How do you solve rational equations?

Question

I have this question...

http://ce.byu.edu/courses/hs/999065041006/secure/imgs/Math%2041/4sb21a.gif

Can someone help me solve it? Thanks!

i have to solve for X...

Answer 1

(3x / x +2) +( 2 / x - 1) = 5 / x ^2 + x - 2
we have : x ^2 + x - 2 = ( x +2)( x - 1) ,
LCD IS ( x + 2)( x - 1)
so
[3x(x -1) + 2(x +2) - 5 ] / (x + 2)( x - 1) = 0
when
3x(x - 1) + 2(x + 2) - 5 = 0
3x ^2 - 3x + 2x + 4 - 5 = 0
3x ^2 - x - 1 = 0
we have , a= 3 , b = - 1 , c = - 1
so
x = { - ( - 1) + square root [( - 1) ^2 - 4(3)( - 1)]} / ( 2)(3)
x = [1 + square root (13)] / 6

x = { - ( - 1) - square root [( - 1) ^2 - 4(3)( - 1)]} / (2)(3)
x = [ 1 - square root ( 13)] / 6
the solution set is
{ [ 1 + square root ( 13)] / 6 , [1 - square root ( 13)] / 6 }

Answer 2

Make sure to have the same denominator. . .

= [3x(x-1)] / (x + 2)(x - 1) + [2(x + 2)] / (x + 2)(x - 1)

Distribute. . .

= (3x^2 - 3x) + (2x + 4) / (x + 2)(x - 1)

Simplify the numerator

= (3x^2 - x + 4) / (x + 2)(x - 1)

Solve the complex trinomial in the numerator and expand the denominator. . .

I think you can take over from here now . . .

Answer 3

3x(x-1) + 2(x+2) = 5
3x^2 - 3x + 2x + 4 - 5 = 0
3x^2 - x - 1 = 0

its a simple quadratic eqn

Answer 4

[(3x)/(x+2)]+[2/(x-1)]=[5/(x^2+x-2)]
take LCM
[(3x)(x-1)+2(x+2)]/[(x+2)(x-1)]=[5/(x^2+x-2)]
[(3x)(x-1)+2(x+2)]/[(x^2+x-2)]=[5/(x^2+x-2)]
[(3x)(x-1)+2(x+2)]=5
[(3x^2-3x)+(2x+4)]=5
[(3x^2-3x+2x+4)]=5
[(3x^2-x+4)]=5
3x^2-x+4-5=0
3x^-x-1=0

Answer 5

irrationally. it spices things up.