`(x+1)/(x+6)+1/x=(2x+1)/(x+6)` Solve the equation by using the LCD. Check for extraneous solutions.

`(x+1)/(x+6)+1/x=(2x+1)/(x+6)`

LCD is `x(x+6)`

Multiply all the terms of the equation by LCD and simplify,

`x(x+6)((x+1)/(x+6))+x(x+6)(1/x)=x(x+6)((2x+1)/(x+6))`

`x(x+1)+(x+6)=x(2x+1)`

`x^2+x+x+6=2x^2+x`

`x^2+2x+6=2x^2+x`

`x^2+2x+6-2x^2-x=0`

`-x^2+x+6=0`

Factorize the above equation,

`-1(x^2-x-6)=0`

`(x^2+2x-3x-6)=0`

`(x(x+2)-3(x+2))=0`

`(x+2)(x-3)=0`

use the zero product property,

`x+2=0` or `x-3=0`

`x=-2` or `x=3`

Let's check the solutions by plugging them in the original equation,

For x=-2,

`(-2+1)/(-2+6)+1/(-2)=((2(-2)+1))/(-2+6)`

`(-1)/4-1/2=(-3)/4`

`-3/4=-3/4`

It's true.

For x=3,

`(3+1)/(3+6)+1/3=(2(3)+1)/(3+6)`

`4/9+1/3=7/9`

`7/9=7/9`

It's true.

So, the solutions of the equation are 3 and -2.

Comments

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