Question

How do you solve \[\dfrac{3}{x+2}=\dfrac{6}{x-1}\]?

Answer 1

Hint: Cross – multiply the numerator of one side of the equation with the denominator of the other side of the equation to simplify the expression. Now, take the terms containing ‘x’ to the left-hand side and the constant terms to the right-hand side. Apply simple mathematical operations like addition, subtraction, multiplication, and division, whichever needed to simplify the equation. Find the value of ‘x’ to get the answer.

Complete step by step answer:
Here, we have been provided with the equation: \[\dfrac{3}{x+2}=\dfrac{6}{x-1}\] and we are asked to solve this equation. That means we have to find the value of x.
Clearly, we can see that the given equation is a linear equation in one variable which is ‘x’, so now cross – multiplying the numerator of one side of the equation with the denominator of the other side of the equation, we get,
\[\Rightarrow 3\left( x-1 \right)=6\left( x+2 \right)\]
Removing the bracket by multiplying the factor with the respective terms, we get,
\[\Rightarrow 3x-3=6x+12\]
Now, taking the terms containing the variable ‘x’ to the left hand side (L.H.S) and the constant terms to the right hand side (R.H.S), we get,
\[\begin{align}
  & \Rightarrow 3x-6x=12+3 \\
 & \Rightarrow -\left( 6x-3x \right)=15 \\
 & \Rightarrow -3x=15 \\
\end{align}\]
Dividing both sides with 3, we get,
\[\Rightarrow \dfrac{-3x}{3}=\dfrac{15}{3}\]
Cancelling the common factors, we get,
\[\Rightarrow -x=5\]
Multiplying both the sides with (-1), we get,
\[\begin{align}
  & \Rightarrow \left( -1 \right)\times \left( -x \right)=\left( -1 \right)\times 5 \\
 & \Rightarrow x=-5 \\
\end{align}\]
Hence, the value of x is -5.

Note:
One may note that we have been provided with a single equation only. The reason is that we have to find the value of only one variable, that is x. So, if we have to solve an equation having ‘n’ number of variables then we should be provided with ‘n’ equations. Now, one can check the answer by substituting the obtained value of ‘x’ in the equation provided in the question. We have to determine the value of L.H.S and R.H.S separately and if they are equal then our answer is correct.