Question

How do you solve the linear equation $\dfrac{240+30x}{16+x}=18$?

Answer 1

Hint: The given equation is a linear equation in the variable $x$. Therefore, it will have only one solution. Since the equation contains the variable terms in the fractional form, we must get rid of the fraction in order to simplify it. For that, we must multiply the given equation $\dfrac{240+30x}{16+x}=18$ by the denominator term $\left( 16+x \right)$ on both sides. Then using the basic algebraic operations on the equation obtained, we will be able to solve it.

Complete step by step solution:
The equation given to us in the above question is written as
$\Rightarrow \dfrac{240+30x}{16+x}=18$
As we can see in the above equation that its LHS contains variable terms in the form of fraction. Due to this the equation is looking cumbersome. So in order to simplify it, we multiply both sides of the above equation by the denominator term $\left( 16+x \right)$ to get
\[\begin{align}
  & \Rightarrow \dfrac{240+30x}{16+x}\left( 16+x \right)=18\left( 16+x \right) \\
 & \Rightarrow 240+30x=18\left( 16+x \right) \\
\end{align}\]
From the distributive property, we know that \[a\left( b+c \right)=ab+ac\]. So we can simplify the RHS of the above equation as
$\begin{align}
  & \Rightarrow 240+30x=18\times 16+18x \\
 & \Rightarrow 240+30x=288+18x \\
\end{align}$
Subtracting $240$ from both the sides, we get
\[\begin{align}
  & \Rightarrow 240+30x-240=288+18x-240 \\
 & \Rightarrow 30x=18x+48 \\
\end{align}\]
Now we subtract \[18x\] from both the sides to get
$\begin{align}
  & \Rightarrow 30x-18x=18x+48-18x \\
 & \Rightarrow 12x=48 \\
\end{align}$
Finally, on dividing both the sides by $12$ we get
$\begin{align}
  & \Rightarrow \dfrac{12x}{12}=\dfrac{48}{12} \\
 & \Rightarrow x=4 \\
\end{align}$
Hence, the solution of the given equation is $x=4$.

Note: Since the given equation involves variable terms in the denominator, we may try to obtain the domain of the given equation by using the condition $\left( 16+x \right)\ne 0$. But we must note that since the given equation is a linear equation, there is no need to determine the domain. It is only required when the degree of the equation is two or more.