Math Help

[Fast GTO]

How do I solve for x in these equations? I am getting ready for the GRE and the GRE study book answers them via plugging in the answer choices instead of doing the math, which I hate.

3x/4 + 10 = x/8 + 15

and

(x+1/x-3) - (x+2/x-4) = 0



I need to see the work. Don't post if you don't have an answer.

 

[DSLethal]

3x/4+10=1x/8 + 15

This equals 3x/4 - 1x/8 = 15 - 10
Then 3x/4 change to 6x/8... so 6x/8 -1x/8 = 15-10
left side you are left with 5x/8=5
Multilpy each side by 8/5 to cancel out the 5/8's on the left side... you are left with x=8.

 

[DSLethal]

Part 2.

X+1/X-3=X+2/X-4

(x+1)(x-4)=(x+2)(x-3)

(x^2-3x-4)=(x^2-x-6)

Then x^2 cancel out.. move the -x from the right side to the left side by adding x to each side. This cancels out the -x on the right side.. and makes the -3x on the left side -2x. Add 4 to both sides.. this cancels out the -4 on the left side and makes the -6 on the right side -2. So you are left with
-2x=-2.. which means x = 1.

If you have any other questions let me know. I taking calculus 4 this fall so this stuff is fun stuff. lol

 

[Fast GTO]

Thanks. I ****ing hate it and am not even sure why this is on the GRE test.

GRE is for liberal arts degree's.

 

[DSLethal]

In the first problem.. the key is just getting "like terms" on the same side. Get the terms with x's in them on one side.. and the constants (plain numbers) on the other. Then recall from long time ago when adding fractions you need to have the same denominators. Once you are left with one fraction term, then you can multiply each side by the reciprocal to leave you with your x term alone.

Problem 2.. same kind of situation..

 

[DSLethal]

If you need any other math help, send me a PM... Take care.

 

[Fast GTO]

Thanks.