How do i solve the equation?

Question

4/(x-3) - 10/(3x-5) = 1

Answer 1

domain D = R \ {3,5/3}

4/(x-3) - 10/(3x-5) = 1

4*(3x-5) - 10*(x-3) = (x-3)(3x-5)

2x + 10 = 3x^2 - 14x + 15

3x^2 - 16x + 5 = 0

x=5 or x=1/3

Answer 2

4(3x-5)-10(x-3)=(x-3)(3x-5)

12x-20-10x+30=3x^2-5x-9x+15

3x^2-16x-15=0

solve x from here

Answer 3

4/(x - 3) - 10/(3x - 5) = 1

[x - 3][3x - 5][4/(x - 3) - 10/(3x - 5)] = 1[x - 3][3x - 5]

4[3x - 5] - 10[x - 3] = [x - 3][3x - 5]

12x - 20 - 10x + 30 = 3x^2 - 9x - 5x + 15

3x + 10 = 2x^2 - 14x + 15

3x^2 - 14x - 2x + 15 - 10 = 0

3x^2 - 16x + 5 = 0

3x^2 - x - 15x + 5 = 0

(3x^2 - x) - (15x - 5) = 0

x(3x - 1) - 5(3x - 1) = 0

(3x - 1)(x - 5) = 0

3x - 1 = 0

3x = 1

x = 1/3

x - 5 = 0

x = 5

∴ x = 1/3 , 5

Answer 4

you have two fractions:

4 / (x - 3)

and

10 / (3x - 5)

You need a common denominator... Which in this case is

(x - 3)(3x - 5)

So... multiply everything by that....

[(x - 3)(3x - 5)][4 / (x-3)] - [(x - 3)(3x - 5)][10 / (3x - 5)]

= [(x - 3)(3x - 5)](1)\

becomes

4(3x - 5) - 10(x - 3) = (x - 3)(3x - 5)

12x - 20 - 10x + 30 = 3x^2 - 9x - 5x + 15

2x + 10 = 3x^2 - 14x + 15

0 = 3x^2 - 16x + 5

0 = (3x - 1)(x - 5)

x = 1/3 and 5

Answer 5

multiply the three terms by (x-3)(3x-5)

then the first term is: 4(3x-5)

second: 10(x-3)

third: (x-3)(3x-5)

now you have 4(3x-5) - 10(x-3) = (x-3)(3x-5)

multiply out and solve like usual