Answer:
Answer:
4.
First you need to assume X not equal to 3 or -3. These values would make the denominator invalid.
Because,
X - 3 must not be 0
X + 3 must not be 0
X^2 - 9 must not be 0
Then, you would have:
1/(X-3) + 2/(X+3) = 1/(X^2-9)
<=> [ X+3 + 2*(X-3) ]/(X^2-9)=1/(X^2-9) (*)
(Note that: (X-3)(X+3) = X^2-9)
Cancel out the denominator of both side of (*), then we have
X+3 + 2*(X-3) = 1
<=> X+3+2X-6=1
<=> 3X =4
<=> X = 4/3 (this value satisfies our first assumption)
=> X = 4/3 is the solution
5.
First you need to assume X not equal to 5 or 0. These values would make the denominator invalid.
Because,
X - 5 must not be 0
2X must not be 0
3X^2-15X = 3X(X-5) must not be 0
Then, you would have:
[2*2X + X-5]/[(X-5)*2X] = 5/[3X*(X-5)]
<=> (5X-5)/[(X-5)*2X] = 5/[3X*(X-5)]
<=> 3*(5X-5)/[(X-5)*6X] =2*5/[6X*(X-5)] (*)
Cancel out the denominator of both side of (*), then we have
3*(5X-5)=2*5
<=> 15X-15=10
<=> 15X =25
<=> X =25/15 =5/3 (this value satisfies our first assumption)
=> X = 5/3 is the solution
