Question

How do you solve $\dfrac{x}{2}+\dfrac{y}{3}=6$ and $\dfrac{x}{3}+\dfrac{y}{2}=12$?

Answer 1

Hint: Multiply $\dfrac{1}{3}$ with the first equation and $\dfrac{1}{2}$ with the second equation to make the coefficient of ‘x’ equal. Then subtract both the equations to get the value of ‘y’. Put that value of ‘y’ in any of the equations to get the value of ‘x’.

Complete step-by-step answer:
The equations we have
$\dfrac{x}{2}+\dfrac{y}{3}=6$……….(1)
$\dfrac{x}{3}+\dfrac{y}{2}=12$……….(2)
Multiplying equation (1) by $\dfrac{1}{3}$, we get
$eq(1)\times \dfrac{1}{3}\Rightarrow \dfrac{1}{3}\left( \dfrac{x}{2}+\dfrac{y}{3} \right)=\dfrac{1}{3}\times 6$
$\Rightarrow \dfrac{x}{6}+\dfrac{y}{9}=2$……….(3)
Multiplying equation (2) by $\dfrac{1}{2}$, we get
$eq(2)\times \dfrac{1}{2}\Rightarrow \dfrac{1}{2}\left( \dfrac{x}{3}+\dfrac{y}{2} \right)=\dfrac{1}{2}\times 12$
$\Rightarrow \dfrac{x}{6}+\dfrac{y}{4}=6$……….(4)
Subtracting equation (4) from equation (3), we get
$\begin{align}
  & \Rightarrow \left( \dfrac{x}{6}+\dfrac{y}{9} \right)-\left( \dfrac{x}{6}+\dfrac{y}{4} \right)=2-6 \\
 & \Rightarrow \dfrac{x}{6}+\dfrac{y}{9}-\dfrac{x}{6}-\dfrac{y}{4}=-4 \\
 & \Rightarrow \dfrac{y}{9}-\dfrac{y}{4}=-4 \\
 & \Rightarrow \dfrac{4y-9y}{36}=-4 \\
 & \Rightarrow -5y=-144 \\
 & \Rightarrow y=\dfrac{-144}{-5} \\
 & \Rightarrow y=\dfrac{144}{5} \\
\end{align}$
Putting the value of ‘y’ in equation (1), we get
$\begin{align}
  & \dfrac{x}{2}+\dfrac{\dfrac{144}{5}}{3}=6 \\
 & \Rightarrow \dfrac{x}{2}+\dfrac{48}{5}=6 \\
 & \Rightarrow \dfrac{x}{2}=6-\dfrac{48}{5} \\
 & \Rightarrow \dfrac{x}{2}=\dfrac{30-48}{5} \\
 & \Rightarrow \dfrac{x}{2}=\dfrac{-18}{5} \\
 & \Rightarrow x=\dfrac{-36}{5} \\
\end{align}$
So, the solution of the equations $\dfrac{x}{2}+\dfrac{y}{3}=6$ and $\dfrac{x}{3}+\dfrac{y}{2}=12$ is $x=\dfrac{-36}{5}$ and $y=\dfrac{144}{5}$.

Note: The equations can also be brought to linear form. Multiplying both the equations by ‘6’ we obtain the equations as $3x+2y=36$ and $2x+3y=72$. Now, the equations can be solved by equating the coefficients of ‘x’ or ‘y’ and then putting the value of one of them in any of the equations to obtain the other one.