Annoying lil' Math Problem. Could Someone Help?

Question

Hello! This is my last math question to complete and it's driving me crazy! Here it is:

10/(x-2) - 1/2 = 3/(x-5)

If someone could help me I'd greatly appreciate it. Thanks!

Answer 1

Multiply through by 2(x-2)(x-5) to get:
10(2)(x-5) - 1(x-2)(x-5) = 3(2)(x-2)
<=> 20x - 100 - x^2 + 7x - 10 = 6x - 12
<=> x^2 - 21x + 98 = 0
<=> (x - 7)(x - 14) = 0
<=> x = 7 or 14.

Answer 2

FRACTIONS! We all hate them, but we seem to have to live with them. Here's how to minimize the pain.

You CANNOT add or subtract fractions unless the denominators are all the same. In your problem we can do it fast, or we can do it slowly. Lets go the slow route.

a) Let's multiply both sides of the equation by 2
2[10/(x-2) -1/2] =2[3/(x-5)
20/(x-2) -1 = 6/(x-5) Looking a bit better.
b) Now let's multiply both sides by (x-2)
(x-2)[20/(x-2) - 1] =(x-2)[6/(x-5)]. We get
20-1(x-2) =6(x-2)/(x-5) Better yet.
c) Now , let's multiply both sides by (x-5)
20(x-5) -1(x-5)(x-2)=6(x-2) Great! No fractions.

Multiplying allthis out will give us
20x-100-1(x^2 -7x+10) =6x-12
20x-100-x^2+7x-10=6x-12
-x^2 +20x+7x-6x-100-10+12=0
-x^2+21x-98=0
x^2-21x+98-0
(x-14)(x-7)=0

Now for (x-14)(x-7) to multiply out to give zero, either
(x-14) has to be zero, or (x-7) has to be zero. Or both have to be zero!
If x-14=0, x=14
If x-7=0, x=7

That's it. The fast method would have been to multiply those 3 terms by their LCM Least Common Multiple) which in this case is[(x-2) 2 (x-5)]. You end up in exactly the same place. I thought going the slow route would be easier on the nerves!.
Good luck

Answer 3

10/(x - 2) - 1/2 = 3/(x - 5)

multiply 2 for both sides
20/(x - 2) - 1 = 6/(x - 5)

multiply (x - 2) for both sides
20 - 1(x - 2) = 6(x-2) / (x - 5)

distriute
20 - x + 2 = (6x - 12) / (x - 5)

combine like terms
22 - x = (6x - 12) / (x - 5)

multily x - 5 for both sides
22(x - 5) - x(x - 5) = 6x - 12

distribute
22x - 110 - x^2 + 5x = 6x - 12

combine like terms
27x - 110 - x^2 = 6x - 12

set the equation equal to 0
x^2 - 21x + 98 = 0

factor
(x - 14) (x - 7) = 0

x = 14 or 7

Answer 4

10/(x-2) - 1/2 = 3/(x-5)
multiply both sides of the equation by the least common multiple of the denominators. Since there are variables, multiply by all three:
2(x-2)(x-5)·[10/(x-2) - 1/2] = 2(x-2)(x-5)·[3/(x-5)]
Distribute:
2·(x-5)·10 - (x-2)(x-5) = 2·(x-2)·3
Distribute again:
[20x-100]-[x²-7x+10] = 6x-12
-x²+27x-110 = 6x-12
x²-27x+110 = -6x+12
x²-21x+98 = 0
Factor:
(x-7)(x-14)=0
x-7=0 or x-14=0
x = 7 or x = 14

Answer 5

10/(x-2) - 1/2 = 3/(x-5)
20 - x + 2/ (2x - 4) = 3 / (x-5)
(22 - x) (x - 5) = 3 (2x - 4)
22x - x^2 - 110 + 5x = 6x - 12
22x + 5x - 6x - 110 + 12 - x^2 = 0
x^2 - 21x + 98 = 0