Question

How do you solve the equation $\dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2{y^2} - 7y - 4}}$ and find the value of y ?

Answer 1

We will factorize the denominator of the right-hand side using the mid-term split method. We are doing a mid-term split to reduce the equation. We will try to simplify the equation by using formulas like ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and find the value of y.

Complete step by step answer:
We have to find the solution of equation $\dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2{y^2} - 7y - 4}}$
$\Rightarrow \dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2{y^2} - 7y - 4}}$
We will factorize the denominator $2{y^2} - 7y - 4$on the right-hand side.
We will use the mid-term split formula
Factorization of $2{y^2} - 7y - 4$ is $2{y^2} - 8y + y - 4$
So, the equation becomes
$\Rightarrow \dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2{y^2} - 8y + y - 4}}$
We have taken 2y and 1 common
$\Rightarrow \dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2y(y - 4) + 1(y - 4)}}$
$\Rightarrow \dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{(2y + 1)(y - 4)}}$
We will multiply the both side by $y - 4$ , we get
$\Rightarrow (1 + 2y) \times (1 + 2y) = 4{y^2} + 5y$
We know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\Rightarrow 4{y^2} + 1 + 4y = 4{y^2} + 5y$
$\Rightarrow y = 1$
Hence, the solution of equation $\dfrac{{1 + 2y}}{{y - 4}} = \dfrac{{4{y^2} + 5y}}{{2{y^2} - 7y - 4}}$ is 1.

Note: We should follow all rules of linear equations. We can also solve this equation directly by cross multiplication but it will become very lengthy so, before solving any equation we should try to reduce the equation. We should follow the BODMAS rule.