r subtract equation (iii) from (ii)

⇒ \(\left ( \frac{5}{2} \scriptsize p \: – \: p \right) + \scriptsize (r \: – \: r) =\left( \scriptsize 1 \: – \: \normalsize \frac{1}{4} \right) \)  

⇒ \(\normalsize \frac{3}{2} \scriptsize p = \normalsize \frac{3}{4}\)

Multiply both sides by 2

⇒ \(\normalsize \frac{3}{2} \scriptsize p \: \times \: 2 = \normalsize \frac{3}{4}\scriptsize \: \times \: 2\)

⇒        3p = \(\frac{3}{2}\)

Divide both sides by 3

⇒   \(\frac{3p}{3} =\frac{ \frac{3}{2}}{3}\)

⇒   \(\frac{3p}{3} = \frac{3}{2} \: \times \: \frac{1}{3}\)

⇒   p = \(\frac{1}{2}\)

Substitute for p in equation (iii)

⇒    \(\frac{1}{2} \scriptsize \: + \: r = \normalsize \frac{1}{4}\)

⇒     r = \(\frac{1}{4}\: -\: \frac{1}{2} \)

⇒    r = \(\: – \frac{1}{4} \)

(b) 3p – 2r = 21;       4p + 5 r = 5

⇒ 3p – 2r = 21       …………..i      × 4     (Multiply by 4)

⇒ 4p + 5r = 5         …………..ii      × 3   (Multiply by 3)

⇒   12p – 8r = 84     ……………(iii)

 ⇒  12p + 15r = 15   ……………(iv)

Subtract equation (iv) from (iii)

⇒   -23r = 69

Divide both sides by -23

⇒ \( \frac {-23r}{-23} = \frac {69}{-23} \)

⇒     r = -3

Substitute for r in equation (i)

⇒     3p – 2 (- 3 ) = 21

          3p + 6 = 21

          3p = 15

          p = \(\frac{15}{3}\)

⇒      p = 5

Check:

3p – 2r = 21

3(5) – 2(-3) = 21

15 + 6 = 21

21 = 21