How do you solve #2/(x+1) + 5/(x-2)=-2#?
1 Answer
Apr 17, 2018
Explanation:
-
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)# -
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)# -
Apply the distributive property:
#(\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)# -
#((x+1)(x-2)(2))/(x+1)+((x+1)(x-2) (5))/(x-2)=\color(orchid)(-2x^2+2x+4)# -
Cancel out like terms and simplify:
#((\cancel(x+1))(x-2)(2))/\cancel(x+1)+((x+1)\cancel((x-2)) (5))/\cancel(x-2)=-2x^2+2x+4# -
#(x-2)(2)+(x+1)(5)=-2x^2+2x+4# -
Re-apply distributive property:
#\color(tomato)((x-2)(2))+\color(seagreen)((x+1)(5))=-2x^2+2x+4# -
#\color(tomato)(2x-4)+\color(seagreen)(5x+5)=-2x^2+2x+4# -
Identify like terms and combine:
#\color(hotpink)(2x)\color(steelblue)(-4)+\color(hotpink)(5x)+\color(steelblue)(5)=-2x^2+\color(hotpink)(2x)+\color(steelblue)(4)# #\color(hotpink)(7x)+\color(steelblue)(1)=-2x^2+\color(hotpink)(2x)+\color(steelblue)(4)# #2x^2+\color(hotpink)(7x-2x)+\color(steelblue)(1-4)=0# #2x^2+\color(hotpink)(5x)+(\color(steelblue)(-3))=0# #2x^2+5x-3=0# And now you have to solve this polynomial.- Solve the polynomial from the last step (see here for details).