Question

Solve :
(x+1) (4x-3) -4x^2+3/(4x-1)​

Answer 1

Step by step explanation:-

Given to solve :-

$\sf\dfrac{(x+1)(4x-3) -4x^2 +3}{4x-1}$

Solution:-

Lets do !

$\sf\dfrac{(x+1)(4x-3) -4x^2 +3}{4x-1}$

$\sf\dfrac{x(4x-3)+1(4x-3) -4x^2+3}{4x-1}$

$\sf\dfrac{4x^2-3x +4x -3 -4x^2+3}{4x-1}$

Keep like terms together

$\sf\dfrac{4x^2 - 4x^2 -3x + 4x +3 -3 }{4x-1}$

$\sf\dfrac{ x}{4x-1}$

So,

$\sf\dfrac{(x+1)(4x-3) -4x^2+3}{4x-1}$ = $\sf\dfrac{ x}{4x-1}$

Know more :-

Transportations :-

If we transpose + after transpose it becomes -

If we tranpose - after tranpose it becomes +

If we tranpose × after tranpose it becomes ÷

If we tranpose ÷ after tranpose it becomes ×

Multiplication of signs:-

(+) ( -) = -

(-) ( +) = -

( + ) ( + ) = +

(-) (-) = +