Question

# In the system of equations $5x - 7y = - 13$ and $2x + 3y = 13$, the values of $x$ and $y$ will beA. $\dfrac{{52}}{{29}},\dfrac{{91}}{{29}}$B. $\dfrac{{31}}{{27}},\dfrac{{64}}{{27}}$C. $\dfrac{{51}}{{29}},\dfrac{{87}}{{29}}$D. $\dfrac{{52}}{{17}},\dfrac{{91}}{{17}}$

Hint: Here we find the value of $x$ from first equation and then substitute the value in second equation, then solve to find the values of $x$ and $y$

We are given two equations in two variables $x$ and $y$.
Consider the equation $5x - 7y = - 13$ as the first equation.
Consider the equation $2x + 3y = 13$ as the second equation.
We will find the value of $x$ from the first equation
$5x - 7y = - 13$
Shift all values other than the ones containing the variable $x$ to the right hand side of the equation.
$\Rightarrow 5x = - 13 + 7y$
Divide both sides of the equations by $5$
$\Rightarrow \dfrac{{5x}}{5} = \dfrac{{ - 13 + 7y}}{5}$
Cancel out the same terms from both the numerator and denominator in the left hand side of the equation.
$\Rightarrow x = \dfrac{{ - 13 + 7y}}{5}$ … (i)
Now we substitute the value of $x = \dfrac{{ - 13 + 7y}}{5}$ in the second equation.
We have second equation as $2x + 3y = 13$
$\Rightarrow 2\left( {\dfrac{{ - 13 + 7y}}{5}} \right) + 3y = 13$
Multiply the terms outside the bracket with terms in the numerator.
$\Rightarrow \left( {\dfrac{{2 \times ( - 13) + 2 \times 7y}}{5}} \right) + 3y = 13 \\ \Rightarrow \left( {\dfrac{{ - 26 + 14y}}{5}} \right) + 3y = 13 \\$
Now we take LCM in left hand side of the equation
$\Rightarrow \dfrac{{ - 26 + 14y + 3y \times 5}}{5} = 13 \\ \Rightarrow \dfrac{{ - 26 + 14y + 15y}}{5} = 13 \\ \Rightarrow \dfrac{{ - 26 + 29y}}{5} = 13 \\$
Now we multiply both sides of the equations by $5$
$\Rightarrow \dfrac{{ - 26 + 29y}}{5} \times 5 = 13 \times 5$
Cancel out the same terms from both numerator and denominator.
$\Rightarrow - 26 + 29y = 65$
Shift all the constant values to the right hand side of the equation.
$\Rightarrow 29y = 65 + 26 \\ \Rightarrow 29y = 91 \\$
Dividing both sides by $29$
$\Rightarrow \dfrac{{29y}}{{29}} = \dfrac{{91}}{{29}}$
Cancel same terms from denominator and numerator
$\Rightarrow y = \dfrac{{91}}{{29}}$
Substituting the value of $y = \dfrac{{91}}{{29}}$ in equation (i)
$\Rightarrow x = \dfrac{{ - 13 + 7\left( {\dfrac{{91}}{{29}}} \right)}}{5}$
Multiply the term outside the bracket to the numerator inside the bracket.
$\Rightarrow x = \dfrac{{ - 13 + \dfrac{{637}}{{29}}}}{5}$
Tale LCM of the term in the numerator and solve.
$\Rightarrow x = \dfrac{{\dfrac{{ - 13 \times 29 + 637}}{{29}}}}{5} \\ \Rightarrow x = \dfrac{{\dfrac{{ - 377 + 637}}{{29}}}}{5} \\ \Rightarrow x = \dfrac{{\dfrac{{260}}{{29}}}}{5} \\$
Now we can write a fraction $\dfrac{{\dfrac{a}{c}}}{b} = \dfrac{a}{{c \times b}}$
$\Rightarrow x = \dfrac{{260}}{{29 \times 5}}$
Cancel out fraction to its simplest form by removing all factors between numerator and denominator.
$\Rightarrow x = \dfrac{{52}}{{29}}$
Thus, value of $x = \dfrac{{52}}{{29}}, y = \dfrac{{91}}{{29}}$

So, the correct answer is “Option A”.

Note: Students are likely to make mistake in calculation part of these questions where there is fraction even in the numerator, always try to solve the complete value in the numerator as fraction first and then using $\dfrac{{\dfrac{a}{c}}}{b} = \dfrac{a}{{c \times b}}$ write in proper form.