3 years ago

Im not quite sure how to solve this.

Mathematics

2 answers:

Tju [1.3M]3 years ago

7 0

The answer is y=0
I am pretty sure that is right and if it’s wrong i am so sorry

Olin [163]3 years ago

3 0

Step-by-step explanation:

yeap I also think so that the answer is 0

You might be interested in

2x + 3/ x - 4 - 2x - 8/ 2× + 1 = 1

Debora [2.8K]

$\dfrac{2x+3}{x-4}-\dfrac{2x-8}{2x+1}=1\ \ \ |the\ domain\ D:x\neq4\ and\ x\neq-\dfrac{1}{2}\\\\\dfrac{(2x+3)(2x+1)}{(x-4)(2x+1)}-\dfrac{(2x-8)(x-4)}{(2x+1)(x-4)}-1=0\\\\\dfrac{4x^2+2x+6x+3}{2x^2+x-8x-4}-\dfrac{2x^2-8x-8x+32}{2x^2+x-8x-4}-\dfrac{2x^2+x-8x-4}{2x^2+x-8x-4}=0\\\\\dfrac{4x^2+8x+3}{2x^2-7x-4}-\dfrac{2x^2-16x+32}{2x^2-7x-4}-\dfrac{2x^2-7x-4}{2x^2-7x-4}=0\\\\\dfrac{4x^2+8x+3-2x^2+16x-32-2x^2+7x+4}{2x^2-7x-4}=0\\\\\dfrac{31x-25}{2x^2-7x-4}=0\iff31x-25=0\\\\31x=25\\\\\boxed{x=\frac{25}{31}}$