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SS1: MATHEMATICS - 2ND TERM

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  1. The Set Theory I | Week 1
    4 Topics
  2. The Set Theory II | Week 2 & 3
    2 Topics
  3. The Set Theory III | Week 4
    2 Topics
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    1 Quiz
  4. Simple Equations & Change of Subject of Formula | Week 5
    1 Topic
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    1 Quiz
  5. Algebraic - Variations | Week 6
    4 Topics
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    1 Quiz
  6. Quadratic Equations I | Week 7
    5 Topics
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    1 Quiz
  7. Logical Reasoning | Week 8
    6 Topics
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    1 Quiz
  8. Construction & Locus I | Week 9
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    1 Quiz




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To transpose a formula means to re-arrange it so that a different letter becomes the subject.

For example, in the formula

\( \scriptsize R = ρ \normalsize \frac{l}{A}\) where R is the subject 

We can also have

\( \scriptsize ρ = \normalsize \frac{RA}{l}\)

Example 4.1.1:

Make x the subject of the formulae in the following:

(a) \(\scriptsize t =\normalsize \frac {3P}{x} \scriptsize \; + \; s \) (WAEC)

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SOLUTION (a)

⇒ \(\scriptsize t =\normalsize \frac {3P}{x} \scriptsize \; + \; s \)

Solution:

Transfer “s” to the left hand side (L.H.S)

⇒ \(\; \; \; \scriptsize t \; – \; s = \normalsize \frac {3P}{x} \)

Invert both sides

⇒ \( \frac {1}{ t \; – \; s}\scriptsize =  \normalsize \frac {x}{3P} \)

Multiply both sides by 3P

⇒ \(\; \; \; \scriptsize 3P \; \times \;  \normalsize \frac {1}{ t \; – \; s} \scriptsize  = 3P \; \times \; \normalsize \frac {x}{3P} \)

⇒ \(\; \; \; \frac {3P}{t \: –  \: s}  = x  \)

Answer: \( \therefore \scriptsize x =  \normalsize \frac {3P}{t \: –  \: s} \)

SOLUTION (b)

\( \frac {1}{y} \;+\; \frac {1}{x} = \frac {1}{z} \)

 

Find the LCM on the left hand side (LHS)….LCM = \( \scriptsize x \; \times \;  y = xy \)

\( \frac {x \:+ \:y}{xy}  = \frac {1}{z} \)

 

Cross multiply

⇒ z(x + y ) = xy

Expand the brackets

⇒ xz + zy = xy

Collect like terms

⇒ xz – xy = -zy

Factorise out x

⇒     x(z – y) = -zy

Divide both sides by (z – y)

⇒ \( \frac {x(z\:  – \: y)}{z\:  – \:  y}  = \frac {-zy}{z \:  – \:  y} \)

Answer: \(\therefore \scriptsize x =  \normalsize \frac {-zy}{z \:  – \:  y} \)

SOLUTION (c)

⇒ \(\scriptsize V = \pi h^2 \left ( x\:  – \:\normalsize \frac {h}{3} \right)\)

 

Divide both sides by \(\scriptsize \pi h^2 \)

⇒ \(\frac{V}{\pi h^2} =\frac { \pi h^2 \left ( x\: – \:\frac {h}{3} \right)}{\pi h^2}\)

⇒ \(\frac{V}{\pi h^2} =\frac { \not{\pi}\not{ h^2} \left ( x\: – \:\frac {h}{3} \right)}{\not{\pi}\not{ h^2}}\)

\(\therefore \frac{V}{\pi h^2} = \left  ( \scriptsize x \: – \:\normalsize\frac {h}{3} \right)\)

 

Transfer \(\frac {h}{3}\)  to the Left Hand Side (LHS)

⇒ \(\frac{V}{\pi h^2} \; + \; \normalsize\frac {h}{3} = \scriptsize x\)

\( \therefore \scriptsize x = \normalsize  \frac{V}{\pi h^2} \scriptsize \; + \; \normalsize\frac {h}{3} \)

 

or \(\scriptsize x = \normalsize  \frac {3V \: +  \: \pi h^3 }{3 \pi h^2} \)

SOLUTION (d)

⇒ \( \scriptsize T = \normalsize \frac{mu^2}{x} \scriptsize \: – \: 5mg \)

 

Transfer – 5mg to the left hand side (LHS)

⇒ \( \scriptsize T + 5mg = \normalsize \frac{mu^2}{x} \)

invert both sides and multiply by \( \scriptsize mu^2 \)

⇒ \( \frac {1}{T  \;+ \;5mg}\scriptsize \; \times \; mu^2  = \normalsize \frac{x}{mu^2} \scriptsize \; \times \; mu^2 \)

 

Answer: \( \scriptsize x = \normalsize \frac{mu^2}{T \;+ \;5mg}\)

SOLUTION (e)

⇒ \( \scriptsize H = \normalsize \frac{m(x^2 \;-\; y^2)}{2gz} \)

 

