Question

Solve the equation
$\frac{4x - 5}{x + 2} - \frac{8x - 1}{2x + 1} = 0$
Plzz

Answer 1

We start solving the expression by simplifying it.

Take the LCM in denominator and form a single fraction. After simplifying it, we have a simpler expression in the numerator and another expression in the denominator.

The RHS is still zero. So we equate the numerator to 0 and that gives the solution for x.

$\sf\displaystyle \frac{4x-5}{x+2}-\frac{8x-1}{2x+1}=0\\\\\\\implies \frac{(4x-5)(2x+1)-(8x-1)(x+2)}{(x+2)(2x+1)}=0\\\\\\\implies\frac{(8x^2+4x-10x-5)-(8x^2+16x-x-2)}{(x+2)(2x+1)}=0\\\\\\\implies\frac{8x^2-6x-5-8x^2-15x+2}{(x+2)(2x+1)}=0\\\\\\\implies\frac{-21x-3}{(x+2)(2x+1)}=0\\\\\\\implies\frac{-3(7x+1)}{(x+2)(2x+1)}=0\\\\\\\implies\frac{7x+1}{(x+2)(2x+1)}=0$

Now, denominator cannot be zero, since division by zero is not defined.

So we have two conditions first:

$\sf x+2\neq 0 \qquad \textsf{AND}\qquad 2x+1\neq 0 \\\\\implies x\neq 2 \qquad \textsf{AND}\qquad x\neq -\dfrac{1}{2}$

Since the RHS is zero, the numerator must be zero.

So,

$\sf 7x+1=0 \\\\\\ \implies 7x = -1 \\\\\\ \implies \huge \boxed{\sf x=-\frac{1}{7}}$

This agrees with the two non-equality conditions we put above, that the value of x cannot be -2 or $-\frac{1}{2}$

Hence, The solution is $\bold{-\dfrac{1}{7}}$