How do you solve #6/(x-4) = 5/2#?
1 Answer
Jan 22, 2016
Explanation:
First of all, your equation is not defined if any denominator is equal to
#x - 4 != 0 <=> x != 4 " "# must hold.
Now, your first step would be to "get rid" of the denominators. To do so, you should multiply both sides of the equation with
#color(white)(x) 6 / (x-4) = 5/2#
#<=> (6* (x-4) * 2)/(x-4) = (5 * (x-4) * 2) / 2#
#<=> (6* cancel((x-4)) * 2)/(cancel(x-4)) = (5 * (x-4) * cancel(2)) / cancel(2)#
#<=> 6* 2 = 5(x-4)#
#<=> 12 = 5x-20# ... add
#20# on both sides...
#<=> 32 = 5x# ... divide by
#5# on both sides...
#<=> 32/5 = x#
So, your solution is