How do you solve rational equations #2/(3x+1) = 1/x - (6x)/(3x+1)#?
1) Check when the equation is defined
First of all, bear in mind that your equation is not defined if any denominator is equal to zero.
Thus, let's first compute which values for
- impossible values for the first and third demoninator:
#3x + 1 = 0 <=> x = -1/3#
- impossible values for the second denominator:
#x = 0 #
Thus, for your equation,
2) Convert into quadratic equation
Now, let's solve the equation!
#2/(3x+1) = 1/x - (6x)/(3x+1)#
... to eliminate the first and the third fraction, multiply both sides of the equation with
... to eliminate the last remaining fraction, multiply both sides of the equation with
... bring all terms to the left side of the equation...
(to do so, add
3) Solve the quadratic equation
At this point, you have a quadratic equation which can be solved e.g. with the quadratic formula.
In your case,
The two solutions for this quadratic equation are
4) Check which solutions are possible
Since we have noted before that
Thus, the only solution for this equation is
Another way of approaching the initial stage. It has to be the same way as given but just looks different!
To make all the denominators the same:
Let some value be k then
Now we have:
As they all have the same denominator the resulting equation would still be a true if we totally removed it.
Then solve in the normal way!
Just another way
Take common terms to same side
Cross multiply both sides
So we have two solutions
x =-1/3 leads to