Question

Question #8e9bf

Answer 1

`a=-3\pm\sqrt{35}`

Explanation:

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}`

Answer 2

`a=2(2\pm\sqrt{7})`

Explanation:

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})`