Question

Solving rational equations: I will type the equation in the comment box

Answer 1

ok

Answer 2

hint-
find the LCM and multiply it through out the equation

Answer 3

x^2-1 = (x+1)(x-1)

Answer 4

you would get a simple linear equation, then you can solve for x easily...

Answer 5

I believe its x(x^2-1)

Answer 6

or may be a quadratic equation...

first start with taking the LCM..

Answer 7

thats right !!

Answer 8

you can write the same as : x(x+1)(x-1)

Answer 9

we got LCM, what will be our next step ?

Answer 10

Yeah but wont that be more complicated when multiplying the numerators

Answer 11

hmm

Answer 12

You have to make the dnominators equal

Answer 13

yeah.. i see... good catch :) lets not factor. keep it as x(x^2-1)

Answer 14

we will multiply the LCM through out the equation

Answer 15

\(\huge \frac{5}{x^2-1}-\frac{2}{x} = \frac{2}{x+1}\)

Answer 16

\(\huge \frac{x(x^2-1).5}{x^2-1}-\frac{x(x^2-1).2}{x} = \frac{x(x^2-1).2}{x+1}\)

Answer 17

fine ?

Answer 18

nothing fancy here.. we just multiplied the entire equation with x(x^2-1)

Answer 19

\(\huge \frac{x(\cancel{x^2-1}).5}{\cancel{x^2-1}}-\frac{\cancel{x}(x^2-1).2}{\cancel{x}} = \frac{x(x-1)\cancel{(x+1)}.2}{\cancel{x+1}} \)

Answer 20

\(\huge 5x - 2(x^2-1) = 2x(x-1)\)

Answer 21

So would it be 5x-2x^2-2= 2x^2-2

Answer 22

\(\huge 5x-2x^2+2 = 2x^2-2x\)

\(\huge 4x^2-7x -2 = 0\)

Answer 23

can you solve the quadratic equation now... try to factor it... ?

Answer 24

Oh! okay! now when factoring does the first and last number have to equal the middle factor

Answer 25

yes,

ac = 4(-2) = -8

sum of two factors of above should equal to : -7