help please :)

Question

Solve the following rational equations; be sure to check for extraneous solutions (please solve all):

A) 2x + 3 = 6
x-3 x-2 x²-5x+6

B) 3 = x+2
x-2 7

Answer 1

If the equation A) seems to be following
2x/(x-3) + 3/(x-2) = 6/(x2-5x+6) then


2x/(x-3) + 3/(x-2) = 6/(x²-5x+6)
=> 2x/(x-3) + 3/(x-2) = 6/(x²-3x-2x+6)
=> 2x/(x-3) + 3/(x-2) = 6/[x(x-3)-2(x-3)]
=> 2x/(x-3) + 3/(x-2) = 6/(x-3)(x-2)
=>2x(x-2)+ 3(x-3)=6 ......... multiplying both sides by (x-3)(x-2)
=>2x²-4x+3x-9+6=0.......... subtracting 6 from both sides
=>2x²-x-3=0..............adjusting operators
=>2x²-3x+2x-3=0 ..............adjusting operators
=>x(2x-3)+(2x-3)=0 ..............adjusting operators
=>(2x-3)(x+1)=0
=>(2x-3)=0 or (x+1)=0
=>2x=3... adding 3 to both sides or x=-1 subtracting 1 from both sides
=>x=3/2... dividing both sides by 2
therefore, x=3/2 or -1
Substituting x=3/2 at Left Hand Side in A), [3/(-3/2)]+[3/(-1/2)]=-2-6=-8
Substituting x=3/2 at Right Hand Side in A),6/[9/4-15/2+6]=24/[9-30+24]=24/3=8 the balance of equation cannot be attained
Substituting x=-1 at Left Hand Side in A), [-2/(-4)]+[3/(-3)]=1/2-1=-1/2
Substituting x=-1 at Right Hand Side in A),6/[1+5+6]=6/12=1/2 the balance of equation cannot be attained

Thus the giveb equation has no real solution

 

If the equation B) seems to be following
3/(x-2) = (x+2)/7 then

 

3/(x-2) = (x+2)/7

=>21= x²-4 ..... multiplying both sides by 7 and (x-2)

=>x²-25=0 ....... subtracting 21 from both sides and adjusting operators

=>(x+5)(x-5)=0 

=>x=-5 or x=5

Substituting x=-5 at left hand side in B), 3/(-5-2)=-3/7 and at Right hand side in B), (-5+2)/7=-3/7

Substituting x=5 at left hand side in B), 3/(5-2)=3/3 =1 and at Right hand side in B), (5+2)/7=7/7=1

thus the equation remains valid for both the values.