I'm stuck at solving a rational equation please help!

[Heanly]

2x+1 - x-4 = -10x+13
x-3 x+5 x+5

= (x-3)(x+5) (2x+1 - x-4) = (x-3)(x+5) (-10x+13)
(x-3 x+5) ( x+5 )

= (2x+1)(x+5)-(x-4)(x-3)=(x-3)(-10x+13)

= 2x²+10x+1x+5-x²+3x-4x+12=-10x²+13x+30x-39

= 2x²+11x+5-x²-1x+12=-10x²+43x-39

= 2x²-x²+10x²+11x-1x-43x+5+12+39=0

= 11x²-33x+56=0

= ?????

any help will be utmostly appreciated

 

[Deleted member 4993]

2x+1 - x-4 = -10x+13
x-3 x+5 x+5

→ (x-3)(x+5) (2x+1 - x-4) = (x-3)(x+5) (-10x+13)
(x-3 x+5) ( x+5 )

→ (2x+1)(x+5)-(x-4)(x-3)=(x-3)(-10x+13)

→ 2x²+10x+1x+5-x²+3x-4x+12=-10x²+13x+30x-39

→ 2x²+11x+5-x²-1x+12=-10x²+43x-39

→ 2x²-x²+10x²+11x-1x-43x+5+12+39=0

→ 11x²-33x+56=0

= ?????

any help will be utmostly appreciated

This is a quadratic equation. Do you know how to solve a quadratic equation?

If not - Google it - you'll find many websites with example solutions....

 

[Ishuda]

2x+1 - x-4 = -10x+13
x-3 x+5 x+5

= (x-3)(x+5) (2x+1 - x-4) = (x-3)(x+5) (-10x+13)
(x-3 x+5) ( x+5 )

= (2x+1)(x+5)-(x-4)(x-3)=(x-3)(-10x+13)

= 2x²+10x+1x+5-x²+3x-4x+12=-10x²+13x+30x-39
...
any help will be utmostly appreciated

Just to make sure I know what the equation is I'll re-write it as
\(\displaystyle \frac{2x+1}{x-3}\, -\, \frac{x-4}{x+5}\, =\, \frac{-10x+13}{x+5}\)
If that is the case, you are good down to the red above. The -4x should be a +4x.