Question

Solve the following: $\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$

Answer 1

Hint: In this question, we have been given a linear equation in one variable y. So, to start the solving of the equation, we should cross multiply and subtract some terms from both sides of the equation to get all the terms with y on one side of the equation and numerical terms on the other side. Finally divide both sides of the equation by the coefficient of y to get the value of y.

Complete step-by-step answer:
The given equation is
 $\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$
Now to convert the equation to the standard form of a linear equation, we will cross-multiply. On doing so, we get
$3\left( 7y+4 \right)=-4\left( y+2 \right)$
Now we will use the distributive property of the multiplication over addition to open the brackets.
$21y+12=-4y-8$
We know that an equation remains valid if we add or subtract the same number both in the Left Hand Side (LHS) and Right Hand Side (RHS). Therefore, we should subtract a number on both sides such that only y containing terms remains on the LHS. So, we will subtract (12-4y) from both sides of the equation. On doing so, we get
$\begin{align}
  & 21y+12-\left( 12-4y \right)=-4y-8-\left( 12-4y \right) \\
 & \Rightarrow 21y+12-12+4y=-4y-8-12+4y \\
 & \Rightarrow 25y=-20 \\
\end{align}$
Now we will divide both sides of the equation by 25. On doing so, we get
$\begin{align}
  & \dfrac{25}{25}y=\dfrac{-20}{25} \\
 & \Rightarrow y=\dfrac{-4}{5} \\
\end{align}$
Therefore, the possible value of y is $y=\dfrac{-4}{5}$ .

Note: For verifying your answer, put the value of the variable you get in the parent equation and check whether the left-hand side of the equation is equal to the right-hand side of the equation or not. For example if we put the value $y=\dfrac{-4}{5}$ in the equation $\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$ , we get $LHS=RHS=\dfrac{-4}{3}$ . You should also be very careful about the domain of the equation, as in the above equation y must not be equal to -2 because this would make the denominator of the left-hand side of the equation 0 making the left-hand side undefined.