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One of the most important applications of algebra is in the use of formula.

A formula is simply an equation which describes the relationship between two or more quantities. 

Let A = \(\frac{1}{2} \scriptsize (a = b )h\! \!\), where “a” and “b” are the parallel sides and “h” is the perpendicular distance between the parallel sides. In this case, A is said to be the subject of the formula.

Example 5

(a) Evaluate \( \frac{ax^2 + bx +c}{mx + c} \)

given that a = 2, x = -3, b = -2, c = 4 and m = -1  (WAEC)

Solution:

Substituting correctly in the expression

i.e. \( \frac{2(-3^2) + (-2)(-3) + 4}{(-1) \; \times \; (-3)\; +\; 4} \)

i.e.   \( \frac{2 \; \times \; 9\; +\; 6 \;+\; 4}{3 \;+ \;4} \\ = \frac{18 \;+ \;10}{7} \\ = \frac{28}{7} \\ = \scriptsize 4\)

(b) Evaluate S20 – S10,

if Sn = \( \frac{4n^3 \; – \; 3n^2 \; + \; 6}{n} \)  (WAEC)

Solution:

When n = 20

S20 = \( \frac{4 (20^3)\; – \; 3 (20^2) \; + \; 6}{20}\)

= \( \frac{4 (8,000) \; – \; 3 (400) + 6}{20}\)

= \( \frac{30,806}{20} \)

= 1,540.3

S10 = \( \frac{4 (10^3)\; – \; 3 (10^2) \;+ \;6}{10}\)

= \( \frac{4 (1,000) \; – \; 3 (100) \;+\; 6}{10}\)

= \( \frac{4000 \; – \; 300 \;+ \;6}{10} \\= \frac{3700 \;+ \;6}{10} \\ = \frac{3706}{10} \)

= 370.6

∴    S20 – S10 = \( \frac{30,806}{20} – \frac{3,706}{10} \)

= 1540.3 – 370.6

= 1169.7

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