Question

solve the Following pair of equation by reducing to a pair of linear equation by substitution method.?????
1. 1/2x+ 1/3y = 2 and
1/3x +1/2y = 13/6​

Answer 1

Given:

$\frac{1}{2x} + \frac{1}{3y} = 2 \\ \\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Solution:

$\texttt{Let } \frac{1}{x} \texttt{ = p } \: \\ \\ \texttt{and} \: \frac{1}{y} \texttt{ = q}$

So our equations become,

$\frac{1}{2} p + \frac{1}{3} q = 2 \: - - (1) \\ \\ \frac{1}{3} p + \frac{1}{2} q = \frac{13}{6} - - (2)$

Taking LCM and simplifying first equation.

$= > \frac{1}{2} p + \frac{1}{3} q = 2 \\ \\ = > \frac{3p + 2q}{2 \times 3} = 2 \\ \\ = > \frac{3p + 2q}{6} = 2 \\ \\ = > 3p + 2q = 12 - - (3)$

Now with second equation

$= > \frac{1}{3} p + \frac{1}{2} q = \frac{13}{6} \\ \\ = > \frac{2p + 3q}{3 \times 2} = \frac{13}{6} \\ \\ = > \frac{2p + 3q}{6} = \frac{13}{6} \\ \\ = > 2p + 3q = \frac{13}{ \cancel6} \times \cancel{6} \\ \\ = > 2p + 3q = 13 \: - - (4)$

Our equations are

$3p + 2q = 12 \: \: ....(3) \\ 2p + 3q = 13 \: \: ....(4)$

From third equation

$3p + 2q = 12 \\ 3p = 12 - 2q \\ \\ p = \frac{12 - 2q}{3} \: \: .......(5)$

Putting the value of p in equation (4)

$= > 2 \left( \frac{12 - 2q}{3} \right) + 3q = 13 \\ \\ = > \frac{24 - 4q}{3} + 3q = 13 \\ \\ \texttt{Taking LCM}\\ \\ = > \frac{24 - 4q + 3q(3)}{3} = 13 \\ \\ = > \frac{24 - 4q + 9q}{3} = 13 \\ \\ = > \frac{24 + 5q}{3} = 13 \\ \\ = > 24 + 5q = 13 \times 3 \\ \\ = > 24 + 5q = 39 \\ \\ = > 5q = 39 - 24 \\ \\ = > 5q = 15 \\ \\ = > q = \frac{15}{5} \\ \\ = > q = 3$

Putting the value of q in equation (5)

$p = \frac{12 - 2q}{3} \\ \\ = \frac{12 - 2 \times 3}{3} \\ \\ = \frac{12 - 6}{3} \\ \\ = \frac{6}{3} \\ \\ = > p = 2$

Hence, the value of p =2 and q =3

$\texttt{But we know that} \\ \frac{1}{x} = p \\ \\ \frac{1}{y} = q$

Therefore,

$\frac{1}{x} = p \\ \\ = > \frac{1}{x} = 2 \\ \\ = > x = \frac{1}{2} \\ \\ \texttt{and} \\ \\ \frac{1}{y} = q \\ \\ = > \frac{1}{y} = 3 \\ \\ = > y = \frac{1}{3}$

$\texttt{So, x} = \frac{1}{2} \: \texttt{and y = } \frac{1}{3}$