Question

How do you solve $11\left( 4-6y \right)+5\left( 13y+1 \right)=9$?

Answer 1

Hint: The equation given in the question is a linear equation. So it will have only one solution. For solving the given equation, we first have to simplify the brackets. Then we have to write all of the equal degree terms together. On simplifying, we will get a simplified form of the linear equation, which can be easily solved.

Complete step by step answer:
The equation given in the question is
$11\left( 4-6y \right)+5\left( 13y+1 \right)=9$
As we can see in the above equation that the highest power of the variable is equal to one, so the given equation is linear. This means that the given equation will have only one solution.
Multiplying $11$ and $5$ with the respective brackets, we get
\[\begin{align}
  & \Rightarrow 11\left( 4 \right)+11\left( -6y \right)+5\left( 13y \right)+5\left( 1 \right)=9 \\
 & \Rightarrow 44-66y+65y+5=9 \\
\end{align}\]
Subtracting $9$ from both the sides of the above equation, we have
\[\begin{align}
  & \Rightarrow 44-66y+65y+5-9=9-9 \\
 & \Rightarrow 44-66y+65y-4=0 \\
\end{align}\]
Taking the equal degree terms together, we get
\[\begin{align}
  & \Rightarrow -66y+65y+44-4=0 \\
 & \Rightarrow -y+40=0 \\
\end{align}\]
Subtracting $40$ from both the sides of the above equation we get
\[\begin{align}
  & \Rightarrow -y+40-40=0-40 \\
 & \Rightarrow -y=-40 \\
\end{align}\]
Finally, multiplying by $-1$ on both the sides of the above equation, we have
$\Rightarrow y=40$

Note: Do not ignore the term present on the right hand side of the given equation. As we can clearly see, the right hand side is non zero. It is a common mistake to consider the right hand side of an equation to be zero. To avoid such mistakes, we must first make the right hand side of the given equation equal to zero, and then solve the equation. Also, the BODMAS rule, which tells us the order of the algebraic operations, must be taken care of while solving an equation.