
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}} \\
$
Answer
615k+ views
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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
Recently Updated Pages
Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 English: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Maths: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 7 English: Engaging Questions & Answers for Success

Trending doubts
Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Who is eligible for RTE class 9 social science CBSE

Which places in India experience sunrise first and class 9 social science CBSE

What is pollution? How many types of pollution? Define it

Name 10 Living and Non living things class 9 biology CBSE

