Lesson 8, Topic 1
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Simultaneous Linear Equations

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Algebraic Solution

This involves:

1. Substitution Method
2. Elimination Method

Substitution Method

Example 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) $$\scriptsize r \; – \; \normalsize \frac {q}{2} \scriptsize + 1 = 0$$

Re-write as   $$\scriptsize r\; -\; \normalsize \frac {q}{2} \scriptsize = -1$$

Multiply both sides by 2

i.e.     2r – q = -2         …………(1)

2r + q = 8          …………(2)

i.e.     q = 8 – 2r         ………….(3)

Substitute for q in equation (1)

i.e.     2r – (8 – 2r) = -2 (Expand brackets)

i.e.     2r – 8 + 4r = -2

i.e.     6r – 8 = -2 (Collect like terms)

i.e.     6r = 6 (Divide both sides by 6)

i.e.     r = 1

Substitute for r in equation (3)

i.e.     q = 8 – 2(1)

i.e.      q = 8 – 2

i.e.      q = 6.

Check:

2r + q = 8

2(1) + 6 = 8

2 + 6 = 8

8 = 8

(b)     $$\scriptsize y\; – \; \normalsize \frac {x}{4} \scriptsize = 3$$

Multiply both sides by 4

4y – x = 12         ……….(i)

3y + x = 23         ………………..(ii)

i.e.          x = 23 – 3y              …………….(iii)

Substitute for x in equation (i)

i.e.       4y – 23 – 3y = 12

i.e.       4y – 23 + 3y = 12

i.e.       7y = 35

i.e.        y = 5

Substitute for y in equation (iii)

i.e.      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

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) 5p + 2r = 2          ………….(i)

Divide both sides by 2

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

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

Subtract equation (iii) from (iii) to eliminate

i.e. $$\left ( \frac{5}{2} \scriptsize p \; – \; p \right) + \scriptsize (r \; – \; r) =\left( \scriptsize 1 \; – \; \normalsize \frac{1}{4} \right)$$

$$\normalsize \frac{3}{2} \scriptsize p = \normalsize \frac{3}{4}$$

Multiply both sides by 2

i.e.        3p = $$\frac{3}{2}$$

Divide both sides by 3

i.e.   p = $$\frac{1}{2}$$

Substitute for p in equation (iii)

i.e.     $$\frac{1}{2} \scriptsize + r = \normalsize \frac{1}{4}$$

i.e.     r = $$\frac{1}{4} – \frac{1}{2}$$

i.e.    r = $$– \frac{1}{4}$$

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

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

i.e.   12p – 8r= 84     ……………(iii)

12p + 15r = 15   ……………(iv)

Subtract equation (iv) from (iii)

i.e.    -23r = 69

Divide both sides by -23

$$\frac {-23r}{23} = \frac {69}{23}$$

i.e.     r = -3

Substitute for r in equation (i)

i.e.     3p – 2 (- 3 ) = 21

3p + 6 = 21

3p = 15

i.e.      p = 5

Check:

3p – 2r = 21

3(5) – 2(-3) = 21

15 + 6 = 21

21 = 21

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