# Thread: How to Solve x/3 - 3 = x/5 - 5 ?

1. ## How to Solve x/3 - 3 = x/5 - 5 ?

I am doing a review packet and have forgotten the formula, evidently....

the problem goes as follows
X/3 - 3 = X/5 - 5

I am absolutely... lost...

any and all replies appreciated thanx

2. Find the least common multiple which is 15. Multiply your whole equation by 15, you should end up with 5x-45 = 3x -75, then it should be a bit easier to isolate X. Subtract 3x from both sides, add 45 to both sides, you'll end up with 2x = -30. Divide by 2, x = -15! Ask if you have any questions.

-Nichole.

3. I'm still stuck... So I multiply x/3 by 15... and the separate three?? and the x/5 as well as the 5???? I'm sorry I'm so handicapped

Think of the problem like this, (x/3)-3=(x/5)-5. I want to get rid of the fraction, the least common multiple of 3, and 5 is 15. So I multiply my whole equation by 15, I end up with. (15x/3)-45=(15x/5)-75. 15x/3=5x. 15x/5=3x. So 5x-45 = 3x -75. I like working with positive X, so I'll subtract 3X from 5X, I'll get 2x. Now I have 2x-45=-75. Isolate 2x, by adding 45 to both sides. We end up with 2x=-30. Then divide by 2, x = -15. !

-Nichole.

5. It's not so much a "formula" as it is a "method". You want to solve
x/3- 3= x/5- 5. That means you want to change this to an equation of the form
"x= something" which just says "x is ...", the answer. And you "change" equations by doing the same thing to both sides. An equation is like a balancing act- the two sides have to be kept balance- if you change one side you must change the other side in exactly the same way so they still ballance.

I want x only on the left side. Right now there is an "x/5" on the right- I need to get rid of that- I need to subtract x/5 from [b]both[/b[] sides:
x/3- 3- x/5= x/5- 5- x/5
(x/3- x/5)- 3= (x/5- x/5)- 5

That "x/5- x/5" on the right is easy- it's equal to 0 so I have, in fact, got rid of the "x/5" as I wanted. On the left to subract x/3- x/5, fractions with different denominators, I need "common denominators" just as I learned to subtract fractions in, what, the third or fourth grade? The "common denominator" is (3)(5)= 15. Multiplying both numerator and denominator of x/5 by 3 I get 3x/15. Multiplying boht numerator and denominator of x/3 by 5 Is get 5x/15. x/3- x/5= 5x/15- 3x/15= 2x/15 so my equation is
2x/15- 3= -5

Since I want x alone on the left, I need to get rid of that "-3" on the left. And I can do that by adding 3 [b]to both sides":
2x/15- 3+ 3= -5+ 3
2x/15= -2

Now I have (2/15) times x on the left, but we still want x alone. Since 2/15 is multiplying x, we get rid of it by dividing both sides by 2/15 (which is the same as multiplying by 15/2).
(2x/15)(15/2)= x= (15/2)(-2)= -15.