Multiply both sides by 2gz

\( \scriptsize H \; \times  \;2gz = \normalsize \frac{m(x^2 \;-\; y^2)}{2gz} \scriptsize \: \times \: 2gz \)

 

\(\scriptsize 2Hgz = m ( x^2 \; – y^2) \)

 

Divide both sides by m

\(  \frac{2Hgz}{m} = \frac{m(x^2 \;-\;y^2)}{m}  \)

 

\( \frac {2Hgz}{m} =  \scriptsize x^2 \; – \; y^2 \)

 

Transfer – y2 to the left hand side (LHS)

⇒ \(\; \; \;  \frac {2Hgz}{m} \; +\;  \scriptsize y^2 =  x^2 \)

Take the square root of both sides

⇒ \( \; \; \; \sqrt{\frac {2Hgz}{m}\scriptsize \: + \: y^2}  = \sqrt{ x^2} \)

⇒ \( \; \; \; \scriptsize x = \normalsize \sqrt{\frac {2Hgz}{m}\scriptsize \: + \: y^2}  \)

or  \( \; \; \;  \scriptsize x =  \normalsize  \sqrt{\frac {2Hgz \: + \; my^2} {m}}  \)

SOLUTION (f)

⇒ \( \scriptsize I = \normalsize \frac{xE}{R \; + \; xr} \)

 

Cross multiply

\( \scriptsize I \; \times \; (R \; + \; xr) = xE \)

 

Expand the bracket

\( \scriptsize I R \; + \; I xr = xE \)

 

Collect like terms (Remember, we are looking for x)

\( \scriptsize I R = xE \; – \; I xr \)

 

Factorise out x on the Right-hand side (RHS)

\( \scriptsize I R = x(E \; – \; I r) \)

 

Divide both sides by (E – Ir)

⇒ \(\; \;  \frac{ I R}{E\; -\; Ir} = \frac {x(E \; – \; I r)}{E \;- \;Ir}\)

∴  \( \; \; \scriptsize  x = \normalsize \frac{ I R}{E \;- \;Ir} \)

SOLUTION (g)

\( \scriptsize I = \normalsize \frac{E}{\sqrt {x^2 \; + \; w^2l^2}} \)

 

Take the square of both sides

\( \scriptsize I^2 = \normalsize \frac{E^2}{x^2 \; + \; w^2l^2} \)

 

Cross multiply

\( \scriptsize I^2  \; \times\; (x^2 \; + \; w^2l^2) = E^2 \)

 

Expand the bracket

\( \scriptsize I^2 x^2 \; + \; I^2w^2l^2 = E^2 \)

 

Transfer I2W2L2 to the right-hand side (RHS)

\( \scriptsize I^2 x^2 = E^2 \;  – \; I^2w^2l^2  \)

 

Divide both sides by I2

\(  \frac{I^2 x^2}{I^2} = \frac{E^2 \;  – \; I^2w^2l^2 }{I^2} \)

 

\(  \scriptsize x^2  = \normalsize \frac{E^2 \;  – \; I^2w^2l^2 }{I^2} \)

 

Take the square root of both sides

\(  \scriptsize \sqrt {x^2} = \normalsize \sqrt{\frac{E^2 \;  – \; I^2w^2l^2 }{I^2}} \)

 

⇒ \(  \scriptsize \sqrt {x^2} = \normalsize \sqrt{\frac{E^2 \;  – \; (Iwl)^2 }{I^2}} \)

⇒ \( \therefore \scriptsize x = \normalsize \sqrt{\frac{E^2 \;  – \; (Iwl)^2 }{I^2}} \)

or \( \scriptsize x = \normalsize \frac {1}{I} \scriptsize \sqrt{(E \;  – \; Iwl)(E \;  + \; Iwl)} \)

SOLUTION (h)

⇒ \( \scriptsize P = \sqrt {Qx \normalsize \frac{1\; + \; 3t}{x} } \)

 

Take the square of both sides

⇒  \( \; \; \; \scriptsize P^2 = Qx \left (1 \; + \; \normalsize \frac{3t}{x}\right ) \)

Find the LCM for the RHS

⇒  \( \; \; \; \scriptsize P^2 = Qx \normalsize\left( \frac{x \; + \; 3t}{x} \right)\)

⇒  \( \; \; \; \scriptsize P^2 = \normalsize \frac {Qx}{x}  \scriptsize \left( {x \; + \; 3t} \right)\)

⇒  \( \therefore \scriptsize P^2 =  Q( x  \; + \; 3t)\)

Expand the bracket

⇒ \( \scriptsize P^2 =  Q x  \; + \; 3Qt\)

Transfer 3Qt to the left hand side (LHS)

⇒ \( \scriptsize P^2 \; – \; 3Qt =  Q x  \)

Divide both sides by Q

⇒ \( \frac {P^2 \; – \; 3Qt}{Q} =  \frac{Q x}{Q}  \)

\( \therefore \scriptsize x = \normalsize \frac {P^2 \; – \; 3Qt}{Q}  \)

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