We look for constants a and b such that
[tex]\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac a{2x-1} + \frac b{x-2}[/tex]
Rewrite all terms with a common denominator and set the numerators equal:
[tex]\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)}{(2x-1)(x-2)} + \frac{b(2x-1)}{(x-2)(2x-1)} \\\\ \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)+b(2x-1)}{(2x-1)(x-2)} \\\\ \implies 7x-8 = a(x-2) + b(2x-1) = (a+2b)x - 2a - b[/tex]
Then
a + 2b = 7
-2a - b = -8
Solve for a and b. Using elimination: multiply the first equation by 2 and add it to the second equation:
2 (a + 2b) + (-2a - b) = 2(7) + (-8)
2a + 4b - 2a - b = 14 - 8
3b = 6
b = 2
Then
a + 2(2) = 7 ==> a = 3
and so
[tex]\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac 3{2x-1} + \frac 2{x-2}[/tex]
