Question

By using the coefficient of variable, solve the following equations:
(1) $3x - 4y = 7;$$5x + 2y = 3$
(2) $5x + 7y = 17;$$3x - 2y = 4$
(3) $x - 2y = - 10;$ $3x - 5y = - 12$
(4) $4x + y = 34;$ $x + 4y = 16$

Answer 1

Hint: If the equation is given in the form as ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ then the solution is given by:
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  {{b_1}}&{{c_1}} \\
  {{b_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$
Now $\left| {\begin{array}{*{20}{c}}
  {{b_1}}&{{c_1}} \\
  {{b_2}}&{{c_2}}
\end{array}} \right|$ represents the determinant

Complete step-by-step answer:
(1) Here we are given two equations like
$3x - 4y = 7;$ $5x + 2y = 3$
So first make it in the form $ax + by + c = 0$
So $3x - 4y -7 = 0;$ $5x + 2y -3 = 0$
So we can solve the solution of two linear lines ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ as follows:
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  {{b_1}}&{{c_1}} \\
  {{b_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$
So we can write it as
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  { - 4}&{ - 7} \\
  2&{ - 3}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  3&{ - 7} \\
  5&{ - 3}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  5&2
\end{array}} \right|}}$
Now upon solving, we get:
$\dfrac{x}{{( - 4)( - 3) - ( - 7)(2)}} = \dfrac{{ - y}}{{3( - 3) + 7(5)}} = \dfrac{1}{{3(2) + 4(5)}}$
$\dfrac{x}{{26}} = \dfrac{{ - y}}{{26}} = \dfrac{1}{{26}}$
So we can write it as
$\dfrac{x}{{26}} = \dfrac{1}{{26}}$ and $\dfrac{{ - y}}{{26}} = \dfrac{1}{{26}}$
So we get on solving these that
$x = 1,y = - 1$
(2) $5x + 7y = 17;$ $3x - 2y = 4$
So again we are given two equations $5x + 7y = 17;$ $3x - 2y = 4$
We apply the formula
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  7&{ - 17} \\
  { - 2}&{ - 4}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  5&{ - 17} \\
  3&{ - 4}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  5&7 \\
  3&{ - 2}
\end{array}} \right|}}$
$\dfrac{x}{{(7)( - 4) - ( - 17)( - 2)}} = \dfrac{{ - y}}{{5( - 4) + 17(3)}} = \dfrac{1}{{5( - 2) - 3(7)}}$
$\dfrac{x}{{ - 62}} = \dfrac{{ - y}}{{31}} = \dfrac{1}{{ - 31}}$
So we can write it as:
$\dfrac{x}{{ - 62}} = \dfrac{1}{{ - 31}}$ and $\dfrac{{ - y}}{{31}} = \dfrac{1}{{ - 31}}$
So we get $x = 2,y = 1$
(3) $x - 2y = - 10;$ $3x - 5y = - 12$
Now again we will solve as we are given two equations $x - 2y = - 10;$ $3x - 5y = - 12$
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  { - 2}&{10} \\
  { - 5}&{12}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  1&{10} \\
  3&{12}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  3&{ - 5}
\end{array}} \right|}}$
$\dfrac{x}{{( - 2)(12) - (10)( - 5)}} = \dfrac{{ - y}}{{12(1) - 10(3)}} = \dfrac{1}{{1( - 5) - ( - 2)(3)}}$
$\dfrac{x}{{26}} = \dfrac{{ - y}}{{ - 18}} = \dfrac{1}{1}$
So we can write it as
$\dfrac{x}{{26}} = \dfrac{1}{1}$ and $\dfrac{{ - y}}{{ - 18}} = \dfrac{1}{1}$
Hence we get that $x = 26,y = 18$
(4) $4x + y = 34;$ $x + 4y = 16$
First we take the equation of the form $ax + by + c = 0$
So we are given two equations $4x + y = 34;$ $x + 4y = 16$
$\dfrac{x}{{\left| {\begin{array}{*{20}{c}}
  1&{ - 34} \\
  4&{ - 16}
\end{array}} \right|}} = \dfrac{{ - y}}{{\left| {\begin{array}{*{20}{c}}
  4&{ - 34} \\
  1&{ - 16}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  1&4 \\
  4&1
\end{array}} \right|}}$
$\dfrac{x}{{( - 16) - ( - 34)(4)}} = \dfrac{{ - y}}{{4( - 16) - ( - 34)}} = \dfrac{1}{{16 - 1}}$
$\dfrac{x}{{120}} = \dfrac{{ - y}}{{ - 30}} = \dfrac{1}{{15}}$
So we can write this as:
$\dfrac{x}{{120}} = \dfrac{1}{{15}}$ and $\dfrac{{ - y}}{{ - 30}} = \dfrac{1}{{15}}$
So we get that $x = 8,y = 2$

Note: We can do this question by the substitution method also like in the
(1) $3x - 4y = 7;$ $5x + 2y = 3$
So first we can find $x$ in the terms of $y$ and then put this in the second equation.
For example: $3x - 4y = 7;$
$x = \dfrac{{7 + 4y}}{3}$ and now put this value in the second equation:
$5(\dfrac{{7 + 4y}}{3}) + 2y = 3$ and now you can get the value of both $x,y.$