Question

Solve the following system of equations by Paravartya method of Vedic mathematics:
\[2x + y = 5,3x - 4y = 2\]

Answer 1

Hint: Here we will directly use the formula for Paravartya method of Vedic mathematics in which roots of the two equations \[{a_1}x + {b_1}y = {c_1}\] and \[{a_2}x + {b_2}y = {c_2}\] are given by:-
\[x = \dfrac{{{b_1}{c_2} - {b_2}{c_{1}}}}{{{a_2}{b_1} - {a_1}{b_2}}}\]and \[y = \dfrac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_2}{b_1} - {a_1}{b_2}}}\]

Complete step-by-step answer:
The given equations are:-
\[2x + y = 5\]…………………………………….. (1)
\[3x - 4y = 2\]………………………………………. (2)
For equation 1:
\[{a_1} = 2;{b_1} = 1;{c_1} = 5\]
For equation 2:
\[{a_2} = 3;{b_2} = - 4;{c_2} = 2\]
Now we know that:
The formula for Paravartya method of Vedic mathematics in which roots of the two equations \[{a_1}x + {b_1}y = {c_1}\] and \[{a_2}x + {b_2}y = {c_2}\] are given by:-
\[x = \dfrac{{{b_1}{c_2} - {b_2}{c_{1}}}}{{{a_2}{b_1} - {a_1}{b_2}}}\]and \[y = \dfrac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_2}{b_1} - {a_1}{b_2}}}\]
Hence applying this formula for the given equations we get:-
\[x = \dfrac{{\left( 1 \right)\left( 2 \right) - \left( { - 4} \right)\left( 5 \right)}}{{\left( 3 \right)\left( 1 \right) - \left( 2 \right)\left( { - 4} \right)}}\]
Solving it further we get:-
\[x = \dfrac{{2 + 20}}{{3 + 8}}\]
\[ \Rightarrow x = \dfrac{{22}}{{11}}\]
\[ \Rightarrow x = 2\]
And, now putting the respective values in y we get:-
\[y = \dfrac{{\left( 5 \right)\left( 3 \right) - \left( 2 \right)\left( 2 \right)}}{{\left( 3 \right)\left( 1 \right) - \left( 2 \right)\left( { - 4} \right)}}\]
Solving it further we get:-
\[y = \dfrac{{15 - 4}}{{3 + 8}}\]
\[ \Rightarrow y = \dfrac{{11}}{{11}}\]
\[ \Rightarrow y = 1\]

Hence the roots are \[x = 2\] and \[y = 1\]

Note: Students can verify their roots by solving the given equation using elimination method.
The given equations are:-
\[2x + y = 5\]…………………………………….. (1)
\[3x - 4y = 2\]………………………………………. (2)
Multiplying equation 1 by 3 we get:-
\[6x + 3y = 15\]………………………… (3)
Multiplying equation 2 by 2 we get:-
\[6x - 8y = 4\]……………………………. (4)
Now subtracting equation 4 from equation 3 we get:-
\[
  {\text{ }}6x + 3y = 15 \\
   - \underline {\left( {6x - 8y = 4} \right)} \\
  {\text{ 11y = 11}} \\
 \]
\[ \Rightarrow y = 1\]
Putting this value in equation 1 we get:-
\[2x + 1 = 5\]
Solving for x we get:-
\[2x = 4\]
\[ \Rightarrow x = 2\]
 Hence the roots are \[x = 2\] and \[y = 1\]