Question

When a certain number $^{'}{{m}^{'}}$ is divided by 5 and added to 8 the result is equal to $3m$ subtracted from 4. Then $m=$\[\]
A.2\[\]
B.$\dfrac{4}{3}$\[\]
C.$\dfrac{-1}{3}$\[\]
D. $\dfrac{-5}{4}$\[\]

Answer 1

Hint: We use the given information and build two algebraic expressions in $m$ which are $\dfrac{m}{5}+8$ and $4-3m$. We equate the expressions as given in the question and the solve resulting in the linear equation in one variable for $m.$\[\]

Complete step-by-step solution
We are given the question that $m$ is a certain number which means $m$ is not variable like $x,y,z$ written in our textbooks and cannot assume any random value. It is fixed. We are also given that it is divided by and then 5 and added to 8. If we divide $m$ by, we can write it as $\dfrac{m}{5}$ and when we add 8 to it we can write it as $\dfrac{m}{5}+8$. We have our first expression in $m$ as
\[\dfrac{m}{5}+8...\left( 1 \right)\]
We are given in the question that $3m$subtracted from 4 which we can write as $4-3m.$ We have our second expression in $m$ as
\[4-3m...\left( 2 \right)\]
We are given the question that the expressions (1) and (2) are equal. So we have
\[\dfrac{m}{5}+8=4-3m....\left( 3 \right)\]
We know that when we solve a linear equation in one variable we can add , subtract , multiply and divide the same number both left and right hand of the equation. We have to separate the terms unknown $m$ and the known numbers in equation (3). In order to do that we first subtract 8 both side of equation (3).We have
\[\begin{align}
  & \dfrac{m}{5}+8-8=4-3m-8 \\
 & \Rightarrow \dfrac{m}{5}=-4-3m \\
\end{align}\]
We add $3m$ both side and have,
\[\begin{align}
  & \Rightarrow \dfrac{m}{5}+3m=-4-3m+3m \\
 & \Rightarrow \dfrac{m}{5}+3m=-4 \\
\end{align}\]
We take $m$ common in the left hand side and add the rational numbers. We have,
\[\begin{align}
  & \Rightarrow m\left( \dfrac{1}{5}+3 \right)=-4 \\
 & \Rightarrow m\left( \dfrac{1+15}{5} \right)=-4 \\
 & \Rightarrow m\times \dfrac{16}{5}=-4 \\
\end{align}\]
We divide both side by $\dfrac{16}{5}$ and have
\[\begin{align}
  & \Rightarrow \dfrac{m\times \dfrac{16}{5}}{\dfrac{16}{5}}=\dfrac{-4}{\dfrac{16}{5}} \\
 & \Rightarrow m=-4\times \dfrac{5}{16}=\dfrac{-5}{4} \\
\end{align}\]
So the correct choice is D.

Note: We note that we can divide both sides of the equation by only a nonzero number. We can also take the same exponent on both sides of the equation. We need at least one linear equation for one variable, two equations for two variables and so on to find the solution.