Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x+6\over{x^2 +4x-12}}\) = [ \(1\over{x-2}\) ]

Q1) \({x-2}\over{x^2 -4}\) = [ \(1\over{x+2}\) ]

Q1) \({4x^2 +29x+30}\over{x+6}\) = [ \(4x+5\) ]

Q2) \({x^2 -6x+8}\over{x-2}\) = [ \(x-4\) ]

Q2) \({x+2}\over{x^2 -4}\) = [ \(1\over{x-2}\) ]

Q2) \({2x^2 -8}\over{x+2}\) = [ \(2x-4\) ]

Q3) \({x^2 -16}\over{x-4}\) = [ \(x+4\) ]

Q3) \({x^2 -4}\over{x-2}\) = [ \(x+2\) ]

Q3) \({2x^2 +2x-24}\over{x+4}\) = [ \(2x-6\) ]

Q4) \({x+7\over{x^2 +9x+14}}\) = [ \(1\over{x+2}\) ]

Q4) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q4) \({3x^2 +15x+12}\over{x+4}\) = [ \(3x+3\) ]

Q5) \({x^2 -7x+12}\over{x-4}\) = [ \(x-3\) ]

Q5) \({x+5}\over{x^2 -25}\) = [ \(1\over{x-5}\) ]

Q5) \({5x^2 -27x-18}\over{x-6}\) = [ \(5x+3\) ]

Q6) \({x^2 +x-12}\over{x+4}\) = [ \(x-3\) ]

Q6) \({x^2 -81}\over{x+9}\) = [ \(x-9\) ]

Q6) \({3x^2 -11x+6}\over{x-3}\) = [ \(3x-2\) ]

Q7) \({x^2 -4x+4}\over{x-2}\) = [ \(x-2\) ]

Q7) \({x^2 -4}\over{x+2}\) = [ \(x-2\) ]

Q7) \({2x^2 +4x-30}\over{x+5}\) = [ \(2x-6\) ]

Q8) \({x^2 -x-6}\over{x+2}\) = [ \(x-3\) ]

Q8) \({x-6}\over{x^2 -36}\) = [ \(1\over{x+6}\) ]

Q8) \({2x^2 +10x+12}\over{x+2}\) = [ \(2x+6\) ]

Q9) \({x-2\over{x^2 +2x-8}}\) = [ \(1\over{x+4}\) ]

Q9) \({x-4}\over{x^2 -16}\) = [ \(1\over{x+4}\) ]

Q9) \({5x^2 -14x+8}\over{x-2}\) = [ \(5x-4\) ]

Q10) \({x^2 -6x+8}\over{x-4}\) = [ \(x-2\) ]

Q10) \({x^2 -9}\over{x-3}\) = [ \(x+3\) ]

Q10) \({2x^2 +11x+12}\over{x+4}\) = [ \(2x+3\) ]