Question #8e9bf

Assuming your question says #a/2-6/(a-2)=(6-3a)/(a-2)#

Steps

• First move all values with denominator #a-2# to the right side.
  #a/2=6/(a-2)+(6-3a)/(a-2)# Simplify this.
  #a/2=(6+6-3a)/(a-2)# → Simplify again → #a/2=(12-3a)/(a-2)#
• Now find the common denominator to make the fractions equivalent.
  #(a(a-2))/(2(a-2))=(2(12-3a))/(2(a-2))# Simplify this.
  #(a^2-2)/(2a-4)=(24-6a)/(2a-4)# We can now take out the denominator as both sides are equivalent.
• Calculate to find #a#: first, set up quadratic. Make one side 0.
  #a^2-2=24-6a#
  #a^2-26+6a=0# reorder this
  #a^2+6a-26=0#
• Solving the quadratic here

Answer
#a=-3\pm\sqrt{35}#

Assuming your question says #a/2-6/(a-2)=6-(3a)/(a-2)#

Steps

• First set all denominators equal; common denominator is #2(a-2)#
  #\therefore\(a(a-2))/(2(a-2))-(2(6))/(a(a-2))=(6(2(a-2)))/(2(a-2))-(2(3a))/(2(a-2))#
  simplifying #(a^2-2a)/(2a-4)-12/(2a-4)=(12a-24)/(2a-4)-(6a)/(2a-4)#
• Now take out the denominator since everything is equivalent...
  #(a^2-2a)-12=(12a-24)-6a#
  #a^2-2a-12=6a-24# adding like terms on right side...
• Make a quadratic (form #ax^2+bx+c#) to solve:
  #a^2-8a-12=0# moving all terms to left side...
  solving here

Answer
#a=2(2\pm\sqrt{7})#