Question

How do you solve `2/(x+1) + 5/(x-2)=-2`?

Answer 1

`x=-3,1/2`

Explanation:

1. Find the least common denominator of the left side. Since we have expressions containing variables, we just multiply both denominators:
   `\color(red)((x+1)(x-2))`
   by the way, if you FOIL this it becomes `\color(orchid)(x^2-x-2)`

2. Multiply everything by that least common denominator.
   #\color(red)((x+1)(x-2))[2/(x+1)+5/(x-2)]=\color(red)((x+1)(x-2)) (-2)#

3. Apply the distributive property:

4. #(\color(red)((x+1)(x-2))(2))/(x+1)+(\color(red)((x+1)(x-2)) (5))/(x-2)=(\color(orchid)(x^2-x-2)) (-2)#

5. #((x+1)(x-2)(2))/(x+1)+((x+1)(x-2) (5))/(x-2)=\color(orchid)(-2x^2+2x+4)#

6. Cancel out like terms and simplify:

7. #((\cancel(x+1))(x-2)(2))/\cancel(x+1)+((x+1)\cancel((x-2)) (5))/\cancel(x-2)=-2x^2+2x+4#

8. `(x-2)(2)+(x+1)(5)=-2x^2+2x+4`

9. Re-apply distributive property:

10. `\color(tomato)((x-2)(2))+\color(seagreen)((x+1)(5))=-2x^2+2x+4`

11. `\color(tomato)(2x-4)+\color(seagreen)(5x+5)=-2x^2+2x+4`

12. Identify like terms and combine:

13. `\color(hotpink)(2x)\color(steelblue)(-4)+\color(hotpink)(5x)+\color(steelblue)(5)=-2x^2+\color(hotpink)(2x)+\color(steelblue)(4)`
14. `\color(hotpink)(7x)+\color(steelblue)(1)=-2x^2+\color(hotpink)(2x)+\color(steelblue)(4)`
15. `2x^2+\color(hotpink)(7x-2x)+\color(steelblue)(1-4)=0`
16. `2x^2+\color(hotpink)(5x)+(\color(steelblue)(-3))=0`
17. `2x^2+5x-3=0` And now you have to solve this polynomial.
18. Solve the polynomial from the last step (see here for details).