Question

Add.4x−3/x2−2x+ −3x+3/x2−2xA) 7x−6/x2−2xB) 1/x-2C) x/x-2D) 1/2

Answer 1

Given:-

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

To find:-

The simplified form.

Since the denominators are same. we add the numerator simpily. so we get,

`\begin{gathered} (4x-3)/(x^2-2x)+(-3x+3)/(x^2-2x)=(4x-3-3x+3)/(x^2-2x) \\ \text{ =}(x)/(x^2-2x) \end{gathered}`

So now we cancel one x from the numerator and denominator. so we get,

`\begin{gathered} (x)/(x^2-2x)=(x)/(x(x-2)) \\ \text{ =}(1)/((x-2)) \end{gathered}`

So the simplified value is,

`(1)/((x-2))`