Question

Solve the following equations and verify your answers:
\[\dfrac{2x-3}{3x+2}=-\dfrac{2}{3}\]

Answer 1

Hint: In this question, we have to find the value of x and for that, we have to cross multiply the numerator and denominator and after finding the value of x, we will put that value in the question, and then we will verify the value of x. After putting the obtained value of x in the equation given in the question, the left-hand side will be proved equal to the right-hand side.

Complete step by step answer:
In mathematics, the fraction is represented in the form of numerator and denominator. The denominator represents the total number and the numerator represents the part of the denominator. Variable and also present in the form of fractions and the value of a variable can be obtained by cross multiplication of the fraction on both sides. A fraction can be a proper fraction or an improper fraction. Proper fractions are the fractions in which the degree of the numerator is less than the denominator and in improper fraction, the degree of the denominator is less than the degree of the numerator.
In the above question, we have to solve the given equation and the equation is as shown below.
\[\dfrac{2x-3}{3x+2}=-\dfrac{2}{3}\]
We have to find the value of x and for that, we have to cross multiply the numerator and denominator of the above equation.
\[3(2x-3)=-2(3x+2)\]
\[3\] will be multiplied with \[(2x-3)\] and \[-2\] will be multiplied with \[(3x+2)\].
\[6x-9=-6x-4\]
So all the terms with variable x will be on one side and the constant terms will be on the other side.
\[\begin{align}
  & 6x+6x=9-4 \\
 & \Rightarrow 12x=5 \\
 & \Rightarrow x=\dfrac{5}{12} \\
\end{align}\]
So the value of x comes out to be \[\dfrac{5}{12}\]. Now we have to verify the value of x. So we will put the value of x in the equation \[\dfrac{2x-3}{3x+2}=-\dfrac{2}{3}\].
In this equation, we have to prove RHS equal to LHS.
We have to check that after putting the value of x in LHS it will be equal to RHS.

LHS is \[\dfrac{2x-3}{3x+2}\].
On putting the value of \[x=\dfrac{5}{12}\].
\[\begin{align}
  & \dfrac{2\left( \dfrac{5}{12} \right)-3}{3\left( \dfrac{5}{12} \right)+2} \\
 & \Rightarrow \dfrac{\dfrac{5}{6}-3}{\dfrac{5}{4}+2} \\
\end{align}\]
On taking the LCM in the denominator.
\[\dfrac{\dfrac{5-18}{6}}{\dfrac{5+8}{4}}\]
\[\Rightarrow \dfrac{\dfrac{-13}{6}}{\dfrac{13}{4}}\]
\[\Rightarrow -\dfrac{2}{3}\]
As we see, LHS becomes equal to RHS after putting the value of \[x=\dfrac{5}{12}\]. Hence it is proved that the value of x is correct.

Note:
 The numbers that are not represented in the form of fractions are known as irrational numbers. If we do the decimal expansion of an irrational number then we will get a repeating and non terminating decimal expansion. A unit fraction is the fraction in which the denominator is one.