Question

PLEASE ANSWER IN UNDER 5mins!!!!!!!!!!!!!!!! pls help i appreciate you!

Answer 1

`\textsf{B)} \quad (x^2+3x-4)/(2x-1)`

Explanation:

Given rational expression:

`((x^2+2x-3)/(x-4))/((2x^2+5x-3)/(x^2-16))`

`\textsf{Factor}\;\;x^2+2x-3:`

`\implies x^2+3x-x-3`

`\implies x(x+3)-1(x+3)`

`\implies (x-1)(x+3)`

`\textsf{Factor}\;\;2x^2+5x-3:`

`\implies 2x^2+6x-x-3`

`\implies 2x(x+3)-1(x+3)`

`\implies (2x-1)(x+3)`

`\textsf{Factor}\;\;x^2-16:`

`\implies x^2-4x+4x-16`

`\implies x(x-4)+4(x-4)`

`\implies (x+4)(x-4)`

Therefore:

`\implies ((x^2+2x-3)/(x-4))/((2x^2+5x-3)/(x^2-16))=(((x-1)(x+3))/(x-4))/(((2x-1)(x+3))/((x+4)(x-4)))`

`\textsf{Apply\:the\:fraction\:rule}:\quad ((a)/(b))/((c)/(d))=(a\cdot \:d)/(b\cdot \:c)`

`\implies ( (x-1)(x+3)(x+4)(x-4))/((x-4)(2x-1)(x+3))`

Cancel the common factors:

`\implies ( (x-1)(x+4))/((2x-1))`

Expand the numerator:

`\implies (x^2+3x-4)/(2x-1)`