Formulae & Substitution

One of the major uses 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 to express mathematical relationships in form of formulae.

A formulae is an equation that describes the relationship between two or more quantities.

Letters are used to represent unknown values in a formulae.

For example the frequency f in cycles per unit time i.e.

F = \(\frac {\omega}{2\pi} \)

This formulae shows the relationship between frequency (f) and circular velocity (\( \scriptsize \omega \))

F is called the subject of the formulae or equation.

Worked Example 8.1.1:

i. If, F = ma, find F when  m = 10 and  a = 6

Solution

F = ma

F = 10 × 6

F = 60N

ii. If y = mx  + c,  find y when m = 5, x = -2,   c  = -3.

Solution

             y = mx   +  c

            y = 5 × (-2) + (-3)

             y = -10 – 3

              y  = -13

iii. If, V = mgh  find V when  m = 10, g = 9.8  and  h = 5

Solution

              V  = mgh

              V = 10 × 9.8 × 5   =  490

iv. If y =  16x²  – 9z²,  find y when x = 7, and  z  =  5

Solution

y =  16 (7)² – 9 (5)²

y = 16 × 49 – 9 × 25

y = 784 – 225

y = 559

v. The sum of terms in a geometric progression is given as: s = \( \frac{a( 1 \; – \; r^n)}{1 \; – \; r} \)
Calculate s when  n = 4, a = 3 and r = 3

Solution

s = \( \frac{a( 1 \; – \; 3^4)}{1 \; – \; 3} \)

s = \( \frac{3( 1 \; – \; 81)}{1 \; – \; 3} \)

s = \( \frac{ 3(-80)}{-2} \)

s = \( \frac{-240}{-2} \)

s = 120

vi. Here’s a formulae for finding the area of a triangle;
A = \( \scriptsize \sqrt{s(s \; – \; a)(s \; – \; b)(s \; – \; c)} \)

Where s = \( \frac{a + b + c}{2} \)  and a, b and c are the three sides.
Find the area of a triangle when a = 5 cm, b = 8 cm, c = 11 cm.

Solution

s = \(\frac{a + b + c}{2} = \frac{5 + 8 + 11}{2} \\ = \frac{24}{2} \\ \scriptsize = 12 \)

A = \( \scriptsize \sqrt{s(s \; – \; a)(s \; – \; b)(s \; – \; c)} \)

A = \( \scriptsize \sqrt{12(12 \; – \; 5)(12 \; – \; 8)(12 \; – \; 11)} \)

A = \( \scriptsize \sqrt{12(7)(4)(1)} \)

A = \( \scriptsize \sqrt{12 \; \times \; 28} \\ \scriptsize = \sqrt{336} \\ \scriptsize = 18.33 \)