algerbraic fractions! how delightful!?

Question

i do them so many different ways .. i don't know what's right... here's the question...

7/(x+2) + 1/(x-1) =4

please don't use a calculator because i am not allowed.. so if you use the qaudratic formula just leave it in its simpliest form thanks.

i ended up with... x=4+rt.192/8 or x=4-rt.192/8.... but i don't know if it's right

i know it's right but
how did you work out....
0 = (2x + 1)(2x - 3)
from 0 = 4x^2 - 4x - 3

Answer 1

you can put the left side over a common denominator thus:

[7(x-1) +1(x+2)]/(x+2)(x-1) = 4
now multiply both sides by the denomiantor:

7(x-1) + (x+2) =4(x+2)(x-1)
now perform indicated operations:

8x -5 = 4 x^2 +4x -8
next subtract 8x-5 from both sides:
0 = 4x^2 -4x -3
you could put this into quadratic formula and solve for x i.e. x = [4 +/- sqrt(16+48)]/8 = {4 +/- sqrt(64)}/8
x = (4+8)/8 and x = (4-8)/8
x=3/2 and x = -1/2

Alternatively you could factor the expression 4x^2-4X-3
0 = (2x-3) (2x+1)
set each factor equal to zero and solve:

x=3/2 and x=-1/2


your second question was how to factor 4x^2-4x-3
if this type of expression can be factored you end up with 2 factors with 2 terms eachsuch as:

(ax +c)(bx +d) in the present case ax*bx must equal 4x^2
bx(c)+ax(d) must equal -4x
c(d) must equal -3
a little trial and error will give you the correct values of a b c and d
in this case 2x, 2x, -3 and 1

Answer 2

If you multiply everything with the LCD (x+2)(x-1), you come up with the equation:

7(x-1) + (x+2) = 4(x+2)(x-1)
7x - 7 + x + 2 = 4(x^2 + x - 2)
8x - 5 = 4x^2 + 4x - 8
0 = 4x^2 - 4x - 3
0 = (2x + 1)(2x - 3)
x = -1/2 or x =3/2