x^2-9 = (x+3)(x-3)
so that is the lcd
1/(x-3) + 1/(x+3) = 10/(x^2-9)
multiply through by x^2-9 and you have
(x+3) + (x-3) = 10
2x = 10
x = 5
For #2 if you mean
5/(2x) - 5/(3(x+5)) = (5/x)
the lcd is 2x(3(x+5)) so
5(3(x+5) - 5(2x) = 5(2)(3(x+5))
15x+75 - 10x = 30x+150
If I got the parentheses wrong, I'm sure you can fix it up.
For #3 I assume
(2x-1)/(x-1)-3x/(2x+1)=(x^2+11)/(2x^2-x-1)
the lcd is (x-1)(2x+1) = 2x^2-x-1, so
(2x-1)(2x+1) - 3x(x-1) = (x^2+11)
4x^2-1-3x^2+3x = x^2+11
...
Solving fractional equations
1)
(1/x-3)+(1/x+3)=(10/x^2-9)
2)
(5/2x)-(5/3(x+5)=(5/x)
3)
(2x-1/x-1)-(3x/2x+1)=(x^2+11/2x^2-x-1)
I just have a problem knowing what the lcd is. For the first one, i know I had to m,multiply the left side by x-3 and x+3, but it had to be the same and you had to multiply all of them by it, so I kind of got lost. Could someone just help me get started for each one and form there I should be able to figure it out.
4 answers
OHHHH, wait, correct me if i'm wrong, but the lcd for problems like this are most likely the denominator of the 2 fraction being combined?
yes, just as with normal fractions
2/3 + 4/7
has an lcd of 4*7
3/4 + 1/6
has an lcd of 12 instead of 24, since 2 is a common factor of both 4 and 6.
LCD(m,n) = mn/GCF(mn)
2/3 + 4/7
has an lcd of 4*7
3/4 + 1/6
has an lcd of 12 instead of 24, since 2 is a common factor of both 4 and 6.
LCD(m,n) = mn/GCF(mn)
Thank you so much
3 answers