Question

Solve: $ \dfrac{x}{3}+\dfrac{y}{4}=11 $ , $ \dfrac{5x}{6}-\dfrac{y}{3}=-7 $ .

Answer 1

Hint: There are two unknowns $ x $ and $ y $ to solve. We simplify the equations and solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable. First, we are applying the process of reduction and then the substitution.

Complete step-by-step answer:
The given equations $ \dfrac{x}{3}+\dfrac{y}{4}=11 $ and $ \dfrac{5x}{6}-\dfrac{y}{3}=-7 $ .
We simplify them to get $ 4x+3y=132 $ and $ 5x-2y=-42 $ which are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We take the equations as $ 4x+3y=132.....(i) $ and $ 5x-2y=-42......(ii) $ .
We multiply 2 to the both sides of the first equation and get
 $ \begin{align}
  & 2\times \left( 4x+3y \right)=132\times 2 \\
 & \Rightarrow 8x+6y=264 \\
\end{align} $
We take the equation as $ 8x+6y=264.....(iii) $ .
We multiply 3 to the both sides of the first equation and get
 $ \begin{align}
  & 3\times \left( 5x-2y \right)=3\times \left( -42 \right) \\
 & \Rightarrow 15x-6y=-126 \\
\end{align} $
We take the equation as $ 15x-6y=-126.....(iv) $ .
Now we add the equation (iii) to equation (iv) and get
 $ \left( 8x+6y \right)+\left( 15x-6y \right)=264-126 $ .
We take the variables together and the constants on the other side.
Simplifying the equation, we get
 $ \begin{align}
  & \left( 8x+6y \right)+\left( 15x-6y \right)=264-126 \\
 & \Rightarrow 23x=138 \\
 & \Rightarrow x=\dfrac{138}{23}=6 \\
\end{align} $
The value of $ x $ is 6. Now putting the value in the equation $ 4x+3y=132.....(i) $ , we get
 $ \begin{align}
  & 4x+3y=132 \\
 & \Rightarrow y=\dfrac{132-4\times 6}{3}=36 \\
\end{align} $ .
Therefore, the values are $ x=6,y=36 $ .
So, the correct answer is “ $ x=6,y=36 $ ”.

Note: We can also find the value of one variable $ y $ with respect to $ x $ based on the equation
 $ 4x+3y=132 $ where $ y=\dfrac{132-4x}{3} $ . We replace the value of $ y $ in the second equation of
 $ 5x-2y=-42 $ and get
\[\begin{align}
  & 5x-2y=-42 \\
 & \Rightarrow 5x-2\left( \dfrac{132-4x}{3} \right)=-42 \\
 & \Rightarrow 15x-264+8x=-126 \\
\end{align}\]
We get the equation of $ x $ and solve
\[\begin{align}
  & 15x-264+8x=-126 \\
 & \Rightarrow 23x=264-126=138 \\
 & \Rightarrow x=\dfrac{138}{23}=6 \\
\end{align}\]
Putting the value of $ x $ we get $ 4x+3y=132\Rightarrow y=\dfrac{132-4\times 6}{3}=36 $ .
Therefore, the values are $ x=6,y=36 $ .