
Solve the following system of equations by elimination method.
11x – 7y = xy and 9x – 4y = 6xy
$
{\text{A}}{\text{. }}\left( {\dfrac{1}{3},\dfrac{1}{2}} \right) \\
{\text{B}}{\text{. }}\left( {\dfrac{1}{2},\dfrac{1}{2}} \right) \\
{\text{C}}{\text{. }}\left( {\dfrac{1}{3},\dfrac{1}{3}} \right) \\
{\text{D}}{\text{. None of these}} \\
$
Answer
614.7k+ views
Hint – To find the values of x and y, we rearrange the equations such that their RHS is equal to zero. Then we multiply them with a number such that they are solvable. We solve them for x and y.
Complete step-by-step answer:
Given data,
11x – 7y = xy and 9x – 4y = 6xy
We re-write the equations, we get
\[ \Rightarrow \dfrac{{{\text{11x - 7y}}}}{{{\text{xy}}}} = 1\]
$ \Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{7}{{\text{x}}} = 1$ -- (1)
And,
\[ \Rightarrow \dfrac{{{\text{9x - 4y}}}}{{{\text{6xy}}}} = 1\]
$ \Rightarrow \dfrac{{\text{3}}}{{{\text{2y}}}} - \dfrac{2}{{{\text{3x}}}} = 1$ -- (2)
As their respective RHS are equal we equate (1) and (2), we get
$
\Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{7}{{\text{x}}} = \dfrac{{\text{3}}}{{{\text{2y}}}} - \dfrac{2}{{{\text{3x}}}} \\
\Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{{\text{3}}}{{{\text{2y}}}} = \dfrac{7}{{\text{x}}} - \dfrac{2}{{{\text{3x}}}} \\
\Rightarrow \dfrac{{11 \times 2{\text{ - 3}}}}{{2{\text{y}}}} = \dfrac{{7 \times 3{\text{ - 2}}}}{{3{\text{x}}}} \\
$
$
\Rightarrow \dfrac{{19}}{{2{\text{y}}}} = \dfrac{{19}}{{3{\text{x}}}} \\
\Rightarrow {\text{3x = 2y}} \\
$
We have obtained a relation between x and y. Now on verifying the given options, we get
Let us check$\left( {\dfrac{1}{3},\dfrac{1}{2}} \right)$,
$
\Rightarrow 3 \times \dfrac{1}{3} = 2 \times \dfrac{1}{2} \\
\Rightarrow 1 = 1 \\
$
Hence the value of (x, y) for the given equations is$\left( {\dfrac{1}{3},\dfrac{1}{2}} \right)$. Option A is the correct answer.
Note – In order to solve this type of problems the key is to convert all the terms in both the equations in such a way that they can be equated, to form a relation. Solving that relation gives us the value of x and y. In this specific question x and y can have so many values satisfying 3x = 2y, which are solutions to the given equation. We intentionally verify the options in order to obtain an answer from the given choices.
Complete step-by-step answer:
Given data,
11x – 7y = xy and 9x – 4y = 6xy
We re-write the equations, we get
\[ \Rightarrow \dfrac{{{\text{11x - 7y}}}}{{{\text{xy}}}} = 1\]
$ \Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{7}{{\text{x}}} = 1$ -- (1)
And,
\[ \Rightarrow \dfrac{{{\text{9x - 4y}}}}{{{\text{6xy}}}} = 1\]
$ \Rightarrow \dfrac{{\text{3}}}{{{\text{2y}}}} - \dfrac{2}{{{\text{3x}}}} = 1$ -- (2)
As their respective RHS are equal we equate (1) and (2), we get
$
\Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{7}{{\text{x}}} = \dfrac{{\text{3}}}{{{\text{2y}}}} - \dfrac{2}{{{\text{3x}}}} \\
\Rightarrow \dfrac{{11}}{{\text{y}}} - \dfrac{{\text{3}}}{{{\text{2y}}}} = \dfrac{7}{{\text{x}}} - \dfrac{2}{{{\text{3x}}}} \\
\Rightarrow \dfrac{{11 \times 2{\text{ - 3}}}}{{2{\text{y}}}} = \dfrac{{7 \times 3{\text{ - 2}}}}{{3{\text{x}}}} \\
$
$
\Rightarrow \dfrac{{19}}{{2{\text{y}}}} = \dfrac{{19}}{{3{\text{x}}}} \\
\Rightarrow {\text{3x = 2y}} \\
$
We have obtained a relation between x and y. Now on verifying the given options, we get
Let us check$\left( {\dfrac{1}{3},\dfrac{1}{2}} \right)$,
$
\Rightarrow 3 \times \dfrac{1}{3} = 2 \times \dfrac{1}{2} \\
\Rightarrow 1 = 1 \\
$
Hence the value of (x, y) for the given equations is$\left( {\dfrac{1}{3},\dfrac{1}{2}} \right)$. Option A is the correct answer.
Note – In order to solve this type of problems the key is to convert all the terms in both the equations in such a way that they can be equated, to form a relation. Solving that relation gives us the value of x and y. In this specific question x and y can have so many values satisfying 3x = 2y, which are solutions to the given equation. We intentionally verify the options in order to obtain an answer from the given choices.
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