LCD of this equation?

Question

4 / (X-5) - (1/3) = -8 / (3X-15)

(Is it 3x-15) If so, I will multiply this by what? I don't get this equation .. If you don't understand it how it is written...


problem 19 of this page is the one that I'm talking about

http://online.math.uh.edu/Math1310/ch2/s21/index.html

Where it says "Exercises" click on it

PLEASE HELP ME!!!

I also want to know if this is right

the Y-intercept of 4xE2 - yE2=9

At the end I got

Square root of -yE2 = Square root of 9

What I did was
get the Square root of 9 which is 3 from the right side and from the left side eliminated the exponent 2 and the square root of -y and I got -y is equal to -3


my final answer was -3

but I am not sure about this.. can somebody telll me please.. dont worry about the first problem, I already got it...

Thank you!!!!

Answer 1

You are right.
3x-15 = 3(x-5) is the LCD.

You need to multiply every term by 3(x-5)

3(x-5)(4/(X-5)) - 3(x-5)(1/3) = 3(x-5)(-8 / (3X-15))
Notice that (x-5) cancels in the first term,
3 cancels in the second term and
3(x-5) cancels in the third term
3(4) - (x-5) = -8
12 - x + 5 = -8
17 - x = -8
Subtract 17 from each side
-x = -25
x = 25

the y-intercept of 4x^2 - y^2=9?
Let x = 0
4(0^2) - y^2 = 9
0 - y^2 = 9
-y^2 = 9
Multiply both sides by -1
y^2 = -9
Take the square root of each side
y = sqrt(-9) or -sqrt(-9)
The square root of -9 is not a real number, so there is no y-intercept.

Answer 2

Yes, your LCD is 3x-15, or 3(x-5). To clear fractions, you would multiply both sides of your equation by 3(x-5):

3(x-5)(4/(x-5)-1/3) = 3(x-5)*-8/(3x-15)

Separate the terms of the left side:

12(x-5)/(x-5) - 3(x-5)/3 = -8
12-(x-5)=-8
17-x=-8
x=25

Note that if x=5 had come up as a solution, there would actually be no solution, because that would make your denominators equal 0.

Answer 3

Yes, the LCD is 3x-15
If you find the LCD then substitute:
(4)(3)/(3x-15) - (1)(x-5)/(3x-15) = -8/(3x-15)
collect terms:
(12 - x + 5 + 8)/(3x-15) = 0
25 - x = 0 (when you multiply both sides by 3x-15)
x = 25
If you substitute back into the original equation, you will find that the values will be the same on both sides of the equation and thus it checks.