Question

Solve the following pairs of equations by reducing them to a pair of linear equations :

$fraction numerator 1 over denominator 2 straight x end fraction plus fraction numerator 1 over denominator 3 straight y end fraction equals 2 fraction numerator 1 over denominator 3 straight x end fraction plus fraction numerator 1 over denominator 2 straight y end fraction equals 13 over 6$

Solution

Let space 1 over straight x equals straight u space and space 1 over straight y equals straight v

Then, the given system of equation becomes

1 half straight u plus 1 third straight v equals 2 rightwards double arrow space space space 3 straight u plus 2 straight v equals 12 and space 1 third straight u plus 1 half straight v equals 13 over 6 rightwards double arrow space fraction numerator 2 straight u plus 3 straight v over denominator 6 end fraction equals 13 over 6 rightwards double arrow space 2 straight u plus 3 straight v equals 13

Thus, we have two equations
3u + 2v = 12 ...(i)
2u + 3v = 13 ...(ii)
From (i), we have
$straight u space equals space fraction numerator 12 minus 2 straight v over denominator 3 end fraction space space space... left parenthesis iii right parenthesis$
Putting the value of ‘u’ in (ii), we get
$2 open parentheses fraction numerator 12 minus 2 v over denominator 3 end fraction close parentheses plus 3 v equals 13 rightwards double arrow space 24 minus 4 v plus 9 v equals 39 rightwards double arrow space 5 v space equals space 39 minus 24 rightwards double arrow space space 5 v equals 15 rightwards double arrow space space space v space equals space 3$
Putting the value of ‘v’ in (iii), we get
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$H e n c e comma space straight x space equals space 1 half comma space space straight y space equals 1 third space space$ is the solution of the given equation.

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Pair Of Linear Equations In Two Variables

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