Answer

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Hint – Using the given equations in the question we transform one equation such that the variable x is in terms of y. Then substitute y in the other equation. Then solve for the value of y and substitute for x.

Complete step-by-step answer:

Given: $\dfrac{{{\text{3x}}}}{2} - \dfrac{{{\text{5y}}}}{3} = - 2$ --- (1)

$\dfrac{{\text{x}}}{3} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6}$ ---- (2)

Rewriting (1),

$

\dfrac{{{\text{3x}}}}{2} = - 2 + \dfrac{{{\text{5y}}}}{3} \\

\Rightarrow \dfrac{{{\text{3x}}}}{2} = - \dfrac{{{\text{ - 6 + 5y}}}}{3} \\

\Rightarrow {\text{3x}} = - \dfrac{{{\text{ - 12 + 10y}}}}{3} \\

\Rightarrow {\text{x}} = - \dfrac{{{\text{ - 12 + 10y}}}}{{\text{9}}}{\text{ - - - - - - - - }}\left( 3 \right) \\

$

Substituting (3) in (2)

$

\dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{9} \times \dfrac{1}{3} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{{27}} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 20{\text{y + 27y}}} \right)}}{{54}} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 47{\text{y}}} \right)}}{{54}} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 47{\text{y}}} \right)}}{9} = 13 \\

\Rightarrow \left( { - 24 + 47{\text{y}}} \right) = 9 \times 13 = 117 \\

\Rightarrow 47{\text{y = 117 + 24}} \\

\Rightarrow {\text{47y = 141}} \\

\Rightarrow {\text{y = 3}} \\

$

Substitute value of y in (3) to find value of x

x = $\dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{9}$

$

\Rightarrow {\text{x = }}\dfrac{{\left( { - 12 + 10 \times 3} \right)}}{9} \\

\Rightarrow {\text{x = }}\dfrac{{\left( { - 12 + 30} \right)}}{9} \\

\Rightarrow {\text{x = }}\dfrac{{18}}{9} \\

\Rightarrow {\text{x = 2}} \\

$

Hence, (x, y) = (2, 3)

Note – This is a question which is of the type in which there are 2 equations and 2 variables. The key is to transform one of the equations such that we have one variable in terms of another. Then the other equation reduces into a single variable equation and becomes easier to solve. On finding the value of one variable the other can be found simply by substituting.

Complete step-by-step answer:

Given: $\dfrac{{{\text{3x}}}}{2} - \dfrac{{{\text{5y}}}}{3} = - 2$ --- (1)

$\dfrac{{\text{x}}}{3} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6}$ ---- (2)

Rewriting (1),

$

\dfrac{{{\text{3x}}}}{2} = - 2 + \dfrac{{{\text{5y}}}}{3} \\

\Rightarrow \dfrac{{{\text{3x}}}}{2} = - \dfrac{{{\text{ - 6 + 5y}}}}{3} \\

\Rightarrow {\text{3x}} = - \dfrac{{{\text{ - 12 + 10y}}}}{3} \\

\Rightarrow {\text{x}} = - \dfrac{{{\text{ - 12 + 10y}}}}{{\text{9}}}{\text{ - - - - - - - - }}\left( 3 \right) \\

$

Substituting (3) in (2)

$

\dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{9} \times \dfrac{1}{3} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{{27}} + \dfrac{{\text{y}}}{2} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 20{\text{y + 27y}}} \right)}}{{54}} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 47{\text{y}}} \right)}}{{54}} = \dfrac{{13}}{6} \\

\Rightarrow \dfrac{{\left( { - 24 + 47{\text{y}}} \right)}}{9} = 13 \\

\Rightarrow \left( { - 24 + 47{\text{y}}} \right) = 9 \times 13 = 117 \\

\Rightarrow 47{\text{y = 117 + 24}} \\

\Rightarrow {\text{47y = 141}} \\

\Rightarrow {\text{y = 3}} \\

$

Substitute value of y in (3) to find value of x

x = $\dfrac{{\left( { - 12 + 10{\text{y}}} \right)}}{9}$

$

\Rightarrow {\text{x = }}\dfrac{{\left( { - 12 + 10 \times 3} \right)}}{9} \\

\Rightarrow {\text{x = }}\dfrac{{\left( { - 12 + 30} \right)}}{9} \\

\Rightarrow {\text{x = }}\dfrac{{18}}{9} \\

\Rightarrow {\text{x = 2}} \\

$

Hence, (x, y) = (2, 3)

Note – This is a question which is of the type in which there are 2 equations and 2 variables. The key is to transform one of the equations such that we have one variable in terms of another. Then the other equation reduces into a single variable equation and becomes easier to solve. On finding the value of one variable the other can be found simply by substituting.

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