Question

How do you solve `\frac { x + 1} { x + 2} + \frac { 1} { x } = \frac { 2x + 1} { x + 2}`?

Answer 1

`x = 2 " or "x =-1`

Explanation:

As soon as you have an equation which has fractions, you can get rid of the fractions altogether.

Multiply EACH term by the LCM of the denominators, which in this case is `color(blue)(x(x+2)`.

`(color(blue)(x(x+2))xx(x+1))/((x+2)) +(color(blue)(x(x+2))xx1)/x = (color(blue)(x(x+2))(2x+1))/((x+2))`

Now cancel the denominators.

`(color(blue)(xcancel((x+2)))xx(x+1))/cancel((x+2)) +(color(blue)(cancelx(x+2))xx1)/cancelx = (color(blue)(xcancel((x+2)))(2x+1))/cancel((x+2))`

This leaves us with an equation without fractions.

`x(x+1) +(x+1) = x(2x+1)" "larr` simplify

`" "x^2 +x +x+1 = 2x^2 +x" "larr` make the quadratic `= 0`

`" "x^2 -x -2 =0" "larr` find the factors

`(x-2)(x+1)=0`

Setting each factor equal to 0 gives:

`x = 2 " or "x =-1`

Answer 2

`color(red)(x=2` or `color(red)(x=-1`

Explanation:

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

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

Multiply L.H.S. and R.H.S. by `x(x+2)` to cancel the denominators

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

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

`:.x^2-2x^2+2x-x+2 =0`

`:.-x^2+x+2=0`

Multiply L.H.S. and R.H.S. by `-1`

`x^2-x-2=0`

`:.(x-2)(x+1)=0`

`:.x-2=0` or `x+1=0`

`:.color(red)(x=2` or `color(red)(x=-1`

substitute `color(red)(x=2`

`:.(color(red)2+1)/(color(red)2+2)+1/color(red)2=(2(color(red)2)+1)/(color(red)2+2)`

`:.(3)/(4)+1/2=(5)/(4)`

`:.(3+2=5)/4`

multiply by 4

`:.(3)/(4)+1/2=(5)/(4)`

`:.cancel4/1*(3+2=5)/cancel4`

`color(red)(5=5`

substitute `color(red)(x=-1`

`:.(color(red)(-1)+1)/(color(red)(-1)+2)+1/color(red)(-1)=(2color(red)(-1)+1)/(color(red)(-1)+2)`

`:.(0)/(1)+1/-1=(-2+1)/(-1+2)`

`:.(0)/(1)+1/-1=(-2+1)/(-1+2)`

`:.0-1=-1`

`:.color(red)(-1=-1`