Question

How do you solve the following system \[\dfrac{1}{6}x - \dfrac{3}{5}y = 1,\dfrac{4}{5}x + \dfrac{2}{3}y = 2\] ?

Answer 1

Hint: In order to determine the solution of a given system of equations having two variables, use the method of elimination of term by eliminating the $y$term by making the mod of coefficient of $y$ in both the equations equal. Then apply the operation of addition or subtraction according to the equations obtained to eliminate $y$term. Solve the result for $x$and put the obtained value of $x$in any of the equations given to get the value of $y$.

Complete step by step solution:
 We are given pair of linear equation in two variables \[\dfrac{1}{6}x - \dfrac{3}{5}y = 1,\dfrac{4}{5}x + \dfrac{2}{3}y = 2\]
 \[\dfrac{1}{6}x - \dfrac{3}{5}y = 1\] ---(1)
 \[\dfrac{4}{5}x + \dfrac{2}{3}y = 2\] ----(2)
In order to solve the system of equations, we have many methods like, substitution, elimination of term, and cross-multiplication.
Here we will be using an elimination method to eliminate the term having $y$ from both the equations.
And to do so we have to first make the mod of coefficient of $y$ in both the equation equal to each.
Multiplying both side of the equation (1) with the number $\dfrac{5}{3}$, we get
 \[\dfrac{5}{3}\left( {\dfrac{1}{6}x - \dfrac{3}{5}y} \right) = 1\left( {\dfrac{5}{3}} \right)\]
 \[\dfrac{5}{{18}}x - y = \dfrac{5}{3}\] ------(3)
 and multiplying both sides of the equation (2) with number $\dfrac{3}{2}$, we get
 \[\dfrac{3}{2}\left( {\dfrac{4}{5}x + \dfrac{2}{3}y} \right) = \dfrac{3}{2}\left( 2 \right)\]
 \[\dfrac{6}{5}x + y = 3\] ----------(4)
Now adding the equation (3) and (4), we get
  \[\dfrac{5}{{18}}x - y + \dfrac{6}{5}x + y = 3 + \dfrac{5}{3}\]
Combining like terms we get
 \[
  \dfrac{5}{{18}}x + \dfrac{6}{5}x = 3 + \dfrac{5}{3} \\
  \dfrac{{5\left( 5 \right) + 6\left( {18} \right)}}{{90}}x\, = \dfrac{{9 + 5}}{3} \\
  \dfrac{{25 + 108}}{{90}}x\, = \dfrac{{14}}{3} \\
  \dfrac{{133}}{{90}}\,x = \dfrac{{14}}{3} \\
  \,x = \dfrac{{14}}{3}\left( {\dfrac{{90}}{{133}}} \right) \\
  x = \dfrac{{420}}{{133}} \\
  x = \dfrac{{60}}{{19}} \;
 \]
Hence, we have obtained the value of \[x = \dfrac{{60}}{{19}}\] .
Now putting this value of $x$ in the equation (1) to get the value of $y$
 \[
  \dfrac{1}{6}\left( {\dfrac{{60}}{{19}}} \right) - \dfrac{3}{5}y = 1 \\
  \dfrac{{10}}{{19}} - \dfrac{3}{5}y = 1 \\
  \dfrac{3}{5}y = \dfrac{{10}}{{19}} - 1 \\
  y = \dfrac{5}{3}\left( {\dfrac{{10 - 19}}{{19}}} \right) \\
  y = \dfrac{5}{3}\left( {\dfrac{{ - 9}}{{19}}} \right) \\
  y = - \dfrac{{15}}{{19}} \;
 \]
Therefore, the solution of system of given equations is $x = \dfrac{{60}}{{19}},y = - \dfrac{{15}}{{19}}$
So, the correct answer is “$x = \dfrac{{60}}{{19}},y = - \dfrac{{15}}{{19}}$”.

Note: Linear Equation in two variable: A linear equation is a equation which can be represented in the form of $ax + by + c$ where $x$ and $y$ are the unknown variables and c is the number known where $a \ne 0,b \ne 0$.
The degree of the variable in the linear equation is of the order 1.
1. One must be careful while calculating the answer as calculation error may come.
2. Solution of two linear equations can be done by using elimination method, substitution method and cross multiplication method.
3.It’s not compulsory to always eliminate the term with $y$. We have to judge according to the equation whose term elimination cost is less.