Evaluating Formula

One of the most important applications of algebraAlgebra is a branch of mathematics that substitutes letters for numbers. Algebra is about finding the unknown or putting real-life variables into equations and then solving them. More 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 1.5.1:

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

⇒ \( \frac{2(-3^2) \:+ \:(-2)(-3) \:+\: 4}{(-1) \: \times \: (-3)\: +\: 4} \)

⇒   \( \frac{2 \: \times \: 9\: +\: 6 \:+\: 4}{3 \:+ \:4} \\ = \frac{18 \:+ \:10}{7} \\ = \frac{28}{7} \\ = \scriptsize 4\)

Example 1.5.2:

Evaluate S₂₀ – S₁₀,

If S_n = \( \frac{4n^3 \: – \: 3n^2 \: + \: 6}{n} \)  (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

S₁₀ = 1169.7