Question

Solve the following simultaneous equations.

`27/(x -2) + 31/(y + 3) = 85, 31/(x -2) + 27/(y + 3) = 89`

Answer 1

Given:

`27(1/(x -2)) + 31(1/(y + 3)) = 85`  ...(1)

`31(1/(x -2)) + 27(1/(y + 3)) = 89`  ...(2)

Let `1/(x - 2) = m and 1/(y + 3) = n`

27m + 31n = 85      ...(3)
31m + 27n = 89      ...(4)

Adding equations (3) and (4), we get,

\[\begin{array}{l}  
\phantom{\texttt{0}}\texttt{27m + 31n = 85}\\ \phantom{\texttt{}}\texttt{+31m + 27n = 89}\\ \hline \end{array}\]
∴ 58m + 58n = 174

∴ m + n = 3   ...(5)    ...[Dividing both sides by 58]

Subtracting equation (4) and (3), we get,
\[\begin{array}{l}  
\phantom{\texttt{0}}\texttt{27m + 31n = 85}\\ \phantom{\texttt{}}\texttt{−31m − 27n = −89}\\ \hline\phantom{\texttt{}}\texttt{  (−) (+)  (+)}\\ \hline \end{array}\]
∴ − 4m + 4n = −4

∴ −m + n = −1  ...(6)   [Dividing both sides by 4]

Adding equations (5) and (6), we get,

\[\begin{array}{l}  
\phantom{\texttt{0}}\texttt{m + n = 3}\\ \phantom{\texttt{}}\texttt{−m + n = −1}\\ \hline \end{array}\]
∴ 2n = 2
n = `2/2` = 1.

∴ n = 1

∴ n + 1 = 3

∴ n = 3 − 1

∴ n = 2

∴ (m, n) = (2, 1)

Resubstitute value of m and n

⇒ `1/(x − 2) = 2`  

⇒ 2(x − 2) = 1
⇒ 2x − 4 = 1
⇒ 2x = 4 + 1
⇒ 2x = 5
⇒ x = `5/2`

∴ `1/(y + 3)`

⇒ `1/(y + 3) = 1` 
⇒ y + 3 = 1
⇒ y = 1 − 3
⇒ y = − 2
Hence, (x, y) = `(5/2, -2)`