solve the equation?

3/x-2 = 1/x-1 + 7/x^2-3x+2

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

Multiply the equation by (x-2)(x-1)

(x-2)(x-1) [ 3/x-2 = 1/x-1 + 7/(x-2)(x-1)]

3(x-1) = 1(x-2) + 7

3x - 3 = x - 2 + 7

3x - 3 = x + 5

3x - 3 + 3 - x = x + 5 - x + 3

2x = 8

x = 4

To check
3/4-2 = 1/4-1 + 7/4^2 -3(4) + 2

3/2 = 1/3 + 7/16 - 12 + 2
3/2 = 1/3 + 7/6

3/2 = 2/6 + 7/6

3/2 = 9/6

3/2 =3/2

x=4

3/(x - 2) = 1/(x - 1) + 7/(x^2 - 3x + 2)
[x - 2][x - 1][3/(x - 2)] = [x - 2][x - 1][1/(x - 1) + 7/(x - 2)(x - 1)]
3(x - 1) = 1(x - 2) + 7
3x - 3 = x - 2 + 7
3x - x = 3 - 2 + 7
2x = 8
x = 8/2
x = 4

3/x-2 = 1/x-1 + 7/x^2-3x+2

Athena here it is:

3/(x - 2) = 1/(x - 1) + 7/ (x² -3x + 2)
3/(x - 2) = 1 /(x - 1) + 7/ (x -1)(x - 2)
LCD = (x -1)(x - 2)
3/(x - 2) = 1(x - 2) /(x - 1)(x - 2) + 7/ (x -1)(x - 2)
3/(x - 2) = [(x - 2) + 7] / (x -1)(x - 2)
Canceling out (x - 2)
3 = x + 5 /(x - 1)
3(x - 1) = x + 5
3x - 3 = x + 5
2x = 8
=============================
Ans::: x = 4
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hope this helps

Believe it or not, this question is meaningless as it stands.

It SHOULD read as :-

3 / (x - 2) = 1 / (x - 1) + 7 / (x² - 3x + 2)
3 / (x - 2) = 1 / (x - 1) + 7 / (x - 2)(x - 1)
3 (x - 1) = (x - 2) + 7
3x - 3 = x + 5
2x = 8
x = 4

PS
Remember to use brackets !!!

3/x --2 = 1/x --1 + 7/x^2 --3x + 2
Or 3(x -- 1) = (x -- 2) + 7
Or 2x = 8
Or x = 4

3 / (x-2) = 1 / (x-1) + 7 (x^2 - 3x + 2)

Heh, x^2 - 3x + 2 is most probably the product of the other two.

So, multiplying both sides of the equation by x^2 - 3x + 2,

3(x-1) = (x-2) + 7
3x - 3 = x + 5
2x = 8
x = 4

get it all dwn to common denominatores

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