Solve the following equation:

If you mean 3 / (z-3) * 5 / (z-1) = 8 / (z-2), then multiply through by the denominators, expand, collect like terms and solve the resulting quadratic using the quadratic formula.

If you didn't leave out brackets, make the substitution x = 1/z, then the equation is:

(3x - 3) (5x - 1) = (8x - 2)
15x^2 - 3x - 15x + 3 = 8x - 2
15x^2 - 26x + 5 = 0

x = ( 26 +/- sqrt (26^2 - 4(15)(5) ) ) / 30
= ( 26 +/- sqrt (376) ) / 30
= ( 26 +/- 2sqrt(94) ) / 30
= ( 13 +/- sqrt(94) ) / 15

Since x = 1/z, the solution is:

z = 15 / ( 13 +/- sqrt(94) )

This one line question contains a number of errors.

Will take a guess at :-

3 / (z - 3) + 5 / (z - 1) = 8 / (z - 2)

You will note subtle use of brackets and + sign !!!!

3(z - 2)(z - 1) + 5(z - 3)(z - 2) = 8 (z - 3)(z - 1)

3(z³ - 3z + 2) + 5(z² - 5z + 6) = 8(z² - 4z + 3)

3z² - 34x + 36 = 8z² - 32z + 24

5z² + 2z - 12 = 0
x = [ - 2 ± √ (4 + 240) ] / 10
x = [ - 2 ± √ (244) ] / 10
x = [ - 2 ± 2√61 ] / 10
x = [ - 1 ± √61 ] / 5

3/(z - 3) + 5/(z - 1) = 8/(z - 2)
[z - 3][z - 1][z - 2][3/(z - 3) + 5/(z - 1)] = [z - 3][z - 1][z - 2][8/(z - 2)]
3[z - 1][z - 2] + 5[z - 3][z - 2] = 8[z - 3][z - 1]
3[z^2 - z - 2z + 2] + 5[z^2 - 3z - 2z + 6] = 8[z^2 - 3z - z + 3]
3z^2 - 9z + 6 + 5z^2 - 25z + 30 = 8z^2 - 32z + 24
3z^2 + 5z^2 - 8z^2 - 9z - 25z + 32z + 6 + 30 - 24 = 0
-2z + 12 = 0
-2z = -12
z = -12/-2
z = 6

(3/z-3) ( 5/z-1) = 8/z-2

15/ (z-3)(z -1) = 8/z -2

To get rid of the denominator, multiply the equation by all the denominators

(z-3)(z-1)(z-2) [15/(z-3)(z-1) = 8/z-2]

15 = 8 [(z-3)(z-1)}

15 = 8 ( z^2 -4z + 3)


15 = 8z^2 - 32z + 24

8z^2 -32z + 9 = 0

Use the quadratic formula

x = (-b +/- sqrt b^2 - 4ac)/2a

a = 8
b = -32
c = 9

x = [- (-32) +/- sqrt (-32)^2 - 4(8)(9)] / 2(8)
x =( 32 +/- sqrt 1024 - 288)/ 16
x =( 32 +/- sqrt 736) /16
x = 32 +/- 4sqrt46)/16

x = 2 +/- sqrt46/4

x = 2 + sqrt 46 / 4

x = 2 - sqrt 46/ 4