Help to do it, Need a solution, not only answer

For the second one....just multiply everything through by  x (x + 1).....this leaves

2(x + 1) + 3x(x)  = 2x + 5     simplify

2x + 2 + 3x^2 = 2x + 5         subtract 2x and 5 from both sides

3x^2 - 3  = 0    factor

3(x^2 -1) = 0        divide both sides by 3

x^2 - 1   = 0         factor again

(x + 1)(x - 1)  = 0        set both factors to 0    and x = -1  and x = 1

Reject x = -1    (it makes two denominators in the original problem = 0)

So....x = 1 is the solution

Be sure and check this in the original problem........

a) The easiest way to do this one is just to cross-multiply......so we have

(2x -1)(x - 1) = (3x + 4)(x +7)     expand both sides

2x^2 - 3x + 1  = 3x^2 + 25x + 28    subtract 2x^2 and 1 from both sides and add 3x to both sides

0 = x^2 + 28x + 27   factor

0 = (x  + 27)(x + 1)      setting both factors to 0, we have x = -1 and x = -27

Neither thing makes a denominator in the original problem = 0, so both solutions are good....

Good job!