 QUESTION

# Solve the following simultaneous equations:$\dfrac{{{\text{27}}}}{{{\text{x - 2}}}}{\text{ + }}\dfrac{{{\text{31}}}}{{{\text{y + 3}}}}{\text{ = 85}} \\ \dfrac{{{\text{31}}}}{{{\text{x - 2}}}}{\text{ + }}\dfrac{{{\text{27}}}}{{{\text{y + 3}}}}{\text{ = 89}} \\$

Hint: In order to solve this problem you need to assume variables for $\dfrac{1}{{x - 2\,\,}}\,\,\,\,{\text{and }}\dfrac{1}{{y + 3}}$. Then solving further for getting the values of variables. Doing this will solve your problem.

The given equations are
$\Rightarrow \dfrac{{{\text{27}}}}{{{\text{x - 2}}}}{\text{ + }}\dfrac{{{\text{31}}}}{{{\text{y + 3}}}}{\text{ = 85}} \\ \Rightarrow \dfrac{{{\text{31}}}}{{{\text{x - 2}}}}{\text{ + }}\dfrac{{{\text{27}}}}{{{\text{y + 3}}}}{\text{ = 89}} \\$
We need to find the value of x and y.
Let us assume $\dfrac{{\text{1}}}{{{\text{x - 2}}}} = {\text{X}}\,\,\,{\text{and }}\dfrac{{\text{1}}}{{{\text{y + 3}}}}{\text{ = Y}}$
Then the above equation will be:
27X + 31Y = 85…………(1)
31X + 27Y = 89…………(2)
The value of X from equation (1) can be written as:
${\text{X = }}\dfrac{{{\text{85 - 31Y}}}}{{{\text{27}}}}$
Putting the value of X in equation (2) and solving we get:
${\text{31}}\left( {\dfrac{{{\text{85 - 31Y}}}}{{{\text{27}}}}} \right){\text{ + 27Y = 89}} \\ \Rightarrow \dfrac{{{\text{85(31) - (31)31Y + 27(27Y)}}}}{{{\text{27}}}}{\text{ = 89}} \\ \Rightarrow \dfrac{{{\text{2635 - 961Y + 729Y}}}}{{{\text{27}}}}{\text{ = 89}} \\ \Rightarrow {\text{2635 - 232Y = 89(27) = 2403}} \\ \Rightarrow {\text{ - 232Y = 2403 - 2635}} \\ \Rightarrow {\text{232Y = 232}} \\ \Rightarrow {\text{Y = 1}} \\$
On putting the value of Y = 1 in equation (1) we get:
${\text{X = }}\dfrac{{{\text{85 - 31(1)}}}}{{{\text{27}}}}{\text{ = }}\dfrac{{{\text{54}}}}{{{\text{37}}}}{\text{ = 2}}$
So, the value of X = 2 and that of Y = 1.
We have assumed $\dfrac{{\text{1}}}{{{\text{x - 2}}}} = {\text{X}}\,\,\,{\text{and }}\dfrac{{\text{1}}}{{{\text{y + 3}}}}{\text{ = Y}}$.
So, on putting the value of X and Y in above equations we get,
$\dfrac{{\text{1}}}{{{\text{x - 2}}}}{\text{ = 2}}\,\,\,{\text{and }}\dfrac{{\text{1}}}{{{\text{y + 3}}}}{\text{ = 1}} \\ \Rightarrow \dfrac{{\text{1}}}{{\text{2}}}{\text{ = x - 2}}\,\,{\text{and }}\dfrac{1}{1} = {\text{y + 3}} \\ \Rightarrow {\text{x = }}\dfrac{{\text{1}}}{{\text{2}}} + 2 = \dfrac{5}{2}\,\,\,{\text{and y = 1 - 3 = - 2}} \\$
Hence, the value of x = $\dfrac{5}{2}$ and that of y = -2.

Note: To solve such problems in which you have asked to solve the equations you need to find the values of variables present in the equations. Remember the number of unknown variables must be equal to the number of equations. We can solve these equations using various methods. Then solving algebraically we get the value of the variables. Doing this will solve your problem.